RD Sharma Solutions Class 10 Probability Exercise 13.1

RD Sharma Class 10 Solutions Chapter 13 Ex 13.1 PDF Free Download

Exercise 13.1

1. The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?

Solution:

Probability that it will rain tomorrow= P(E) = 0.85

Probability that it will not rain tomorrow= \(P(\bar{E})\)

Since the sum of the probability of occurrence of an event and probability of non occurrence of an event is 1,

We know that,

P(E) + \(P(\bar{E})\)= 1

0.85 + \(P(\bar{E})\) = 1

\(P(\bar{E})\) = 1 – 0.85

\(P(\bar{E})\) = 0.15

Therefore, the probability that it will not rain tomorrow is = 0.15

 

2. A die is thrown. Find the probability of getting:

(i) a prime number  

(ii) 2 or 4

(iii) a multiple of 2 or 3  

(iv) an even prime number

(v) a number greater than 5           

(vi) a number lying between 2 and 6

Solution:

Total number on a dice is 6.

(i) Probability of getting a prime number

Prime numbers on a dice =2,3 and 5

i.e., The total number of prime numbers on dice =3

We know that,

Probability = \(\frac{ Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting  a prime number = \(\frac{3}{6} = \frac{1}{2}\)

 

(ii) Probability of getting 2 or 4

The favorable outcome of getting 2 and 4 =2

We know that,

Probability = \(\frac{ Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting 2 or 4 = \(\frac{2}{6}= \frac{1}{3}\)

 

(iii) Probability of getting a multiple of 2 or 3.

Multiple of 2 or 3 on the dice are 2, 3, 4 and 6

i.e., the favorable outcome = 4

We know that,

Probability = \(\frac{ Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting a multiple of 2 or 3 on a dice= \(\frac{4}{6}= \frac{2}{3}\)

 

(iv) Probability of getting an even prime number

2 is the only even prime number.

i.e, the favorable outcome = 1

We know that,

probability= \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting an even prime number \(\frac{1}{6}\)

 

(v) Probability of getting a number greater than five.

On a dice, the number greater than 5 = 6

i.e., the favorable outcome =1

We know that,

probability= \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting a number greater than 5 = \(\frac{1}{6}\)

 

(vi) Probability of a number lying between 2 and 6

On a dice, the number lying between 2 and 6 =3, 4 and 5

i.e., total number of numbers lying between 2 and 6 =3

We know that,

probability= \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting a number lying between 2 and 6 = \(\frac{3}{6} = \frac{1}{2}\)

 

3. Three coins are tossed together. Find the probability of getting:

(i) exactly two heads

(ii) at most two heads

(iii) at least one head and one tail

(iv) no tails

Solution:

When three coins are tossed then the possible outcomes will be

TTT, THT, TTH, THH. HTT, HHT, HTH, HHH.

i.e., the total number of outcome = 8.

(i) exactly two heads

i.e., the possible outcomes are THH, HHT ,HTH

Therefore, total number of favorable outcomes i.e. exactly two head= 3

We know that PROBABILITY = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting exactly two head is \(\frac{3}{8}\)

(ii) at most two heads

In case of at most two heads, the possible outcomes are HTT, THT, TTH ,TTT, HHT, HTH, THH

Therefore, total number of favorable outcomes i.e. at most two head=7

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting at most two head when three coins are tossed simultaneously =\(\frac{7}{8}\)

(iii) at least one head and one tail

In case of at least one head and one tail, the possible outcomes are THT, TTH, THH. HTT, HHT, HTH,

Therefore, total number of favorable outcome i.e. at least one tail and one head is 6

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting at least one head and one tail is equal to \(\frac{6}{8} = \frac{3}{4}\)

(iv) no tails

The possible outcome for no tail= HHH

Therefore, total number of favorable outcome =1

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

Probability of getting no tails is \(\frac{1}{8}\)

 

4. A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number

Solution:

All possible outcomes of the event are,

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),

i.e., total number of events is 62 = 36

The possible outcomes of the event of getting the total of numbers on the dice greater than 9 are as follows.

(5,5), (5,6), (6,4), (4,6), (6,5) and (6,6),

Then, the total number of favorable results= 6

We know that,

probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting the total of numbers on the dice greater than 9 is \(\frac{6}{36} = \frac{1}{6}\)

 

5. Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.

Solution:

All possible outcomes of the event are,

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),

Then, the total number of events = 62 = 36

The possible outcomes of the event of getting the total of numbers on the dice greater than 10 are

(5, 6), (6, 5) and (6, 6)

Then, the total number of favorable results= 3

We know that,

probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting the total of numbers on the dice greater than 10 is \(\frac{3}{36} = \frac{1}{12}\)

 

6. A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:

(i) a black king         

(ii) either a black card or a king

(iii) black and a king

(iv) a jack, queen or a king

(v) neither a heart nor a king

(vi) spade or an ace

(vii) neither an ace nor a king

(viii) neither a red card nor a queen

(ix) the seven of clubs

(x) a ten

(xi) a spade

(xii) a black card

(xiii) a seven of clubs

(xiv) jack

(xv) the ace of spades

(xvi) a queen

(xvii) a heart

(xviii) a red card

Solution:

Total number of cards = 52

(i) a black king         

Number of cards in a deck of cards which has black king = 2

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a black king is equal to \(\frac{2}{52} = \frac{1}{26}\)

(ii) either a black card or a king

Total number of black cards in a deck of cards = 26

Total numbers of kings in a deck of cards= 4

Total number of cards which has king and are black = 2

Now, we can conclude that the total number of black cards or king will be 26+2 = 28

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a black cards or a king = \(\frac{28}{52} = \frac{7}{13}\)

 

(iii) black and a king

Total number of cards which has king and are black = 2

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a black cards and a king is \(\frac{2}{52} = \frac{1}{26}\)

 

(iv) a jack, queen or a king

A jack, queen or a king are 3 from each 4 sets

Therefore, the total number of a jack, queen and king are 3×4=12

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a jack, queen or a king is \(\frac{12}{52} = \frac{3}{13}\)

 

(v) neither a heart nor a king

Total number of heart cards in a deck of cards=13

Total number of king cards in a deck of cards=4

Card with king of heart =1

Total number of cards that are a heart and a king = 13 + (4-1)=13+3 = 16

Therefore,

The total number of cards in a deck that are neither a heart nor a king = 52 – 16 = 36

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting cards neither a heart nor a king = \(\frac{36}{52} = \frac{9}{13}\)

 

(vi) spade or an ace

Total number of spade cards in a deck of cards=13

Total number of aces cards in a deck of cards=4

Card with ace of spade =1

Hence total number of card which are spade or ace = 13 + (4-1) = 16

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting cards that is spade or an ace = \(\frac{16}{52} = \frac{4}{13}\)

 

(vii) neither an ace nor a king

Total number of ace cards in a deck of cards=4

Total number of king cards in a deck of cards=4

Then, total number of cards that are a ace and a king = 4 + 4 = 8

Therefore, Total number of cards that are neither an ace nor a king is 52 – 8 = 44

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting cards neither an ace nor a king = \(\frac{44}{52} = \frac{11}{13}\)

 

(viii) neither a red card nor a queen

Total number of red cards in a deck of cards=26

Total number of queens cards in a deck of cards=4

Then, total number of cards that are red and queen = 2

Then, total number of red card or queen will be 26+2 = 28

Therefore, total number of cards that are neither a red nor a queen= 52 -28 = 24

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting neither a red card nor a queen is equal to \(\frac{24}{52} = \frac{6}{13}\)

 

(ix) the seven of clubs

Total number of card other than ace in a deck of cards= 52-4 = 48

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting other than ace is \(\frac{48}{52} = \frac{12}{13}\)

 

(x) a ten

Total number of ten in a deck of cards= 4

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a ten is \(\frac{4}{52} = \frac{1}{13}\)

 

(xi) a spade

Total number of spade in a deck of cards=13

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a spade = \(\frac{13}{52} = \frac{1}{4}\)

 

(xii) a black card

Total number of black cards in a deck of cards=26

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting black cards is \(\frac{26}{52} = \frac{1}{2}\)

 

(xiii) a seven of clubs

Total number of 7 of club in a deck of cards=1

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a 7 of club is equal to = \(\frac{1}{52}\)

 

(xiv) jack

Total number of jacks in a deck of cards=4

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting jack \(\frac{4}{52} = \frac{1}{13}\)

 

(xv) the ace of spades

Total number of ace of spade in a deck of cards=1

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a ace of spade = \(\frac{1}{52}\)

 

(xvi) a queen

Total number of queen in a deck of cards=4

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a queen is \(\frac{4}{52} = \frac{1}{13}\)

 

(xvii) a heart

Total number of heart cards in a deck of cards=13

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a heart cards = \(\frac{13}{52} = \frac{1}{4}\)

 

(xviii) a red card

Total number of red cards in a deck of cards=26

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a red cards = \(\frac{26}{52} = \frac{1}{2}\) 

 

7. In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability that the drawn ticket bears a prime number.

Solution:

Total number of lottery tickets = 5.

Prime numbered tickets= 1,3,5,7,11,13,17,19,23,29,31,37,43,47,49

i.e., total number of Prime numbered tickets = 15

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a prime number on the ticket is \(\frac{15}{50} = \frac{3}{10}\)

 

8. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.

Solution:

Total number of balls in the urn= 10 + 8 = 18

Total number of white balls in the urn= 8

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a white ball is \(\frac{8}{18}= \frac{4}{9}\)

 

9. A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:

(i) white?                                              

(ii) red?

(iii) black?

(iv) not black

Solution:

From the question, we know that,

Total number of balls 3+5+4 =12

(i) White ball

Total number white balls in a bag= 4

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting white ball = \(\frac{4}{12} = \frac{1}{3}\)

(ii) Red ball

Total number red balls in a bag= 3

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting red ball is \(\frac{3}{12} = \frac{1}{4}\)

(iii) Black ball

Total number of black balls in a bag= 5

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting black ball = \(\frac{5}{12}\)

(iv)Not black balls

Total number of non red balls in a bag= 4 white balls +5 black balls= 4+5 = 9

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting no red ball = \(\frac{9}{12} = \frac{3}{4}\)

 

10. What is the probability that a number selected from the numbers 1, 2, 3, …, 15 is a multiple of 4?

Solution:

One number is to be selected from the numbers 1 to 15.

Then, total number= 15

Numbers that are multiple of 4 within the numbers 1 to 15= 4, 8, 12,

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of selecting a multiple of 4 is \(\frac{3}{15}= \frac{1}{5}\) 

 

11. A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that the ball drawn is white?

Solution:

Total number of balls in a bag= 6+8+4 = 18

Total number of black balls in a bag= 8

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The of getting a black ball P(E)\(\frac{8}{18} = \frac{4}{9}\)………… (1)

Since the sum of the probability of occurrence of an event and probability of non occurrence of an event = 1.

We get, P(E) + \(P(\bar{E})\) = 1

\(\frac{4}{9}+P(\bar{E}) = 1\)

\(P(\bar{E}) = 1 – \frac{4}{9}\)

\(P(\bar{E}) = \frac{5}{9}\)

  

12. A bag contains 5 white balls and 7 red balls. One ball is drawn at random. What is the probability that the ball drawn is white?

Solution:

Total number of balls in a bag= 7 + 5 = 12

Total number of white balls in a bag= 5

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a white ball \(P(\bar{E}) = \frac{5}{12}\)

 

13. Tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7?

Solution:

Cards marked multiple of 3 or 7 in the 20 tickets=3, 6, 7, 9, 12, 14, 15 and 18

i.e., the total number of cards marked multiple of 3 or 7 = 8

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a, multiple of 3 or 7 is \(\frac{8}{20} = \frac{2}{5}\)

 

14. In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?

Solution:

Number of lottery tickets = 10 + 25 = 35

Total number of prize carrying tickets =10

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of winning a prize is \(\frac{10}{30} = \frac{2}{7}\)

 

15. If the probability of winning a game is 0.3, what is the probability of losing it?

Solution:

Since the sum of probability of occurrence of an event and probability of non occurrence of an event = 1.

We have,

P(E) + \(P(\bar{E})\) = 1

0.3 + \(P(\bar{E})\) = 1

\(P(\bar{E})\) = 1 – 0.3

\(P(\bar{E})\) = 0.7

∴, According to the question,

The probability of losing the game is \(P(\bar{E})\) = 0.7

  

16. A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

(i) red

(ii) black or white

(iii) not black

Solution:

Total number of balls 7+5+3 =15

(i) Total number red balls =7

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a red ball is equal to = \(\frac{7}{15}\)

 

(ii) Total number of black or white balls = 5+3 = 8

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting white or black ball = \(\frac{8}{15}\)

 

(iii) Total number of black balls =5

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting black ball P(E) = \(\frac{5}{15} = \frac{1}{3}\)

Since the sum of probability of occurrence of an event and probability of non occurrence of an event =1,

We have,

P(E) + \(P(\bar{E})\) = 1

\(\frac{1}{3} + P(\bar{E})\) = 1

\( P(\bar{E}) = 1 -\frac{1}{3} \)

\( P(\bar{E}) = \frac{2}{3} \)

Therefore, the probability of getting non black ball \( P(\bar{E}) = \frac{2}{3} \)

 

17. A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

(i) White                  

(ii) Red         

(iii) Not black

(iv) Red or White

Solution:

Total number of balls in a bag= 4+5+6=15

(i) Total number white balls =6

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting white a ball is \(\frac{6}{15} = \frac{2}{5}\)

(ii) Total number of red =4

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting red a ball is equal to = \(\frac{4}{15}\)

(iii) Total number of black balls =5

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting black ball P(E) = \(\frac{5}{15} = \frac{1}{3}\)

Since sum of probability of occurrence of an event and probability of non occurrence of an event =1,

We get,

P(E) + \(P(\bar{E})\) = 1

\(\frac{1}{3} + P(\bar{E})\) = 1

\( P(\bar{E}) = 1 -\frac{1}{3} \)

\( P(\bar{E}) = \frac{2}{3} \)

Therefore, the probability of getting non black ball is \( P(\bar{E}) = \frac{2}{3} \)

 

(iv) Total number of red or white balls= 4 + 6 = 10

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting white or red ball \(\frac{10}{15} = \frac{2}{3}\)

 

18. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:

(i) A king of red suit

(ii) A face card

(iii) A red face card

(iv) A queen of black suit

(v) A jack of hearts

(vi) A spade

Solution:

Total number of cards = 52

(i) Cards with a king of red suit in a deck of card = 2

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting cards which are king of red suit is \(\frac{2}{52} = \frac{1}{26}\)

 

(ii)Each suit has 3 face cards.

One deck of cards has 4 suits.

Then, total number of face cards in a deck of card =3×4 = 12

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting face cards is \(\frac{12}{52} = \frac{3}{13}\)

 

(iii) Total number of red face cards in a deck of card= 6

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting red face cards is \(\frac{6}{52} = \frac{3}{26}\)

 

(iv) Total number of queen of black suit cards in a deck of card = 2

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting cards which are queen of black suit cards is \(\frac{2}{52} = \frac{1}{26}\)

 

(v) Total number of jack of hearts in a deck of card=1

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting cards which are jack of hearts is equal to = \(\frac{1}{52}\)

 

(vi) Total number of spades cards in a deck of card=13

Total numbers of favorable events i.e. total number of queen of black suit cards are 13

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting spade cards is equal to = \(\frac{13}{52} = \frac{1}{4}\)

 

19. Five cards – ten, jack, queen, king, and an ace of diamonds are shuffled face downwards. One card is picked at random.

(i) What is the probability that the card is a queen?

(ii) If a king is drawn first and put aside, what is the probability that the second card picked up is the (a) ace? (b) king?

Solution:

Total number of cards = 5

(i) Cards which is a queen

Total number of Cards which are queen is 1

Number of favorable event i.e. Total number of Cards which are queen is 1

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting cards which are queen = \(\frac{1}{5}\)

(ii) If a king is drawn first and put aside then

Total number of cards is 4

Number of favorable event i.e. Total number of ace card is 1

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting ace cards = \(\frac{1}{4}\)

 

20. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:

(i) Red

(ii) Back

Solution:

Total number of balls 3+5 .8

(i) Total number red balls are 3

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting red ball is equal to = \(\frac{3}{8}\)

 

(ii) Total number of black ball are 5

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting black ball = \(\frac{5}{8}\)

 

21. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number, 1, 2, 3, …., 12 as shown in figure what is the probability that it will point to:

(i) 10?

(ii) an odd number?

(iii) a number which is multiple of 3?

(iv) an even number?

Solution:

Total number on the spin is 12

(i) Favorable event i.e. to get 10 is 1

Total number of Favorable event i.e. to get 10 is 1

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a 10 = \(\frac{1}{12}\)

 

(ii) Favorable event i.e. to get an odd number are 1,3,5,7,9,11,

Total number of Favorable event i.e. to get a prime number is 6

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a prime number is = \(\frac{6}{12} = \frac{1}{2}\)

 

(iii) Favorable event i.e. to get an multiple of 3 are 3,6,9, 12

Total number of Favorable event i.e. to get a multiple of 3 is 4

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting multiple of 3 = \(\frac{4}{12} = \frac{1}{3}\)

 

(iv) Favorable event i.e. to get an even number are 2, 4, 6, 8, 10, 12

Total number of Favorable events i.e. to get an even number is 6

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting an even number is \(\frac{6}{12} = \frac{1}{2}\)

 

22. In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for class monitor. What she does, she writes the name of each pupil on a card and puts them into a basket and mixes thoroughly. A child is asked to pick one card from the basket. What is the probability that the name written on the card is:

(i) The name of a girl

(ii) The name of a boy?

Solution:

Total number of students in the class 18 + 16 = 34

(i) The name of a girl are 18 hence favorable cases are 18

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a name of girl on the card is equal to = \(\frac{18}{34} = \frac{9}{17}\)

 

(ii) The name of a boy are 16 hence favorable cases are 16

We know that,

Probability = \(\frac{Number of favorable event }{Total number of event}\)

∴, According to the question,

The probability of getting a name of boy on the card is equal to = \(\frac{16}{34} = \frac{8}{17}\)

 

23. Why is tossing a coin considered to be a fair way of deciding which team should choose ends in a game of cricket?

Solution:

Possible outcomes=1 head & 1 tail

∴, No. of possible outcomes while tossing a coin = 2

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P( getting head) = \(\frac{1}{2}\)

P( getting tail) = \(\frac{1}{2}\)

Since, probability of two events are equal, these are called equally like events. Hence, tossing a coin is considered to be a fair way of deciding which team should choose ends in a game of cricket.

 

24. What is the probability that a number selected at random from the number 1,2,2,3,3,3, 4, 4, 4, 4 will be their average?

Solution:

Possible outcomes= 1, 2, 2, 3, 3, 3, 4, 4, 4, 4

∴, Total no. of possible outcomes =10

Average of the no’s = \(\frac{sum of numbers}{total numbers}\)

\(\frac{1+2+2+3+3+3+4+4+4+4}{10}\)

= \(\frac{30}{10}\)

= 3

E = event of getting 3

Possible outcomes=3,3,3

No. of favorable outcomes = 3

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{3}{10}\)

 

25. There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3.

Solution:

Possible outcomes= 1,2,3,….30

Total no. of possible outcomes – 30

E = event of getting number divisible by 3

Possible outcomes= 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Number of favorable outcomes = 10

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{10}{30}\)

=\(\frac{1}{3}\)

\(\overline{E}\) = event of getting number not divisible by 3

P(\(\overline{E}\)) = 1 – P(E)

P(\(\overline{E}\)) = 1 – \(\frac{1}{3}\)

= \(\frac{2}{3}\)

 

26. A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is

(i)  red or white

(ii) not black

(iii) neither white nor black.

Solution:

Possible outcomes=5 red, 8 white & 7 black

Total number of possible outcomes = 20

(i)  E = event of drawing red or white ball

possible outcomes= 5 red, 8 white

No. of favorable outcomes = 13 {5 red, 8 white}

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{13}{20}\)

(ii) Let E = event of getting black ball

Possible outcomes= 7 black balls

No. of favorable outcomes =7

P(E) = \(\frac{7}{20}\)

\(\overline{E}\) = event of not getting black ball

P(\(\overline{E}\)) = 1 – P(E)

P(\(\overline{E}\)) = 1 – \(\frac{7}{20}\)

P(\(\overline{E}\)) = \(\frac{13}{20}\)

(iii) Let E = event of getting neither white nor black ball

Possible outcomes= total balls – no. of white balls – no. of black balls

No. of favorable outcomes = 20 – 8 – 7 = 5

P(E) =  \(\frac{5}{20}\)

P(E) = \(\frac{1}{4}\)

 

27. Find the probability that a number selected from the number 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected.

Solution:

Total no. of possible outcomes = 25 {1, 2, 3…. 25}

E = event of getting a prime no.

Possible outcomes= 2, 3, 5, 7, 11, 13, 17, 19, 23

No. of favorable outcomes = 9

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{9}{25}\)

(\(\overline{E}\)) = event of not getting a prime

P(\(\overline{E}\)) = 1— P(E)

P(\(\overline{E}\)) = 1 – \(\frac{9}{25}\)

P(\(\overline{E}\)) = \(\frac{16}{25}\)

 

28. A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is

(i) Red or white

(ii) Not black

(iii) Neither white nor black

Solution:

Total no. of possible outcomes = 8 + 6 + 4 =18 {8 red, 6 white, 4 black)

(i)  E = event of getting red or white ball

Possible outcomes=8 red balls, 6 white balls

No. of favorable outcomes = 14

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{14}{18}\)

P(E) =  \(\frac{7}{9}\)

(ii)  E = event of getting a black ball

Possible outcomes=4 black balls

Number of favorable outcomes = 4

P(E) = \(\frac{4}{18}\)

P(E) = \(\frac{2}{9}\)

(\(\overline{E}\)) = event of not getting a black ball

P(\(\overline{E}\)) = 1 – P(E)

P(\(\overline{E}\)) = 1 – \(\frac{2}{9}\)

P(\(\overline{E}\)) = \(\frac{7}{9}\)

(iii)  E = event of getting neither white nor black

No. of favorable outcomes = 18 – 6 – 4

= 8(Total balls – no. of white balls – no. of black balls)

P(E) = \(\frac{8}{18}\)

P(E) = \(\frac{4}{9}\)

 

29. Find the probability that a number selected at random from the numbers 1, 2, 3…. 35 is a:

(i) Prime number

(ii) Multiple of 7

(iii) Multiple of 3 or 5

Solution:

Total no. of possible outcomes = 35 {1, 2, 3….. 35}

(i) E = event of getting a prime no.

Possible outcomes= 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

No. of favorable outcomes =11

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{11}{35}\)

(ii)  E = event of getting no. which is multiple of 7

Possible outcomes= 7, 14, 21, 28, 35

No. of favorable outcomes = 5

P(E) = \(\frac{5}{35}\)

P(E) = \(\frac{1}{7}\)

(iii)  E = event of getting no which is multiple of 3 or 5

Possible Outcomes= 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 5, 10, 20, 25, 35

No. of favorable outcomes = 16

P(E) = \(\frac{16}{35}\)

 

30. From a pack of 52 playing cards Jacks, queens, kings and aces of red color are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is

(i) A black queen

(ii) A red card

(iii) A black jack

(iv) a picture card (Jacks. queens and kings are picture cards)

Solution:

Total no. of cards = 52

All jacks, queens & kings, aces of red color are removed.

Total no. of possible outcomes = 52 – 2 – 2 – 2 – 2 = 44 {remaining cards}

(i) E = event of getting a black queen

favorable outcomes = queen of spade & club

No. of favorable outcomes = 2

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{2}{44}\)

P(E) = \(\frac{1}{22}\)

(ii)   E = event of getting a red card

No. of favorable outcomes = 26-8

= 18 (total red cards jacks – queens, kings, aces of red color)

P(E)= \(\frac{18}{44}\)

P(E) = \(\frac{9}{22}\)

(iii)  E = event of getting a black jack

favorable outcomes = jack of club & spade

No. of favorable outcomes = 2

P(E) = \(\frac{2}{44}\)

P(E) = \(\frac{1}{22}\)

(iv)  E = event of getting a picture card

favorable outcomes = 2 jacks, 2 kings & 2 queens of black color

No. of favorable outcomes = 6

P(E) = \(\frac{6}{44}\)

P(E) = \(\frac{3}{22}\)

 

31. A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out :

(i) an orange flavored candy

(ii) a lemon flavored candy

Solution:

(i)  The bag contains lemon flavored candies only.

Therefore, the event that Malini will take out an orange flavored candy is an impossible event.

Since, probability of impossible event is 0= P(an orange flavored candy) = 0

(ii)  The bag contains lemon flavored candies only.

Therefore, the event that Malini will take out a lemon flavored candy is sure event.

We know that, probability of sure event is 1.

P(a lemon flavored candy) = 1

 

32. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

Let E be the event that 2 students have same birthday

P(E) = 0.992 (given)

Let, (\(\overline{E}\)) be the event that 2 students do not have same birthday.

We know that,

P(\(\overline{E}\)) + P(E) = 1

P(\(\overline{E}\)) = 1 — P(E)

P(\(\overline{E}\))  = 1 – 0.992 = 0.008

 

33. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) red

(ii) not red

Solution:

possible outcomes = 3 red, 5 black

Total no. of possible outcomes = 8

(i) E = event of getting red ball.

favorable outcomes = 3 red

No. of favorable outcomes = 3

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{3}{8}\)

(ii)  E = event of getting no red ball.

P(E) + P(\(\overline{E}\)) = 1

P(\(\overline{E}\)) = 1 – P(E)

P(\(\overline{E}\)) = 1 – \(\frac{3}{8}\)

P(\(\overline{E}\)) = \(\frac{5}{8}\)

 

34. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be:

(i) red

(ii) not green

Solution:

possible outcomes = 5 red, 8 white, 4 green

Total no. of possible outcomes = 17

(i) E = event of getting a red marble

favorable outcomes = 5 red marbles

Number of favorable outcomes = 5

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{5}{17}\)

(ii) E= event of getting a green marble

favorable outcomes = 4 green marbles

Number of favorable outcomes = 4

P(E) = \(\frac{4}{17}\)

(\(\overline{E}\)) = event of getting not a green marble

P(\(\overline{E}\)) = 1 – P(E)

P(\(\overline{E}\)) = 1 – \(\frac{4}{17}\)

P(\(\overline{E}\)) = \(\frac{13}{17}\)

 

35. A lot consists of 144 ball pens of which 20 are defective and others good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i)She will buy it

(ii) She will not buy it

Solution:

No. of good pens =144 —20 = 24

No. of detective pens = 20

Total no. of possible outcomes =144 (total no. of pens)

(i) E = event of buying pen which is good.

No. of favorable outcomes =124 (124 good pens)

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{124}{144}\)

P(E) = \(\frac{31}{36}\)

(ii)  E = event of not buying a pen = bad

P(E) + P(\(\overline{E}\)) = 1

P(\(\overline{E}\)) = 1 – P(E)

P(\(\overline{E}\)) = 1 – \(\frac{31}{36}\)

P(\(\overline{E}\)) = \(\frac{5}{36}\)

 

36. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is good one.

Solution:

No. of good pens =132

No. of defective pens = 12

Total no. of possible outcomes = 132 + 12 {total no of pens}=144

E = event of getting a good pen.

No. of favorable outcomes =132 {132 good pens}

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{132}{144}\)

P(E) = \(\frac{11}{12}\)

 

37. Five cards— the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is

a) an ace?

b)a queen?

Solution:

possible outcomes = 5 cards

Total no. of possible outcomes = 5

(i) E = event of getting a queen.

possible outcomes = 1 queen

No. of favorable outcomes = 1

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{1}{5}\)

(ii)  If queen is drawn & put aside.

Total no. of remaining cards = 4

(a) E = events of getting a queen.

No. of favorable outcomes = 3

possible outcomes = 4 remaining cards

Total no. of possible outcomes = 4

P(E) = \(\frac{3}{4}\)

(b) E = event of getting a good pen.

favorable outcomes = no queen

No. of favorable outcomes = 0

P(E) = 0

Therefore, E is known as impossible event.

 

38. Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that he gets at least one head?

Solution:

possible outcomes = HT, HH, TT, TH

Total no. of possible outcomes = 4

E = event of getting at least one head

favorable outcomes = HT, HH, TH

No. of favorable outcomes = 3

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{3}{4}\)

 

39. Cards marked with numbers 13, 14, 15…. 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the card drawn is:

(i) divisible by 5

(ii) a number is a perfect square

Solution:

possible outcomes = 13, 14, 15, ….,60

Total no. of possible outcomes = 48

(i) E = event of getting no divisible by 5

favorable outcomes = 15, 20, 25, 30, 35, 40, 45, 50, 55, 60

No. of favorable outcomes =10

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{10}{48}\)

P(E) = \(\frac{5}{24}\)

(ii) E = event of getting a perfect square.

favorable outcomes = 16, 25, 36, 49

No. of favorable outcomes = 4

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{4}{68}\)

P(E) = \(\frac{1}{16}\)

 

40.  A bag contains tickets numbered 11, 12……. 30. A ticket is taken out from the bag at random. Find the probability that the number on the drawn ticket

(i) is a multiple of 7

(ii) is greater than 15 and a multiple of 5.

Solution:

possible outcomes = 11, 12, 13…… 30

Total no. of possible outcomes = 20

(i) E = event of getting no. which is multiple of 7

favorable outcomes = 14, 21, 28

No. of favorable outcomes = 3

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{3}{20}\)

(ii) E = event of getting no. greater than 15 & multiple of 5

favorable outcomes = 20, 25, 30

No. of favorable outcomes = 3 {20, 25, 30}

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{3}{20}\)

 

41. Fill in the blanks:

(i) Probability of a sure event is …….

(ii)Probability of an impossible event is ……..

(iii)The probability of an event (other than sure and impossible event) lies between …… 

(iv)Every elementary event associated to a random experiment has ………. probability.

(v) Probability of an event A + Probability of event ‘not A’ = …………….. 

(vi) Sum of the probabilities of each outcome in an experiment is ………….

Solution:

(i) 1, P(sure event) =1

(ii) 0, P (impossible event) = 0

(iii) 0 & 1, \(0< P(E)< 1\)

(iv)Equal

(v) 1, P(E) + P(\(\overline{E}\)) = 1

(vi)1

 

42. Examine each of the following statements and comment:

(i) If two coins are tossed at the same time. There are 3 possible outcomes—two heads, two tails, or one of each. Therefore, for each outcome, the probability of occurrence is \(\frac{1}{3}\).

(ii) If a die is thrown once, there are two possible outcomes – an odd number or an even number. Therefore, the probability of obtaining an odd number is \(\frac{1}{2}\) and the probability of obtaining an even number is \(\frac{1}{2}\).

Solution:

(i)  Given statement is incorrect. If 2 coins are tossed at the same time.

possible outcomes = HH, HT, TH, TT

Total no. of possible outcomes = 4 {HH, HT, TH, TT}

P(HH)= P(HT) = P(TH) = P(TT) = \(\frac{1}{4}\)

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

for each outcome, probability of occurrence is  \(\frac{1}{4}\)

Outcomes can be classified as (2H, 2T, 1H & 1T)

P(2H) = \(\frac{1}{4}\), P(2T) = \(\frac{1}{4}\),

P(1H & 1T)= \(\frac{1}{2}\)

Events are not equally likely because the event ‘one head & one tail’ is twice as likely to occur as remaining two.

(ii)  This statement is true.

When a die is thrown; possible outcomes = 1, 2, 3, 4, 5, 6

total no. of possible outcomes = 6

These outcomes can be taken as even number & odd number.

P(even no.) = P(2, 4, 6) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

P(odd no.) = P(1, 3,5) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Therefore, two outcomes are equally likely

 

43. A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be

(i) a blue card

(ii) not a yellow card

(iii) Neither yellow nor a blue card.

Solution:

Total no. of possible outcomes = 100 + 200 + 50 = 350 {100 red, 200 yell= & SO blue}

(i)  E = event of getting blue card.

No. of favorable outcomes = 50 (50 blue cards}

P(E) = \(\frac{50}{350}\)

P(E) = \(\frac{1}{7}\)

(ii)  E = event of getting yellow card

No. of favorable outcomes = 200 (200 yellow)

P(E) = \(\frac{200}{350}\)

P(E) = \(\frac{4}{7}\)

(\(\overline{E}\)) = event of not getting yellow card

P(\(\overline{E}\)) = 1 – P(E)

P(\(\overline{E}\)) = 1 –  \(\frac{4}{7}\)

= \(\frac{3}{7}\)

(iii)  E =  getting neither yellow nor a blue card

removing 200 yellow & 50 blue cards

No. of favorable outcomes = 350 – 200 – 50 = 100

P(E) =\(\frac{100}{350}\)

P(E) = \(\frac{2}{7}\)

 

55. The faces of a red cube and a yellow cube are numbered from 1 to 6. Both cubes are rolled. What is the probability that the top face of each cube will have the same number.

Solution:

Total no. of outcomes when both cubes are rolled = 6 x 6 = 36 which are

{(1,1) (1, 2) (1, 3) (1, 4) (1, 5)(1, 6)

(2,1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}

E = event of getting same no. on each cube

No. of favorable outcomes = 6 which are

{ (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{6}{36}\)

= \(\frac{1}{6}\)

 

56. The probability of selecting a green marble at random from a jar that contains only green, white and yellow marbles is \(\frac{1}{4}\). The probability of selecting a white marble at random from the same jar is \(\frac{1}{3}\).If this jar contains 10 yellow marbles. What is the total number of marbles in the jar?

Solution:

Let the no. of green marbles = x

The no. of white marbles = y

No. of yellow marbles =10

Total no. of possible outcomes = x+ y+ 10 (total no. of marbles)

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

Probability (green marble) = \(\frac{1}{4}\)= \(\frac{x}{x+ y+ 10}\)

x + y+ 10 = \(4\times x\)

\(3\times x\) – y – 10 = 0 …….. (i)

Probability (White marble) = \(\frac{1}{3}\) = \(\frac{y}{x + y + 10}\)

x + y+ 10 = \(3\times y\)

x – \(2\times y\) + 10 = 0 …………. (ii)

Multiplying (ii) by 3,

\(3\times x\)\(6\times y\)  + 30 = 0 ……… (iii)

Sub (i) from (iii), we get

\(-6\times y+ y + 30 + 10 = 0\)

\(-5\times y\) + 40=0

\(5\times y\) = 40

y= 8

Substitute y=8 in (i),

\(3\times x\) – 8 -10=0

\(3\times x\)-18=0

x = \(\frac{18}{3}\) = 6

Total number of marbles in jar= x +y + 10 = 6+ 8 + 10=24

57.

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and not replaced. Now bulb is drawn at random from the rest. What Is the probability that this bulb is not defective?

Solution:

Total no. of possible outcomes = 20 {20 bulbs}

(i)  E = event of getting defective bulb.

No. of favorable outcomes = 4 (4 defective bulbs)

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{4}{20}\)

P(E) = \(\frac{1}{5}\)

(ii)  Bulb drawn in is not detective & is not replaced

Remaining bulbs = 15 good + 4 bad bulbs = 19

Total no. of possible outcomes =19

E = event of getting defective

No. of favorable outcomes = 15 (15 good bulbs)

P(E)= \(\frac{15}{19}\)

 

58. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears:

(i) a two digit number

(ii) a perfect square number

(iii) a number divisible by 5.

Solution:

Possible outcomes =1, 2, 3 ……. 90

Total no. of possible outcomes -90

(i) E = event of getting 2 digit no.

Favorable outcomes = 10, 11, 12, …. 90

No. of favorable outcomes = 81

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) =  \(\frac{81}{90}\)

P(E) =  \(\frac{9}{10}\)

(ii) E = event of getting a perfect square.

Favorable outcomes = 1, 4, 9, 16, 25, 26, 49, 64, 81

No. of favorable outcomes = 9

P(E) = \(\frac{9}{90}\)

P(E) = \(\frac{1}{10}\)

(iii) E = event of getting a no. divisible by 5.

Favorable outcomes = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90

No. of favorable outcomes = 18

P(E) = \(\frac{18}{90}\)

P(E) = \(\frac{1}{5}\)

 

59. Two dice, one blue and one grey, are thrown at the same time. Complete the following table:

Class 10 maths chapter 13 probability exercise 13.1-1From the above table a student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7,  8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\) . Do you agree with this argument?

Solution:

Total no. of possible outcomes when 2 dice are thrown = 6 x 6 = 36 which are

{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5)(1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}

E = event of getting 2 as the sum

Favorable outcomes: (1, 1)

No. of favorable outcomes = 1

We know that,

Probability P(E) = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{1}{36}\)

E = event of getting sum 3 as the sum

Favorable outcomes: (1, 2) (2, 1)

No. of favorable outcomes = 2

P(E) = \(\frac{2}{36}\)

P(E) = \(\frac{1}{18}\)

 E = event of getting sum 4 as the sum

Favorable outcomes: (3, 1) (2, 2) (1, 3)

No. of favorable outcomes = 3

P(E) = \(\frac{3}{36}\)

P(E) = \(\frac{1}{12}\)

 E = event of getting 5 as the sum

Favorable outcomes: (1,  4) (2, 3) (3, 2) (4, 1)

No. of favorable outcomes = 4

P(E) = \(\frac{4}{36}\)

P(E) = \(\frac{1}{9}\)

 E = event of getting 6 as the sum

Favorable outcomes: (1, 5) (2, 4) (3, 3) (4, 2) (5, 1)

No. of favorable outcomes = 5

P(E) = \(\frac{5}{36}\)

E = event of getting 7 as the sum

Favorable outcomes: (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)

No. of favorable outcomes = 6

P(E) = \(\frac{6}{36}\)

P(E) = \(\frac{1}{6}\)

 E = event of getting 8 as the sum

Favorable outcomes: (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)

No. of favorable outcomes = 5

P(E) =  \(\frac{5}{36}\)

E = event of getting  9 as the sum

Favorable outcomes: (3, 6) (4, 5) (5, 4) (6, 3)

No. of favorable outcomes = 4

P(E) = \(\frac{4}{36}\)

P(E) = \(\frac{1}{9}\)

E = event of getting 10 as the sum

Favorable outcomes: (4, 6) (5, 5) (6, 4)

No. of favorable outcomes = 3

P(E) = \(\frac{3}{36}\)

P(E) = \(\frac{1}{12}\)

 

60. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag.

Solution:

No of red balls in the bag = 6

Let the no. of blue balls in the bag be x

Then, the total no. of possible outcomes= total no. of balls = 6+ x

We know that,

Probability = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(Blue ball) = 2P(red ball)

\(\frac{x}{x+6}= \frac{2\times 6}{x+6}\)

\(x= 2\times 6\)

x = 12

Therefore, number of blue balls = 12

 

61. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of  

(i) heart

(ii) queen

 (iii) clubs. 

Solution:

Total no. of remaining cords = 52 – 3 = 49

(i) E = event of getting hearts

There are 13 heart cards in a deck of cards.

No. of favorable outcomes = 13

We know that,

Probability = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{13}{49}\)

(ii)  E= event of getting queen

There are 4 heart cards in a deck of cards.

Since queen of clubs is removed, the no. of favorable outcomes = 3 {4 – 1}

We know that,

Probability = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{3}{49}\)

(iii)  E = event of getting clubs

There are total 13 club cards in a deck of cards.

Since 3 club cards are removed, no. of favorable outcomes =13 – 3 = 10

P(E) =\(\frac{10}{49}\)

E = event of getting sum 11

No. of favorable outcomes = 2 {(5, 6) (6, 5)}

We know that,

Probability = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{2}{36}\)

P(E) = \(\frac{1}{18}\)

E = event of getting sum 12

No. of favorable outcomes =1 {(6, 6)}

P(E) = \(\frac{1}{36}\)

 

62. Two dice are thrown simultaneously. What is the probability that:

(i) 5 will not come up on either of them?

(ii) 5 will come up on at least one?

(iii) 5 will come up at both dice?

Solution:

The possible outcomes are:

{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}

Total no. of possible outcomes when 2 dice are thrown = 6 x 6 = 36

(i) E = event of 5 not coming up on either of them

Favorable outcomes are:

{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}

Total no. of favorable outcomes = 25

We know that,

Probability = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{25}{36}\)

 

(ii) E = event of 5 coming up at least once

Favorable outcomes are:

{(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6) (6, 5)}

Total number of favorable outcomes = 11

We know that,

Probability = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{11}{36}\)

 

(iii)  E = event of getting 5 on both dice

Favorable outcome is:

(5, 5)

No. of favorable outcomes =1

We know that,

Probability = \(\frac{Number of favorable outcomes}{Possible outcomes}\)

∴, According to the question,

P(E) = \(\frac{1}{36}\)

 

63. A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.

Solution:

Total no. of possible outcome = 1, 2, 3 …. 50 = 50

Multiples of 3 and 4 in numbers from 1 to 50 = 12, 24, 36, 48

∴, No. of favorable outcomes = 4

P(E) = \(\frac{Favorable outcomes}{Possible outcomes}\)

P(E) = \(\frac{4}{50}\)

P(E) = \(\frac{2}{25}\)

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