*1-Two dice are thrown simultaneously. What is the probability that the sum of the numbers is *

*a) *8?

*b) *2?

*c) *Prime number?

**Sol: **When two dice are thrown simultaneously, the maximum number of possibilities in the sample

space is (\(6\times 6\)=36).

**a)**Sum of numbers is 8 {(2,6),(3,5),(4,4),(6,2),(5,3)}

Therefore, P(sum 8) = \(\frac{5}{36}\)

** b)**Sum of numbers is 2 {(1,1)}

Therefore, P(sum 2) = \(\frac{1}{36}\)

**c)**Sum is a Prime number {(1,1),(1,2),(1,4),(2,1),(2,3),(3,2),(4,1),(1,6),(2,5),(3,4),(4,3),(5,2),(5,6),(6,1),(6,5)}

(Because, prime numbers within 12 are, 2, 3, 5, 7, 11)

Therefore, P(sum is a prime number) = \(\frac{15}{36}\)= \(\frac{5}{12}\)

*2- Two dice are thrown together. Find the probability that the product of the numbers on the top**of the dice is:*

**i) 8**

**ii) 12**

**iii) 5**

**Sol: ** When two dice are thrown simultaneously, the maximum number of possibilities in the sample

space is (\(6\times 6\)=36).

i) Product of the numbers on the top is 8 {(2,4), (4,2)}

Therefore, P(product is 8) = \(\frac{2}{36}\)= \(\frac{1}{18}\)

ii) Product of on the top is 12 {(2,6), (3, 4),(4,3),(6,2)}

Therefore, P(product is 12) = \(\frac{4}{36}\)= \(\frac{1}{9}\)

iii) Product of numbers on the top is 5 {(1,5),(5,1)}

Therefore, P(product is 5) = \(\frac{2}{36}\)= \(\frac{1}{18}\)

*3- A bag contains 24 balls of which A are red, 2A are white and 3A are blue. A ball is selected at**random. What is the probability that it is:*

**i)Not blue?**

**ii)Red?**

**Sol: ** Total number of balls in the bag = 24

number of red balls in the bag = A

number of white balls in the bag = 2A

number of blue balls in the bag = 3A

Total number of balls = \(A+2A+3A\)= 6A which is actually 24.

Therefore, 6A = 24 and A = 4.

number of red balls in the bag = 4

number of white balls in the bag = 8

number of blue balls in the bag = 12

i)Not blue?

P(not blue) = \(\frac{12}{24}\)= \(\frac{1}{2}\)

ii)Red?

P(Red) = \(\frac{4}{24}\)= \(\frac{1}{6}\)

*4- At a gala, cards bearing numbers 1 to 500, one number on one card are put in a box. Each player**selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 250, the player wins a prize. What is the probability that:*

*i)The first player wins a prize? *

*ii)The second player wins a prize if the first has won?*

**Sol:** Total number of cards – 500

i) The first player wins a prize.

Perfect squares greater than 250 are 256, 289, 324, 361, 400, 441 and 484 (7 outcomes only).

P(first player winning) = \(\frac{7}{500}\)

ii) The second player wins a prize, if the first has won.

If the first has won, the number of perfect squares available will be 6 and

The total number of available outcomes becomes 499.

P(second player winning after first has won) = \(\frac{6}{499}\)

*5- All the jacks, queens and kings are removed from a deck of 52 cards. The remaining cards are**well shuffled and then one card is drawn at random. Giving ace a value of 1 and similar values for**other cards, find the probability that the card has a value:*

*i) 6*

*ii) Greater than 6 *

*iii) Less than 6*

**Sol: ** When all jacks (4) queens (4) and kings (4) are removed then, sample space = 52 -12 = 40 cards.

i)P(6) = \(\frac{4}{40}\)= \(\frac{1}{10}\)

ii)P(>6) = \(\frac{16}{40}\)= \(\frac{2}{5}\)

iii)P(<6) = \(\frac{20}{40}\)= \(\frac{1}{2}\)

** **

*6- A box contains 10 red marbles, 5 blue marbles and 7 green marbles. One marble is taken out of at random. What is the probability that the marble taken out will be:*

*i) Neither green nor blue*

*ii) Blue*

*iii Not green*

**Sol: ** Number of red marbles = 10

Number of blue marbles = 5

Number of green marbles = 7

Total number of marbles = \(10+5+7= 22\)

i)P(Neither green nor blue) = P(Blue) = \(\frac{10}{22}\)= \(\frac{5}{11}\)

ii)P(Not green) = \(\frac{15}{22}\)

*7- Two dice are thrown at the same time. Find the probability of getting:*

*i)Same number on both dice*

*ii)Different numbers on both dice*

**Sol:** When two dice are thrown. then sample space contains 36 outcomes(\(6\times 6\)).

i)P(Same number on both dice) = \(\frac{6}{36}\)= \(\frac{1}{6}\)

{The six outcomes are [(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)]}

ii)P(Different numbers on both dice) = \(\frac{30}{36}\)= \(\frac{5}{6}\)

*8- A lot consists of 50 mobile phones of which 43 good, 3 have only minor defects and 4 have major defects. Mani will buy a phone only if it is good. And the trader will buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is:** *

*i)Acceptable to Mani?*

*ii)Acceptable to Trader?*

**Sol: **

i) Phones are acceptable to Mani when it is good

P(Acceptable to Mani) = \(\frac{43}{50}\)

ii) Phones are acceptable to the trader only if it has no major defect

P(Acceptable to the trader) = \(\frac{46}{50}\)= \(\frac{23}{25}\)

*9- Box A contains 25 slips of which 20 are marked $1 and the rest are marked $5 each. Box B contains 50 slips of which 45 are marked $1 each and the rest are marked $13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than $5?*

**Sol: **Total number of slips = \(25+50\)= 75

Number of slips marked $1 = 65

Number of slips marked $5 = 5

Number of slips marked $13 = 5

P( Slip marked other than $5) = \(\frac{70}{75}\)= \(\frac{14}{15}\)

* *

*10- A carton of 30 bulbs contains 5 defective bulbs. One bulb is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?*

**Sol: **Total number of bulbs = 30

Number of non-defective bulbs = 30-5 = 25

P(Non-defective bulb) = \(\frac{25}{30}\)= \(\frac{5}{6}\)

If the bulb is defective and removed, the number of defective bulbs becomes 4.

Therefore, P(Defective bulb) = \(\frac{4}{29}\)

11- *In a game, the entry fee is $5. The game consists of tossing a coin 3 times. If one or two heads show, Shiva gets his entry fee back. If he throws 3 heads, he receives double the entry fee. Otherwise, he will lose. For tossing a coin three times, find the probability that he: (i) Loses the entry fee (ii) Gets double the entry fee (iii) Just gets his entry fee*

**Sol: **When a coin is tossed thrice, there are 8 possible outcomes.

{(HHH),(HHT),(HTT),(HTH),(THH),(TTT),(THT),(TTH)}

i)P(Loses his entry fee) = \(\frac{1}{8}\)

ii)P(Gets double the entry fee) = \(\frac{1}{8}\)

iii)P(Just gets his entry fee) = \(\frac{6}{8}\)= \(\frac{3}{4}\)

*12- The probability of getting a rotten egg from a lot of 400 eggs is 0.07. Find the number of rotten**eggs in the lot.*

Sol: Number of rotten eggs = \(400×0.07\)= 28

*13- A bag contains 3 red balls, 5 white balls and 7 black balls. What is the probability that a ball**drawn from the bag at random will be neither red nor white?*

Sol: Total number of balls = \(3+5+7\)= 15

P(Neither red nor white) = \(\frac{7}{15}\)

*14- If the probability of an event is 0.75, find the probability of its complimentary event.*

Sol: P(complimentary) = \(1-0.75\)= 0.25

*15- A card is selected from a deck of 52 playing cards. Find the probability of a black face card.*

Sol: Total number of cards = 52

Number of black face cards = 6

P( Black face card) = \(\frac{6}{52}\)= \(\frac{3}{26}\)

*16- Find the probability that a leap year selected at random will contain 53 Mondays.*

Sol: A leap year contains 366 days (52 weeks and 2 days). The possible outcomes of the two days are

(Mon, Tue), (Tue, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat), (Sat, Sun), (Sun,Mon)

i.e., there are 7 outcomes. Out of them, favourable outcomes = 2 {(Mon, Tue) and (Sun, Mon)}

P(53 Mondays) = \(\frac{2}{7}\)

*17- When a die is thrown once, find the probability of getting an odd number less than 5.*

Sol: When a die is thrown once, the possible outcomes are 1,2,3,4,5 and 6.

Out of them, odd numbers less than 5 are 1 and 3.

P(Odd no less than 5) = \(\frac{2}{6}\)= \(\frac{1}{3}\)

*18- A card is drawn from a deck of 52 cards. The event E is that the card is not an ace of Spades. Find**the number of outcomes favourable to E.*

Sol: Number of ace of spades = 1

Number of outcomes favourable to E = \(52-1\)= 51

*19- A girl calculates that the probability of her winning the lottery is 0.08. if 5000 tickets were sold,**how many tickets has she bought?*

Sol: Number of tickets bought = \(0.08×5000\)= 400

*20- One ticket is drawn at random from a bag containing tickets numbered 1 to 80. Find the**probability that the selected ticket has a number which is a multiple of 4.*

Sol: Numbers between 1 and 80 which are multiples of 4 = 20

P(Multiple of 4) = \(\frac{20}{80}\)= \(\frac{1}{4}\)

*21- Mani is asked to take a number from 1 to 50. Find the probability that it is a prime number.*

Sol: The prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37,41,43 and 47.

There are 15 favourable outcomes. Therefore, P(Prime number) = \(\frac{15}{50}\)= \(\frac{3}{10}\)

*22- A school has five houses A, B, C, D and E. A class has 50 students, 8 from house A, 16 from house**B, 10 from house C, 4 from house D and rest from house E. A single student is selected at**random to be the class monitor. Find the probability that the selected student is not from A, D**and E.*

Sol: Number of students in A = 8

Number of students in D = 4

Number of students in E = 50 – (\(8+16+10+4\)) = 12

Number of students not from A,D and E = \(50-(8+4+12)\)= 26

P(Student not from A,D and E) = \(\frac{26}{50}\)= \(\frac{13}{25}\)