RS Aggarwal Class 10 Solutions Chapter 15 - Probability Ex 15B (15.2)

RS Aggarwal Class 10 Chapter 15 - Probability Ex 15B (15.2) Solutions Free PDF

The RS Aggarwal Class 10 Solutions Chapter 15 Probability Ex 15B is provided here. With the help of these solutions, students can clear their doubts that may come up while solving the exercise questions from the RS Aggarwal Maths textbook. The solutions are prepared by subject experts in accordance to the latest CBSE syllabus. It explains the difficult concepts in a simple language for can be easily understood. The solutions will not only help in class 10 board exam preparation but also prepares student for competitive exams. These solutions are highly convenient.

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QUESTION 1:A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card was drawn is

(i) a card of a spade or an Ace

(ii) a red king

(iii) either a king or queen

(iv) neither a king nor the queen.

Solution:

Total number of all possible outcomes = 52

(i) Number of space card = 13

Number of aces = 4 (including 1 of spade)

Therefore, number of spade cards and aces = (13 + 4 â€“ 1) = 16

Therefore, P(getting a spade or an ace card) = $\frac{16}{52}$ = $\frac{4}{13}$

(ii) Number of red kings = 2

Therefore, P(getting a red king) = $\frac{2}{52}$ = $\frac{1}{26}$

(iii) Total number of kings = 4

Total number of queens = 4

Let E be the event of getting either a king or a queen.

Then, the favorable outcomes = 4 + 4 = 8

Therefore, P(getting a king or a queen) = P(E) = $\frac{8}{52}$ = $\frac{2}{13}$

(iv) Let E be the event of getting either a king or a queen. Then, (not E) is the event that drawn card is neither a king nor a queen.

Then, P(getting a king or a queen) = $\frac{2}{13}$

Now, P(E) + P(not E) = 1

Therefore, P(getting neither a king nor a queen) = 1 â€“ $\frac{2}{13}$ = $\frac{11}{13}$

QUESTION 2:A box contains 25 cards numbers from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) divisible by 2 or 3,

(ii) a prime number.

Solution:

Total number of outcomes = 25

(i) Let E1 be the event of getting a card divisible by 2 or 3.

Out of given numbers, numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24.

Out of the given numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24.

Out of the given numbers, numbers divisible by both 2 and 3 are 6, 12, 18 and 24.

Number of favorable outcomes = 16

Therefore, P(getting a card divisible by 2 or 3) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{16}{25}$

Thus, the probability that the number on the drawn card is divisible by 2 or 3 is $\frac{16}{25}$.

(ii) Let E2 be the event of getting a prime number.

Out of the given numbers, prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

Number of favorable outcomes = 9

Therefore, P(getting a prime number) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{9}{25}$

Thus, the probability that theÂ number on the drawn card is a prime number is $\frac{9}{25}$.

QUESTION 3:A box contains cards numbered 3, 5, 7, 9 â€¦â€¦, 35, 37. A card is drawn at random from the box. Find the probability that the number on the card is a prime number.

Solution:

Given numbers 3, 5, 7, 9 â€¦â€¦, 35, 37 form an AP with a = 3 and d = 2.

Let Tn = 37. Then,

3 + (n â€“ 1)2 = 37

$\Rightarrow$ 3 + 2n â€“ 2 = 37

$\Rightarrow$ 2n = 36

$\Rightarrow$ n = 18

Thus, total number of outcomes = 18.

Let E be the event of getting a prime number.

Out of these numbers, the prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37.

The number of favorable outcomes = 11.

Therefore, P(getting a prime number) = P(E) = $\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}$ = $\frac{11}{18}$

Thus, the probability that the number on the card is a prime number is $\frac{11}{18}$.

QUESTION 4:Card numbered 1 to 30 are put in a bag. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) not divisible by 3,

(ii) a prime number greater than 7,

(iii) not a perfect square number.

Solution:

The total number of outcomes = 30.

(i) Let E1 be the event of getting a number not divisible by 3.

Out of these numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30.

Number of favorable outcomes = 30 â€“ 10 = 20

Therefore, Â P(getting a number not divisible by 3) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{20}{30}$ = $\frac{2}{3}$.

Thus, the probability that the number on the card is not divisible by 3 is $\frac{2}{3}$.

(ii) Let E2 be the event of getting a prime number greater than 7.

Out of these numbers, prime numbers greater than 7 are 11, 13, 17, 19, 23 and 29.

Number of favorable outcomes = 6

Therefore, Â P(getting a prime number greater than 7) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{6}{30}$ = $\frac{1}{5}$.

Thus, the probability that the number on the card is a prime number greater than 7 is $\frac{1}{5}$.

(iii) Let E3 be the event of getting a number which is not a perfect square number.

Out of these numbers, perfect square numbers are 1, 4, 9, 16 and 25.

Number of favorable outcomes = 30 â€“ 5 = 25

Therefore, P(getting non-perfect square number) = P(E3) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{3}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{25}{30}$ = $\frac{5}{6}$

Thus, the probability that the number on the card is not a perfect square number is $\frac{5}{6}$.

QUESTION 5:Cards bearing numbers 1, 3, 5, . . . . . . . , 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing

(i) a prime number less than 15,

(ii) a number divisible by 3 and 5.

Solution:

Given numbers 1, 3, 5, . . . . . . . , 35 form an AP with a = 1 and d = 2.

Let Tn = 35. Then,

1 + (n â€“ 1)2 = 35

$\Rightarrow$ 1 + 2n â€“ 2 = 35

$\Rightarrow$ 2n = 36

$\Rightarrow$ n = 18

Thus, total number of outcomes = 18.

(i) Let E1 be the event of getting a prime number less than 15.

Out of these numbers, prime numbers less than 15 are 3, 5, 7, 11 and 13.

The number of favorable outcomes = 5.

Therefore, P(getting a prime number less than 15) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{5}{18}$

Thus, the probability of getting a card bearing a prime number less than 5 is $\frac{5}{18}$.

(ii) Let E2 be the event of getting a number divisible by 3 and 5.

Out of these numbers, the number divisible by 3 and 5 means number divisible by 15 is 15.

The number of favorable outcomes = 1.

Therefore, P(getting a number divisible by 3 and 5) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{1}{18}$

Thus, the probability of getting a card bearing a number divisible by 3 and 5 is Â $\frac{1}{18}$.

QUESTION 6:A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, find the probability that it bears

(i) a 1 digit number,

(ii) a number divisible by 5,

(iii) an odd number less than 30,

(iv) a composite number between 50 and 70.

Solution:

Given numbers 6, 7, 8, â€¦â€¦., 70 form an AP with a = 6 and d = 1.

Let Tn = 70. Then,

6 + (n â€“ 1)1 = 70

$\Rightarrow$ 6 + n â€“ 1 = 70

$\Rightarrow$ n = 65

Thus, total number of outcomes = 65.

(i) Let E1 be the event of getting a one-digit number.

Out of these numbers, one-digit numbers are 6, 7, 8 and 9.

The number of favorable outcomes = 4.

Therefore, Â P(getting a one-digit number) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{4}{65}$

Thus, the probability that the card bears a one-digit number is $\frac{4}{65}$.

(ii) Let E2 be the event of getting a number divisible by 5.

Out of these numbers, numbers divisible by 5 are 10, 15, 20, â€¦â€¦ , 70.

Given number 10, 15, 20, â€¦â€¦. , 70 form an AP with a = 10 and d = 5.

Let Tn = 70. Then,

10 + (n â€“ 1)5 = 70

$\Rightarrow$ 10 + 5n â€“ 5 = 70

$\Rightarrow$ 5n = 65

$\Rightarrow$ n = 13

Thus, number of favorable outcomes = 13.

Therefore, Â P(getting a number divisible by 5) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{13}{65}$ = $\frac{1}{5}$

Thus, the probability that the card bears a number divisible by 5 is $\frac{1}{5}$.

(iii) Let E3 be the event of getting an odd number less than 30.

Out of these numbers, odd numbers less than 30 are 7, 9, 11, â€¦â€¦ , 29.

Given number 7, 9, 11, â€¦â€¦ , 29 form an AP with a = 7 and d = 2.

Let Tn = 29. Then,

7 + (n â€“ 1)2 = 29

$\Rightarrow$ 7 + 2n â€“ 2 = 29

$\Rightarrow$ 2n = 24

$\Rightarrow$ n = 12

Thus, number of favorable outcomes = 12.

Therefore, P(getting a odd number less than 30) = P(E3) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{3}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{12}{65}$

Thus, the probability that the card bears an odd number less than 30 is $\frac{12}{65}$.

(iv) Let E4 be the event of getting a composite number between 50 and 70.

Out of these numbers, composite numbers between 50 and 70 are 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68 and 69.

Number of favorable outcomes = 15.

Therefore, P(getting a composite number between 50 and 70) = P(E4) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{4}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{15}{65}$= $\frac{3}{13}$

Thus, the probability that the card bears composite number between 50 and 70 is $\frac{3}{13}$.

QUESTION 7:Cards marked with numbers 1, 3, 5, . . . . . . . , 101 are placed in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) less than 19,

(ii) a prime number less than 20.

Solution:

Given number 1, 3, 5, . . . . . . . . , 101 form an AP with a = 1 and d = 2.

Let Tn = 101. Then,

1 + (n â€“ 1)2 = 101

$\Rightarrow$ 1 + 2n â€“ 2 = 101

$\Rightarrow$ 2n = 102

$\Rightarrow$ n = 51

Thus, total number of outcomes = 51.

(i) Let E1 be the event of getting a number less than 19.

Out of these numbers, numbers less than 19 are 1, 3, 5, â€¦â€¦. , 17.

Given number 1, 3, 5, â€¦â€¦. , 17 form an AP with a = 1 and d = 2.

Let Tn = 17. Then,

1 + (n â€“ 1)2 = 17

$\Rightarrow$ 1 + 2n â€“ 2 = 17

$\Rightarrow$ 2n = 18

$\Rightarrow$ n = 9

Thus, number of favorable outcomes = 9.

Therefore, Â P(getting a number less than 19) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{9}{51}$= $\frac{3}{17}$

Thus, the probability that the number on the drawn card is less than 19 is $\frac{3}{17}$.

(ii) Let E2 be the event of getting a prime number less than 20.

Out of these numbers, prime numbers less than 20 are 3, 5, 7, 11, 13, 17 and 19.

Thus, the number of favorable outcomes = 7.

Therefore, Â P(getting a prime number less than 20) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{7}{51}$

Thus, the probability that the number on the drawn card is a prime number less than 20 is $\frac{7}{51}$.

QUESTION 8:Tickets numbered 2, 3, 4, 5, . . . . . . , 100, 101 are placed in a box and mix thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is

(i) an even number

(ii) a number less than 16

(iii) a number which is a perfect square

(iv) a prime number less than 40.

Solution:

All possible outcomes are 2, 3, 4, 5, â€¦â€¦â€¦â€¦.. , 101.

Number of all possible outcomes = 100

(i) Out of these the numbers that are even = 2, 4, 6, 8, â€¦â€¦â€¦â€¦â€¦â€¦â€¦. , 100

Let E1 be the event of getting an even number.

Then, number of favorable outcomes = 50

[Tn = 100 $\Rightarrow$ 2 + (n â€“ 1) x 2 = 100, $\Rightarrow$ n = 50]

Therefore, P(getting an even number) = $\frac{50}{100}$ = $\frac{1}{2}$

(ii) Out of these, the numbers that are less than 16 = 2, 3, 4, 5, â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. , 15.

Let E2 be the event of getting a number less than 16.

Then, number of favorable outcomes = 14

Therefore, P(getting a number less than 16) = $\frac{14}{100}$ = $\frac{7}{50}$

(iii) Out of these, the numbers that are perfect squares = 4, 9, 16, 25, 36, 49, 64, 81 and 100

Let E3 be the event of getting a number that is a perfect square.

Then, number of favorable outcomes = 9

Therefore, P(getting a number that is a perfect square) = $\frac{9}{100}$

(iv) Out of these, prime numbers less than 40 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Let E4 be the event of getting a prime number less than 40.

Then, number of favorable outcomes = 12

Therefore, P(getting a prime number less than 40) = $\frac{12}{100}$ = $\frac{3}{25}$

QUESTION 9:Â A box contains 80 discs, which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect square number.

Solution:

The total number of outcomes = 80.

Let E1 be the event of getting a perfect square number.

Out of these numbers, perfect square numbers are 1, 4, 9, 16, 25, 36, 49 and 64.

Thus, the number of favorable outcomes = 8.

Therefore, Â P(getting a perfect square number) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{8}{80}$ = $\frac{1}{10}$

Thus, the probability that the disc bears a perfect square number is $\frac{1}{10}$.

QUESTION 10:A piggy bank contains hundred 50-p coins, seventy Rs. 1 coin, fifty Rs. 2 coins and thirty Rs. 5 coins. If it is equally likely that one of the coins will fall out when the blank is turned upside down, what is the probability that the coin

(i) will be a Rs. 1 coin?

(ii) will not be a Rs. 5 coin

(iii) will be 50-p or a Rs. 2 coin?

Solution:

Number of 50-p coins = 100.

Number of Rs. 1 coins = 70.

Number of Rs. 2 coins = 50.

Number of Rs. 5 coins = 30.

Thus, the total number of outcomes = 250.

(i) Let E1 be the event of getting a Rs. 1 coin.

The number of favorable outcomes = 70.

Therefore, P(getting a Rs. 1 coin) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{70}{250}$ = $\frac{7}{25}$

Thus, the probability that the coin will be a Rs. 1 coin is $\frac{7}{25}$.

(ii) Let E2 be the event of not getting a Rs. 5.

Number of favorable outcomes = 250 â€“ 30 = 220

Therefore, P(not getting a Rs. 5 coin) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{220}{250}$ = $\frac{22}{25}$

Thus, probability that the coin will not be a Rs. 5 coin is $\frac{22}{25}$.

(iii) Let E3 be the event of getting a 50-p or a Rs. 2 coins.

Number of favorable outcomes = 100 + 50 = 150

Therefore, P(getting a 50-p or a Rs. 2 coin) = P(E3) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{3}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{150}{250}$ = $\frac{3}{5}$

Thus, probability that the coin will be a 50-p or a Rs. 2 coin is $\frac{3}{5}$.

QUESTION 11:The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is $\frac{1}{4}$. The probability of selecting a blue ball at random from the same jar is $\frac{1}{3}$.If the jar contains 10 orange balls, find the total number of balls in the jar.

Solution:

It is given that,

P(getting a red ball) = $\frac{1}{4}$ and P(getting a blue ball) = $\frac{1}{3}$

Let P(getting an orange ball) be x.

Since, there are only 3 types of balls in the jar, the sum of probabilities of all the three balls must be 1.

Therefore, $\frac{1}{4}+\frac{1}{3}+x=1$

$\Rightarrow$ $x=1-\frac{1}{4}-\frac{1}{3}$

$\Rightarrow$ x = $\frac{12-3-4}{12}$

$\Rightarrow$ x = $\frac{5}{12}$

Therefore, Â P(getting an orange ball) = $\frac{5}{12}$.

Let the total number of balls in the jar be n.

Therefore, Â P(getting an orange ball) = $\frac{10}{n}$

$\Rightarrow$ $\frac{10}{n}$ = $\frac{5}{12}$

$\Rightarrow$ n = 24

Thus, the total number of balls in the jar is 24.

Key Features of RS Aggarwal Class 10 Solutions Chapter 15 â€“Â Probability Ex 15B (15.2)

• It improves mathematical problem-solving skills.
• It explains every solution in a proper stepwise manner.
• The RS Aggarwal maths solutions provide students with different types of questions to practice.
• It helps in boosting their self-confidence level.

Practise This Question

Carbon has a valency of four. It is capable of forming single bonds with four other carbon atoms or atoms of some other monovalent element, etc. What is this property of carbon known as?