# A box contains 20 balls bearing numbers 1,2,3,…,20 respectively. A ball is drawn at random from the box, what is the probability that the number on the ball is

$(i)$An odd number

$(ii)$Divisible by 2 or 3

$(iii)$A prime number

$(iv)$Not divisible by 10

$

{\text{A}}{\text{.}} \\

(i)\dfrac{1}{3} \\

(ii)\dfrac{{12}}{{19}} \\

(iii)\dfrac{2}{5} \\

(iv)\dfrac{9}{{10}} \\

$ $

{\text{B}}{\text{.}} \\

(i)\dfrac{1}{2} \\

(ii)\dfrac{{13}}{{20}} \\

(iii)\dfrac{2}{5} \\

(iv)\dfrac{9}{{10}} \\

$ $

{\text{C}}{\text{.}} \\

(i)\dfrac{1}{3} \\

(ii)\dfrac{{18}}{{37}} \\

(iii)\dfrac{2}{5} \\

(iv)\dfrac{2}{5} \\

$ $

{\text{D}}{\text{.}} \\

(i)\dfrac{1}{2} \\

(ii)\dfrac{{15}}{{29}} \\

(iii)\dfrac{2}{5} \\

(iv)\dfrac{2}{5} \\

$

Last updated date: 25th Mar 2023

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Answer

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Hint- Here, we will proceed by using the general formula for finding the probability of occurrence of an event. Here, the favourable cases will be evaluated by knowing all the odd numbers out of 1 to 20, all the numbers divisible by 2 or 3 out of 1 to 20, all the prime numbers out of 1 to 20 and all the numbers that are not divisible by 10 out of 1 to 20.

Complete step-by-step answer:

Given, we are having a box containing 20 balls having numbers 1,2,3,….,20 respectively on them.

As we know that the general formula for the probability is given by

Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable cases}}}}{{{\text{Total number of possible cases}}}}{\text{ }} \to {\text{(1)}}$

Here, a ball is selected at random from the box so any one of the 20 balls can occur.

Total number of possible cases = 20

$(i)$ Here, the favourable case is that the number on the selected ball is an odd number.

As, odd numbered balls are the balls with numbers 1,3,5,7,9,11,13,15,17,19 (total 10 balls)

Number of favourable cases = 10

Using the formula given by equation (1), we get

Probability that the number on the selected ball is an odd number$ = \dfrac{{{\text{10}}}}{{{\text{20}}}} = \dfrac{1}{2}$.

$(ii)$ Here, the favourable case is that the number on the selected ball is divisible by 2 or 3.

The balls that have numbers divisible by 2 or 3 are the balls with numbers 2,3,4,6,8,9,10,12,14,15,16,18,20 (total 13 balls)

Number of favourable cases = 13

Using the formula given by equation (1), we get

Probability that the number on the selected ball is divisible by 2 or 3$ = \dfrac{{{\text{13}}}}{{{\text{20}}}}$.

$(iii)$ Here, the favourable case is that the number on the selected ball is a prime number.

As, prime numbered balls are the balls with numbers 2,3,5,7,11,13,17,19 (total 8 balls)

Number of favourable cases = 8

Using the formula given by equation (1), we get

Probability that the number on the selected ball is a prime number$ = \dfrac{{\text{8}}}{{{\text{20}}}} = \dfrac{2}{5}$.

$(iv)$ Here, the favourable case is that the number on the selected ball is not divisible by 10.

As, odd numbered balls are the balls with numbers 1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,19 (total 18 balls)

Number of favourable cases = 18

Using the formula given by equation (1), we get

Probability that the number on the selected ball is not divisible by 10 $ = \dfrac{{{\text{18}}}}{{{\text{20}}}} = \dfrac{9}{{10}}$.

Hence, option B is correct.

Note- In this particular problem, out of numbers from 1 to 20 there are 10 odd numbers (i.e., 1,3,5,7,9,11,13,15,17,19) and 10 even numbers (i.e., 2,4,6,8,10,12,14,16,18,20). Also, numbers out of 1 to 20 that are not divisible by 10 can be determined as (20-2)=18 where 20 represents the total numbers and 2 represents the numbers that are divisible by 10 (which are 10 and 20).

Complete step-by-step answer:

Given, we are having a box containing 20 balls having numbers 1,2,3,….,20 respectively on them.

As we know that the general formula for the probability is given by

Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable cases}}}}{{{\text{Total number of possible cases}}}}{\text{ }} \to {\text{(1)}}$

Here, a ball is selected at random from the box so any one of the 20 balls can occur.

Total number of possible cases = 20

$(i)$ Here, the favourable case is that the number on the selected ball is an odd number.

As, odd numbered balls are the balls with numbers 1,3,5,7,9,11,13,15,17,19 (total 10 balls)

Number of favourable cases = 10

Using the formula given by equation (1), we get

Probability that the number on the selected ball is an odd number$ = \dfrac{{{\text{10}}}}{{{\text{20}}}} = \dfrac{1}{2}$.

$(ii)$ Here, the favourable case is that the number on the selected ball is divisible by 2 or 3.

The balls that have numbers divisible by 2 or 3 are the balls with numbers 2,3,4,6,8,9,10,12,14,15,16,18,20 (total 13 balls)

Number of favourable cases = 13

Using the formula given by equation (1), we get

Probability that the number on the selected ball is divisible by 2 or 3$ = \dfrac{{{\text{13}}}}{{{\text{20}}}}$.

$(iii)$ Here, the favourable case is that the number on the selected ball is a prime number.

As, prime numbered balls are the balls with numbers 2,3,5,7,11,13,17,19 (total 8 balls)

Number of favourable cases = 8

Using the formula given by equation (1), we get

Probability that the number on the selected ball is a prime number$ = \dfrac{{\text{8}}}{{{\text{20}}}} = \dfrac{2}{5}$.

$(iv)$ Here, the favourable case is that the number on the selected ball is not divisible by 10.

As, odd numbered balls are the balls with numbers 1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,19 (total 18 balls)

Number of favourable cases = 18

Using the formula given by equation (1), we get

Probability that the number on the selected ball is not divisible by 10 $ = \dfrac{{{\text{18}}}}{{{\text{20}}}} = \dfrac{9}{{10}}$.

Hence, option B is correct.

Note- In this particular problem, out of numbers from 1 to 20 there are 10 odd numbers (i.e., 1,3,5,7,9,11,13,15,17,19) and 10 even numbers (i.e., 2,4,6,8,10,12,14,16,18,20). Also, numbers out of 1 to 20 that are not divisible by 10 can be determined as (20-2)=18 where 20 represents the total numbers and 2 represents the numbers that are divisible by 10 (which are 10 and 20).

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