QJEEM23Ma7881: JEE Main 2023 Mathematics - Class 12 JEEM23Ma7881 Solution: A bag contains 6 white and 4 black balls. A die is rolled once…

Let $X$ be the number rolled on the die, and let $W$ be the event that all balls drawn are white. We want to find the probability $P(W)$ , which can be calculated using the law of total probability as follows :

$P(W) = \sum\limits_{x=1}^{6} P(W|X=x)P(X=x)$

The probability of rolling any number from 1 to 6 on the die is equal, so $P(X=x) = \frac{1}{6}$ for all $x \in \{1, 2, 3, 4, 5, 6\}$ .

Now let's calculate the conditional probabilities $P(W|X=x)$ for each possible value of $x$ :

1. $P(W|X=1) = {{{}^6{C_1}} \over {{}^{10}{C_1}}}= \frac{6}{10} = \frac{3}{5}$ , since there are 6 white balls out of a total of 10 balls.

2. $P(W|X=2) = {{{}^6{C_2}} \over {{}^{10}{C_2}}} = \frac{15}{45} = \frac{1}{3}$ , since there are 15 ways to choose 2 white balls out of 6, and 45 ways to choose 2 balls out of 10.

3. $P(W|X=3) = {{{}^6{C_3}} \over {{}^{10}{C_3}}} = \frac{20}{120} = \frac{1}{6}$ , since there are 20 ways to choose 3 white balls out of 6, and 120 ways to choose 3 balls out of 10.

4. $P(W|X=4) = {{{}^6{C_4}} \over {{}^{10}{C_4}}} = \frac{15}{210} = \frac{1}{14}$ , since there are 15 ways to choose 4 white balls out of 6, and 210 ways to choose 4 balls out of 10.

5. $P(W|X=5) = {{{}^6{C_5}} \over {{}^{10}{C_5}}} = \frac{6}{252} = \frac{1}{42}$ , since there are 6 ways to choose 5 white balls out of 6, and 252 ways to choose 5 balls out of 10.

6. $P(W|X=6) = {{{}^6{C_6}} \over {{}^{10}{C_6}}} = \frac{1}{210}$ , since there are 1 ways to choose 6 white balls out of 6, and 210 ways to choose 6 balls out of 10.

Using the law of total probability, we have :

$P(W) = \frac{1}{6} \left(P(W|X=1) + P(W|X=2) + P(W|X=3) + P(W|X=4) + P(W|X=5) + P(W|X=6)\right)$

$P(W) = \frac{1}{6} \left(\frac{3}{5} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14} + \frac{1}{42} + \frac{1}{210}\right)$

To simplify this expression, find a common denominator :

$P(W) = \frac{1}{6} \left(\frac{126}{210} + \frac{70}{210} + \frac{35}{210} + \frac{15}{210} + \frac{5}{210} + \frac{1}{210}\right)$

Add the fractions :

$P(W) = \frac{1}{6}\left(\frac{126+70+35+15+5+1}{210}\right)=\frac{42}{210}=\frac{1}{5}$