JEE MAIN - Mathematics (2023 - 15th April Morning Shift - No. 9)
A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is :
$\frac{1}{4}$
$\frac{9}{50}$
$\frac{1}{5}$
$\frac{11}{50}$
Explanation
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 :
$$P(W) = \frac{1}{6}\left(\frac{126+70+35+15+5+1}{210}\right)=\frac{42}{210}=\frac{1}{5}$$
$$P(W) = \sum\limits_{x=1}^{6} P(W|X=x)P(X=x)$$
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 :
$$P(W) = \frac{1}{6}\left(\frac{126+70+35+15+5+1}{210}\right)=\frac{42}{210}=\frac{1}{5}$$
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