Logo
Jeeify
MCQEasyJEE 2024Conditional Probability & Bayes Theorem

JEE Mathematics 2024 Question with Solution

Bag A contains 3 white and 7 red balls, and bag B contains 3 white and 2 red balls. One bag is selected at random, and a ball is drawn. The probability of drawing the ball from bag A, given the ball drawn is white, is:

  • A

    14\frac{1}{4}

  • B

    19\frac{1}{9}

  • C

    13\frac{1}{3}

  • D

    310\frac{3}{10}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Bag AA contains 33 white and 77 red balls, bag BB contains 33 white and 22 red balls. One bag is chosen at random and a white ball is observed.

Find: The probability that the selected bag was AA, given that the drawn ball is white.

Define events: E1E_1 = Bag AA is selected, E2E_2 = Bag BB is selected, and EE = a white ball is drawn.

The probabilities of selecting the bags are:

P(E1)=12,P(E2)=12P(E_1)=\frac{1}{2}, \qquad P(E_2)=\frac{1}{2}

The conditional probabilities of drawing a white ball are:

P(E∣E1)=310,P(E∣E2)=35P(E\mid E_1)=\frac{3}{10}, \qquad P(E\mid E_2)=\frac{3}{5}

Using Bayes' theorem:

P(E1∣E)=P(E∣E1)×P(E1)P(E∣E1)×P(E1)+P(E∣E2)×P(E2)P(E_1\mid E)=\frac{P(E\mid E_1)\times P(E_1)}{P(E\mid E_1)\times P(E_1)+P(E\mid E_2)\times P(E_2)}

Substituting the values:

P(E1∣E)=310×12310×12+35×12P(E_1\mid E)=\frac{\frac{3}{10}\times \frac{1}{2}}{\frac{3}{10}\times \frac{1}{2}+\frac{3}{5}\times \frac{1}{2}}

Simplifying:

P(E1∣E)=320320+310=320320+620=39=13P(E_1\mid E)=\frac{\frac{3}{20}}{\frac{3}{20}+\frac{3}{10}}=\frac{\frac{3}{20}}{\frac{3}{20}+\frac{6}{20}}=\frac{3}{9}=\frac{1}{3}

Therefore, the correct option is C.

Using total probability first

Given: P(A1)=P(A2)=12P(A_1)=P(A_2)=\frac{1}{2}, P(W∣A1)=310P(W\mid A_1)=\frac{3}{10}, and P(W∣A2)=35P(W\mid A_2)=\frac{3}{5}.

Find: P(A1∣W)P(A_1\mid W).

First compute the total probability of drawing a white ball:

P(W)=P(W∣A1)⋅P(A1)+P(W∣A2)⋅P(A2)P(W)=P(W\mid A_1)\cdot P(A_1)+P(W\mid A_2)\cdot P(A_2)

Substitute the values:

P(W)=(310×12)+(35×12)P(W)=\left(\frac{3}{10}\times \frac{1}{2}\right)+\left(\frac{3}{5}\times \frac{1}{2}\right)

Simplify:

P(W)=320+310=320+620=920P(W)=\frac{3}{20}+\frac{3}{10}=\frac{3}{20}+\frac{6}{20}=\frac{9}{20}

Now apply Bayes' theorem:

P(A1∣W)=P(W∣A1)⋅P(A1)P(W)P(A_1\mid W)=\frac{P(W\mid A_1)\cdot P(A_1)}{P(W)}

Substitute again:

P(A1∣W)=(310×12)920=320920=320×209=13P(A_1\mid W)=\frac{\left(\frac{3}{10}\times \frac{1}{2}\right)}{\frac{9}{20}}=\frac{\frac{3}{20}}{\frac{9}{20}}=\frac{3}{20}\times \frac{20}{9}=\frac{1}{3}

Thus, the probability of drawing the ball from Bag AA, given that the drawn ball is white, is 13\frac{1}{3}.

Common mistakes

  • Using P(A∣W)=P(W∣A)P(A\mid W)=P(W\mid A). These two conditional probabilities are different. Use Bayes' theorem to reverse the condition correctly.

  • Ignoring that each bag is selected with probability 12\frac{1}{2}. The prior probability of selecting each bag must be included in the numerator and denominator.

  • Taking the probability of white from Bag BB as 310\frac{3}{10} instead of 35\frac{3}{5}. Bag BB has only 55 balls in total, so its white-ball probability is computed out of 55, not 1010.

Practice more Conditional Probability & Bayes Theorem questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Conditional Probability & Bayes Theorem

Related questions