A bag contains 4 balls. Let E be an event in which two balls are drawn at random and found to be white , then
Match List-I with List-II
| List-I | List-II |
| (A) Probability that all balls are white | (I) \( \displaystyle\frac{1}{6} \) |
| (B) Probability of drawing 2 white ball at random given that there are 3 white ball | (II) \( \displaystyle\frac{3}{5} \) |
| (C) P(E) | (III) \( \displaystyle\frac{1}{2} \) |
| (D) Probability of drawing 2 white ball at random given that there is only 2 white ball | (IV) \( \displaystyle\frac{5}{9} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
Since two white balls were drawn, the bag has 2, 3 or 4 white balls. As is usual in this problem, these three cases \( E_2,E_3,E_4 \) are taken as equally likely, each with probability \( \frac{1}{3} \). Two balls can be drawn in \( {}^{4}C_{2}=6 \) ways.
\( \displaystyle P(E\mid E_2)=\frac{1}{6} \)
\( \displaystyle P(E\mid E_3)=\frac{{}^{3}C_{2}}{6}=\frac{1}{2} \)
\( \displaystyle P(E\mid E_4)=1 \)
By the theorem of total probability:
\( \displaystyle P(E)=\frac{1}{3}\left(\frac{1}{6}+\frac{1}{2}+1\right) \)
\( \displaystyle =\frac{1}{3}\cdot\frac{5}{3}=\frac{5}{9} \)
(A) Probability that all balls are white
This matches (II) 3/5. By Bayes’ theorem:
\( \displaystyle P(E_4\mid E)=\frac{\frac{1}{3}\times 1}{\frac{5}{9}}=\frac{3}{5} \)
(B) Probability of drawing 2 white ball at random given that there are 3 white ball
This matches (III) 1/2. This is \( P(E\mid E_3)=\frac{1}{2} \).
(C) P(E)
This matches (IV) 5/9, the total probability \( P(E)=\frac{5}{9} \).
(D) Probability of drawing 2 white ball at random given that there is only 2 white ball
This matches (I) 1/6. This is \( P(E\mid E_2)=\frac{1}{6} \).
So the matching is (A) - (II), (B) - (III), (C) - (IV), (D) - (I).