Question

A bag contains four balls. Two balls are drawn randomly and found them to be white. The probability that all the balls in the bag are white is

• A. \(\frac{1}{2}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{2}{3}\)

Answer 1

The Correct Option is B

To find the probability that all 4 balls in a bag are white given that two balls drawn randomly are white, we proceed with:

1. Defining the Random Variable:
Let $X$ be the number of white balls in the bag. Since the bag contains 4 balls, $X$ can take values 0, 1, 2, 3, or 4. 
Assuming each value is equally likely, so the probability is:
$ P(X=k) = \frac{1}{5} $
for $k = 0, 1, 2, 3, 4$.

2. Defining the Event:
Let $E$ be the event that two balls drawn randomly from the bag are both white. 
We need to find $P(X=4 | E)$, the probability that all 4 balls are white given that the two drawn balls are white.

3. Applying Bayes' Theorem:
By Bayes' Theorem, we have:

$ P(X=4 | E) = \frac{P(E | X=4) P(X=4)}{P(E)} $

4. Calculating $P(X=4)$:
Since each value of $X$ is equally likely:

$ P(X=4) = \frac{1}{5} $

5. Calculating $P(E | X=4)$:
If $X=4$, all 4 balls in the bag are white. Therefore, if two balls are drawn, both must be white, so:

$ P(E | X=4) = 1 $

6. Calculating $P(E)$:
The total probability of drawing two white balls is given by summing over all possible values of $X$:

$ P(E) = \sum_{k=0}^4 P(E | X=k) P(X=k) $
The probability of drawing two white balls given $X=k$ is the number of ways to choose 2 white balls from $k$ white balls divided by the number of ways to choose 2 balls from 4 balls:

$ P(E | X=k) = \frac{\binom{k}{2}}{\binom{4}{2}} $
Since $\binom{k}{2} = 0$ for $k=0, 1$ (as we cannot choose 2 white balls if there are fewer than 2), the sum simplifies to:

$ P(E) = \sum_{k=2}^4 \frac{\binom{k}{2}}{\binom{4}{2}} P(X=k) $
Substituting $P(X=k) = \frac{1}{5}$ and $\binom{4}{2} = 6$, we get:

$ P(E) = \frac{1}{6} \cdot \frac{1}{5} \left[ \binom{2}{2} + \binom{3}{2} + \binom{4}{2} \right] $
Now compute the binomial coefficients:

$ \binom{2}{2} = 1, \quad \binom{3}{2} = 3, \quad \binom{4}{2} = 6 $
So:

$ P(E) = \frac{1}{6} \cdot \frac{1}{5} \cdot [1 + 3 + 6] = \frac{1}{6} \cdot \frac{1}{5} \cdot 10 = \frac{10}{30} = \frac{1}{3} $

7. Computing the Final Probability:
Now substitute into Bayes' Theorem:

$ P(X=4 | E) = \frac{P(E | X=4) P(X=4)}{P(E)} = \frac{1 \cdot \frac{1}{5}}{\frac{1}{3}} = \frac{1}{5} \cdot 3 = \frac{3}{5} $

Final Answer:
The probability that all 4 balls are white given that two drawn balls are white is $ \frac{3}{5} $.