PUNE MCA Exam 2015 Question Paper with answers for online practice

Concept:We need the conditional probability that the bag contains all white balls, given that 10 draws with replacement all yielded white.Explanation:Let event $A$: all 10 balls in the bag are white (probability = $1/2^{10}$).Let event $B$: drawing 10 balls with replacement, all are white.If $A$ is true, $P(B|A)=1$ because every draw is white.Thus $P(A \cap B) = P(A) = 1/2^{10}$.To find $P(B)$, sum over all possible numbers $k$ of white balls in the bag (0 to 10):$$P(B) = \sum_{k=0}^{10} \binom{10}{k} \left(\frac{1}{2}\right)^{10} \left(\frac{k}{10}\right)^{10} = \frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} \left(\frac{k}{10}\right)^{10}.$$Then the desired probability is$$P(A|B) = \frac{P(A)}{P(B)} = \frac{1}{\sum_{k=0}^{10} \binom{10}{k} \left(\frac{k}{10}\right)^{10}}.$$Compute the sum exactly:$\displaystyle \sum_{k=1}^{10} \binom{10}{k} k^{10} = 142{,}469{,}413{,}360$.So $\displaystyle \sum_{k=0}^{10} \binom{10}{k} \left(\frac{k}{10}\right)^{10} = \frac{142{,}469{,}413{,}360}{10^{10}} = 14.246941336$.Hence $P(A|B) = \frac{1}{14.246941336} \approx 0.070188$, which is not equal to any of the given fractions $\frac{1}{2^9}, \frac{10}{2^9}, \frac{9}{2^9}, \frac{2}{2^9}$.Answer:E. None of Above