A bag contains $10$ different balls. Five balls are drawn simultaneously and then replaced and then seven balls are drawn. The probability that exactly three balls are common to the two drawn is $p$ then the value of $10p$ is ______.
Step-by-Step Solution
Key Concept: The probability requires counting favorable outcomes where exactly 3 of the 7 balls drawn in the second draw are from the original 5 balls drawn first, using the hypergeometric distribution principle: P = C(5,3)·C(5,2)/C(10,7).
The second draw of 7 balls has $\binom{10}{7}$ possible outcomes. For exactly 3 balls to be common with the first draw of 5 balls, we need 3 from the original 5 and 2 from the remaining 5. The number of ways is $\binom{5}{3} \cdot \binom{5}{2}$. Thus the probability is $\frac{\binom{5}{3} \cdot \binom{5}{2}}{\binom{10}{7}} = \frac{10 \cdot 10}{120} = \frac{5}{12}$.
Correct Answer: 4.17