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 draws is P, then the value of 12P is_____

Five balls are drawn  from 10 balls and replaced in first draw.

n(S) = For second  draw no of ways of selecting 7 balls =¹⁰C₇

E = For second draw 3 balls should be common to the balls of first draw.

therefore 3 must be taken from 5 balls of first draw and 4 must be taken from the other  5 balls.

n(E)=Thus for each set of 5 balls of the first  draw there are ⁵C₃ × ⁵C₄ ways of making the second  draw so that there are 3 balls common hence the probability having 3 balls in common

$\begin{array}{r}\mathrm{Probability}= P=\frac{{ }^5{\mathrm{C}}_3{\times }^5{\mathrm{C}}_4}{{ }^{10}{\mathrm{C}}_7}=\frac{5}{12}\\ \Rightarrow 12\mathrm{p}=5\end{array}$