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 

The no. of ways of drawing 7 balls $={}^{10}\mathrm{C}_7$

For each set of 7 balls of the second draw, 3 must be common to the set of 5 balls of the draw, 

i.e., 2 other balls can be drawn in $={}^3\mathrm{C}_2$ ways thus, for each set of 7 balls of the second  draw, 

there are ${}^7\mathrm{C}_3\times {}^3\mathrm{C}_2$ ways of making the first draw so that there are 3 balls common. Hence, the
probability of having three balls in common $\frac{{}^7\mathrm{C}_3\times {}^3\mathrm{C}_2}{{}^{10}\mathrm{C}_7}=\frac{5}{12}.$