Mathematics - Probability Question with Solution | TestHub

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

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 ${}^3{C}_2$ ways thus, for each set of 7 balls of the second

draw, there are ${}^7{C}_3\times {}^3{C}_2$ ways of making the first draw so that there are 3 balls common.

Hence, the probability of having three balls in common $\frac{{}^7{C}_3\times {}^3{C}_2}{{}^{10}{C}_7}=\frac{5}{12}$.