Question

Sol. [Key Concept: Hypergeometric overlap: $P(exactly\ 3\ common)$ between the two draws of the 10-ball bag.]

There are 10 distinct balls.

First drawing: 7 balls are selected and replaced.

Second drawing: 5 balls are selected.

We need exactly 3 balls common to both drawings.

After the first drawing, 7 balls are in the first set and 3 balls are outside it.

For the second drawing to have exactly 3 common balls:

Choose 3 from the 7 common balls: C(7, 3)

Choose remaining 2 from the 3 non-common balls: C(3, 2)

Total ways to choose 5 balls from 10: C(10, 5)

Therefore,

P = [C(7,3) × C(3,2)] / C(10,5)

= (35 × 3)/252

= 105/252

= 5/12