Given, from a lot of 10 items, 3 are defective
∴ Good items = 10 – 3 = 7
Let x represents the number of defective items.
Clearly values of X are 0, 1,2, 3.
P(x = 0) = P(GGGG)
= p (good items) = 7/10 × 6/9 × 5/8 × 4/7 = 1/6
P (x = 1) = P(one good and three defective)