Question

The eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg in the lot of 10 eggs.

Answer 1

Let X denote the number of defective eggs in the 10 eggs drawn. Since the drawing is done with replacement, the trials are Bernoulli trials.

Probability of success = `1/10`

∴ `"p" = 1/10, "q" = 1 - "p" = 1 - 1/10 = 9/10`

n = 10.

X ∼ B(n, p)

`X ∼ B (10, 1/10)`

P(X = x) = `""^"n""C"_x"P"^x"q"^("n" - x)`

P(X = x) = `""^"10""C"_x(1/10)^x (9/10)^(10 - x)`

Here, X ≥ 1.

P(X ≥ 1)

= 1 − P(X < 1)

= 1 − P(X = 0)

= `1 - {""^"10""C"_0(1/10)^0 (9/10)^(10 - 0)}`

= `1 - 1 xx 1 xx (9/10)^10`

= `1 - (9/10)^10`