Given that, 10 eggs are drawn successively.

Defective eggs in the lot = 10%

Let p be the probability that the eggs drawn from the lot are defective.

⇒ p = 10%

Then, q is the probability that the eggs drawn from the lot is not defective.

And, p + q = 1

⇒ q = 1 – p

Let X be a random variable that represents defective eggs picked out of n eggs from the lot.

Then, the probability of taking r defective eggs out of n eggs from the lot is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 10

Putting the values of n, p, and q in the above formula, we get

…(i)

We need to find the probability of getting at least one defective egg from the lot.

This is represented as,

Probability = P (X ≥ 1)

This is also written as,

P (X ≥ 1) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)

Or

P (X ≥ 1) = 1 – P (X < 1)

⇒ P (X ≥ 1) = 1 – P (X = 0)

Putting r = 0 in P (X = r) formula in (i), we get

∴, the probability of getting at least one defective egg from the lot is .