A factory produce

Given that, the probability that one bulb is defective is 1/50.

The bulbs are packed in boxes of 10.

Let p be the probability of bulb being defective.

Then,

Let q be the probability of the bulb not being defective.

Also, we know that

p + q = 1

⇒ q = 1 – p

Let X be a random variable representing a number of defective bulbs out of n bulbs.

Then, the probability of r bulbs to be defective out of n bulbs 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 equation. We get

…(A)

(i). We need to find the probability that none of the bulbs is defective.

The probability is given by,

Probability = P (X = 0)

Put r = 0 in equation (A),

∴, the probability that none of the bulbs is defective is (49/50)¹⁰.

(ii). We need to find the probability that exactly two bulbs are defective.

The probability is given by,

Probability = P (X = 2)

Put r = 2 in equation (A),

∴, the probability that none of the bulbs is defective is .

(iii). We need to find the probability of more than 8 bulbs working properly.

This can also be interpreted as, the probability that at most 2 bulbs are defective.

This can be represented as,

Probability = P (X ≤ 2)

Or

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

Put r = 0, 1, 2 to find P (X = 0), P (X = 1) and P (X = 2) and then, substitute in the above equation.

∴, the probability of more than 8 bulbs working properly is .