A factory produces bulbs. The probability that one bulb is defective is 1/50, and they are packed in boxes of 10. From a single box, find the probability that
i. none of the bulbs is defective
ii. exactly two bulbs are defective
iii. more than 8 bulbs work properly.
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) = nCrprqn-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 .