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Natalka [10]
4 years ago
9

A factory tests 100 light bulbs for defects. The probability that a bulb is defective is 0.02. The occurrences of defects among

the light bulbs are mutually independent events. Calculate the probability that exactly two are defective given that the number of defective bulbs is two or fewer.
Mathematics
1 answer:
Kobotan [32]4 years ago
6 0

Answer:

The probability that exactly two are defective given that the number of defective bulbs is two or fewer is 0.4040.

Step-by-step explanation:

Let X be the number of defective bulbs.

X\sim B(n,p)

Where n is sample size and p is probability of success.

According to the given information,

n=100

p=0.02

q=1-p=1-0.02=0.98

According to binomial distribution the probability of exactly r success from n is

P(X=r)=^nC_rp^rq^{n-r}

The probability that exactly two are defective is

P(X=2)=^{100}C_2(0.02)^2(0.98)^{98}\approx 0.2724

The probability that the number of defective bulbs is two or fewer is

P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)

P(X\leq 2)=^{100}C_0(0.02)^0(0.98)^{100}+^{100}C_1(0.02)^1(0.98)^{99}+^{100}C_2(0.02)^2(0.98)^{98}

P(X\leq 2)=0.1326+0.2707+0.2734

P(X\leq 2)=0.6767

We have to find the probability that exactly two are defective given that the number of defective bulbs is two or fewer.

P(x=2|x\leq 2)

According to the conditional probability

P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}

P(x=2|x\leq 2)=\frac{P(x=2\cap x\leq 2)}{P(x\leq 2)}

P(x=2|x\leq 2)=\frac{P(x=2)}{P(x\leq 2)}

P(x=2|x\leq 2)=\frac{0.2734}{0.6767}=0.4040

Therefore the probability that exactly two are defective given that the number of defective bulbs is two or fewer is 0.4040.

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