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A sampling plan states that if 20 incoming transistors are checked and 2 or less defects are found, the lot is accepted. If an incoming lot is 10 percent defective, what is the probability of accepting the lot of transistors

Sagot :

A sampling plan states that if 20 incoming transistors are checked and 2 or less defects are found, the lot is accepted. The probability of accepting the lot of transistors is 0.68

What is the probability?

Probability refers to a possibility that deals with the occurrence of random events. The probability of all the events occurring need to be 1.

Assuming a binomial probability distribution, the formula for binomial distribution is expressed as

[tex]P(x = r) = nC_r \times q^{n - r} p^r[/tex]

Where

n = number of samples

p = probability that an event will happen.

q = probability that an event will not happen.

From the information given,

n = 20

If an incoming lot is 10 percent defective, it means that

p = 10/100 = 0.1

q = 1 - p = 1 - 0.1 = 0.9

The probability of rejecting the lot would be

P(x lesser than or equal to 2)

P(x = 0) + P(x = 1) + P(x =2)

[tex]P(x = 0) = 20C_0 \times 0.9^(20 - 0) \times 0.1^0 = 0.12\\(x = 1) = 20C_1 \times 0.9^(20 - 1) \times 0.1^1 \\= 0.12 = 0.27\\(x = 2) = 20C_2 \times 0.9^(20 - 2) \times 0.1^2 \\= 0.12 = 0.29[/tex]

P(x lesser than or equal to 2) = 0.12 + 0.27 + 0.29 = 0.68

Learn more about probability here;

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