Suppose that for a very large shipment of integrated-circuit chips, the probability of failure for an one chip is 0.20. Assuming that the assumptions underlying the binomial distributions are met, find the probability that at most 4 chips fail in a random sample of 17

Answer:the probability that at most 4 chips fail in a random sample of 17 is 0.76

Step-by-step explanation:

The formula for binomial distribution is expressed as

P (x = r) = nCr * q^ (n - r) * p^r

Where

x represents the outcome,

n represents the number of samples.

p represents the probability that an outcome will happen.

q represents the probability that an outcome will not happen

From the information given,

p = 0.2

q = 1 - 0.2 = 0.8

n = 17

We want to find P (x lesser than or equal to 4)

P (x lesser than or equal to 4) = P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4)

P (x = 0) = 17C0 * 0.8^ (17 - 0) * 0.2^0

P (x = 0) = 1 * 0. 023 * 1 = 0.023

P (x = 1) = 17C1 * 0.8^ (17 - 1) * 0.2^1

P (x = 1) = 17*0.028 * 0.2 = 0.0952

P (x = 2) = 17C2 * 0.8^ (17 - 2) * 0.2^2

P (x = 2) = 136 * 0.035 * 0.04 = 0.1904

P (x = 3) = 17C3 * 0.8^ (17 - 3) * 0.2^3

P (x = 3) = 680 * 0.044 * 0.008 = 0.24

P (x = 4) = 17C4 * 0.8^ (17 - 4) * 0.2^4

P (x = 4) = 2380 * 0.055 * 0.0016 = 0.21

P (x lesser than or equal to 4) = 0.023 + 0.0952 + 0.1904 + 0.24 + 0.21 = 0.76

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