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

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

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: