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janeou17xn
10.12.2019 •
Mathematics
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
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Ответ:
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
Ответ:
D.
Step-by-step explanation:
AAS triangle congruence theorem. These triangle have the same side, angle and other angle