Question #6397

Records in a beanie babies factory show that for every 100 newly produced toys, 20 are defective. If the quality control takes a sample of 10 newly manufactured stuffed toys, what is the chance that she will find 5 or more defective?

Expert's answer

The probability of producing defective toy is

p = 20/100 = 1/5 = 0.2.

Let's use the Bernoulli's formula:

P(n,k) = C(n,k)*p^k*(1-p)^(n-k),

where n=10, k=5, p=0.2.

We have

P(10,5) = 10!/(5!*(10-5)!) *0.2^5*(1-0.2)^(10-5) = 10!/(5!)² *(0.2*0.8)^5 = 10!/(5!)² *0.16^5 = 252*0.16^5 = 0.0264241152.

p = 20/100 = 1/5 = 0.2.

Let's use the Bernoulli's formula:

P(n,k) = C(n,k)*p^k*(1-p)^(n-k),

where n=10, k=5, p=0.2.

We have

P(10,5) = 10!/(5!*(10-5)!) *0.2^5*(1-0.2)^(10-5) = 10!/(5!)² *(0.2*0.8)^5 = 10!/(5!)² *0.16^5 = 252*0.16^5 = 0.0264241152.

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