This is a binomial distribution case "P(X=x) ={n\\choose x} p^x q^{n-x}"
The probability of obtaining defective chips is p=0.03, which implies that q=1-p=1-0.03=0.97
n=20
Probability that at least 2 chips are defective
"P(x\\geq 2)"
"P(X \\geq 2)=\\sum_{x=2}^{20} 0.03^x 0.97^{20-x}"
=0.12
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