This is a binomial probability case.
P(X=x) ="\\binom {n} {k} p^n q^{n-x}"
n=100,p=0.4"\\implies" q=0.6
i) at least 44 defectives
P(X"\\geq" 44) ="\\displaystyle \\sum_{x=44}^{100} \\binom {100}{x} 0.4^x 0.6^{100-x}"
=0.2365
ii) exactly 44
P(x=44) ="\\binom {100} {44} *0.4^{44}*0.6^{56}"
=0.0576
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