n=7,r=3,p=0.25 Pr= 7!/3!(7-3)!=1(0.25)^3(1-0.25)^4 =0.004943848. Why is this answer wrong ? The correct answer is 0.173035
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Originally Posted by chaosx n=7,r=3,p=0.25 Pr= 7!/3!(7-3)!=1(0.25)^3(1-0.25)^4 =0.004943848. Why is this answer wrong ? The correct answer is 0.173035 For the Binomial distribution: $\displaystyle Pr(X=r)={n \choose r}p^r(1-p)^{n-r}$ so in this case: $\displaystyle Pr(X=3)={7 \choose 3}\,0.25^3\,0.75^4=\frac{7!}{3!\,4!}\,0.25^3\,0.75 ^4=0.173035$ RonL
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