[SOLVED] binomial distribution problem

chaosx · February 2nd 2007, 11:53 AM

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

**CaptainBlack** · February 2nd 2007, 12:02 PM

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:

$Pr(X=r)={n \choose r}p^r(1-p)^{n-r}$

so in this case:

$Pr(X=3)={7 \choose 3}\,0.25^3\,0.75^4=\frac{7!}{3!\,4!}\,0.25^3\,0.75 ^4=0.173035$

RonL

Thread: [SOLVED] binomial distribution problem

LinkBack

Thread Tools

Display

[SOLVED] binomial distribution problem

Similar Math Help Forum Discussions

[SOLVED] Binomial distribution problem

[SOLVED] Y binomial r.v, show that the distribution of n-y is binomial r.v

[SOLVED] Mean of Binomial Distribution

[SOLVED] Binomial Distribution Problem

[SOLVED] Statistics - Binomial Distribution

Search Tags