# Urgent help for statistics please!

• Mar 7th 2009, 01:07 PM
Kalyn
30% of chicken is contaminated, if a person buys 12, what's the possibility that 6 of them are contaminated? thanks
• Mar 7th 2009, 01:16 PM
running-gag
Hi

Let X be the random variable which counts the number of chickens contaminated out of 12.

X follows a binomial distribution (with parameters n=12 and p=0.30) because it is the number of "yes" in a sequence of 12 independent yes/no experiments (contaminated / not contaminated).

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

Therefore $P(X=6) = \binom{12}{6} 0.30^6\: (1-0.30)^{12-6}$

$P(X=6) = \binom{12}{6} 0.30^6 \: 0.70^6$
• Mar 7th 2009, 01:27 PM
Kalyn
I don't get this
How do you solve (12 over 6)? how do you put that in the equation?

Quote:

Originally Posted by running-gag
Hi

Let X be the random variable which counts the number of chickens contaminated out of 12.

X follows a binomial distribution (with parameters n=12 and p=0.30) because it is the number of "yes" in a sequence of 12 independent yes/no experiments (contaminated / not contaminated).

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

Therefore $P(X=6) = \binom{12}{6} 0.30^6\: (1-0.30)^{12-6}$

$P(X=6) = \binom{12}{6} 0.30^6 \: 0.70^6$

• Mar 7th 2009, 01:31 PM
running-gag
Quote:

Originally Posted by Kalyn
How do you solve (12 over 6)? how do you put that in the equation?

$\binom{12}{6}$ is the binomial coefficient

$\binom{12}{6} = \frac{12!}{6! \:(12-6)!}$