Question #305868

In math club there are 7 girls and 5 boys. A team of 6 students must be formed to coMpete In math contest.let x be the number of boys in the team.

1. What are the posible value of x?

2. Construct a probability distribution for the random variable.

Expert's answer

Let's denote B - boys, G - girls.

Sample space S is all possible outcomes.

Let XX be the random variables representing the number of number of boys in the team.

1. The possible values of xx are 0,1,2,3,4,5.0,1,2,3,4,5.


2. There are (7+56)=924\dbinom{7+5}{6}=924 possible outcomes.


P(X=0)=(50)(76)(126)=1(7)924=1132P(X=0)=\dfrac{\dbinom{5}{0}\dbinom{7}{6}}{\dbinom{12}{6}}=\dfrac{1(7)}{924}=\dfrac{1}{132}

P(X=1)=(51)(75)(126)=5(21)924=544P(X=1)=\dfrac{\dbinom{5}{1}\dbinom{7}{5}}{\dbinom{12}{6}}=\dfrac{5(21)}{924}=\dfrac{5}{44}

P(X=2)=(52)(74)(126)=10(35)924=2566P(X=2)=\dfrac{\dbinom{5}{2}\dbinom{7}{4}}{\dbinom{12}{6}}=\dfrac{10(35)}{924}=\dfrac{25}{66}

P(X=3)=(53)(73)(126)=10(35)924=2566P(X=3)=\dfrac{\dbinom{5}{3}\dbinom{7}{3}}{\dbinom{12}{6}}=\dfrac{10(35)}{924}=\dfrac{25}{66}

P(X=4)=(54)(72)(126)=5(21)924=544P(X=4)=\dfrac{\dbinom{5}{4}\dbinom{7}{2}}{\dbinom{12}{6}}=\dfrac{5(21)}{924}=\dfrac{5}{44}

P(X=5)=(55)(71)(126)=1(7)924=1132P(X=5)=\dfrac{\dbinom{5}{5}\dbinom{7}{1}}{\dbinom{12}{6}}=\dfrac{1(7)}{924}=\dfrac{1}{132}

Construct the probability distribution of the random variable

x012345p(x)1/1325/4425/6625/665/441/132\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline p(x) & 1/132 & 5/44 & 25/66 & 25/66 & 5/44 & 1/132 \end{array}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS