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.

1
Expert's answer
2022-03-08T03:12:06-0500

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}


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