Given:
There are 8 women and 5 men in a game show.
Producer of the show going to choose 5 people.
Required:
Find the probability that the producer chooses 3 women and 2 men.
Formula:
[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]
Explanation:
We can use Combination for solving the problem.
[tex]\begin{gathered} Total\text{ number of members to be choosen from for gameshow=8+5} \\ =13 \end{gathered}[/tex][tex]\begin{gathered} Total\text{ number of ways that we made without any restriction =13C}_5 \\ =\frac{13!}{5!(13-5)!} \\ =\frac{13!}{5!(8!)} \\ =\frac{13\times12\times11\times10\times9\times8!}{5!(8!)} \\ =\frac{13\times12\times11\times10\times9}{1\times2\times3\times4\times5} \\ =13\times11\times9 \\ =1287 \end{gathered}[/tex][tex]\begin{gathered} No\text{ of ways to select 3 women and 2 men=8C}_3\times5C_2 \\ =\frac{8\times7\times6}{1\times2\times3}\times\frac{5\times4}{1\times2} \\ =560 \end{gathered}[/tex]
We can take the total number of ways as sample space(S).
[tex]n(S)=1287[/tex]
Let A be the even that no of ways to select 3 women and 2 men.
[tex]n(A)=560[/tex][tex]\begin{gathered} P(A)=\frac{n(A)}{n(S)} \\ =\frac{560}{1287} \\ =0.435 \end{gathered}[/tex]
Final Answer:
Probability that the producer choosing 3 women and 2 men is 0.435
