Question

Hello I seem to be having trouble on this problem it is requiring a decimal place answer please help

Answer 1

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:

`nC_r=(n!)/(r!(n-r)!)`

Step-by-step explanation:

We can use Combination for solving the problem.

`\begin{gathered} Total\text{ number of members to be choosen from for gameshow=8+5} \\ =13 \end{gathered}`
`\begin{gathered} Total\text{ number of ways that we made without any restriction =13C}_5 \\ =(13!)/(5!(13-5)!) \\ =(13!)/(5!(8!)) \\ =(13*12*11*10*9*8!)/(5!(8!)) \\ =(13*12*11*10*9)/(1*2*3*4*5) \\ =13*11*9 \\ =1287 \end{gathered}`
`\begin{gathered} No\text{ of ways to select 3 women and 2 men=8C}_3*5C_2 \\ =(8*7*6)/(1*2*3)*(5*4)/(1*2) \\ =560 \end{gathered}`

We can take the total number of ways as sample space(S).

`n(S)=1287`

Let A be the even that no of ways to select 3 women and 2 men.

`n(A)=560`
`\begin{gathered} P(A)=(n(A))/(n(S)) \\ =(560)/(1287) \\ =0.435 \end{gathered}`

Final Answer:

Probability that the producer choosing 3 women and 2 men is 0.435