Chapter CSB, Problem 4.13E

To determine

Find the probability $P\left(5girls,1boyor5boys,1girl\right).$

Answer to Problem 4.13E

  $0.1278$.

Explanation of Solution

Given information:

In a video store, $6$ girls and $8$ boys enter at the same time.

Number of available salespeople is $6.$

Calculation:

As there are $6$ girls and $8$ boys, then the total number of choices are $14$ out of which $6$ persons are chosen.

Calculate the total number of outcomes as,

  $\begin{array}{l}{}_{14}{C}_6=\frac{14!}{6!\left(8!\right)}\\ =3003\end{array}$

So, the total number of outcomes is $3003$.

Probability that the salesperson will help $5$ girls and $1$ boy is shown below,

Possible number of outcomes when $5$ girls are chosen from $6$ girls is ${}_6{C}_5.$

Possible number of outcomes when $1$ boy is chosen from $8$ boys is ${}_8{C}_1.$

The required probability is given by,

  $\begin{array}{l}P\left(5girls,1boy\right)=\frac{{}_6{C}_5{\cdot }_8{C}_1}{{}_{14}{C}_6}\\ =\frac{\frac{6!}{5!\left(1!\right)}\cdot \frac{8!}{1!\left(7!\right)}}{3003}\\ =\frac{6\times 8}{3003}\\ =\frac{48}{3003}\end{array}$

Probability that the salesperson will help $5$ boys and $1$ girl as shown below,

Possible number of outcomes when $1$ girl are chosen from $6$ girls is ${}_6{C}_1.$

Possible number of outcomes when $5$ boys are chosen from $8$ boys is ${}_8{C}_5.$

The required probability is given by,

  $\begin{array}{l}P\left(5boys,1girl\right)=\frac{{}_8{C}_5{\cdot }_6{C}_1}{{}_{14}{C}_6}\\ =\frac{\frac{8!}{5!\left(3!\right)}\cdot \frac{6!}{1!\left(5!\right)}}{3003}\\ =\frac{56\times 6}{3003}\\ =\frac{336}{3003}\end{array}$

The probability that the salespeople will help $5$ girls and $1$ boy or $5$ boys and $1$ girl is,

  $\begin{array}{l}P\left(5girls,1boyor5boys,1girl\right)=P\left(5girls,1boy\right)+P\left(5boys,1girl\right)\\ =\frac{48}{3003}+\frac{336}{3003}\\ =\frac{128}{1001}\\ =0.1278\end{array}$

Hence, the probability that the salespeople will help $5$ girls and $1$ boy or $5$ boys and $1$ girl is about $0.1278$.