Chapter CSB, Problem 4.15E

To determine

Find the probability $P\left(atleast4boys\right).$

Answer to Problem 4.15E

  $\mathrm{0.47086.}$

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 $3003$.

The probability that salespeople will first help at least $4$ boys is as follows,

That is, the probability that the salespeople will help $4$ boys and $2$ girls as,

Possible number of outcomes of choosing $2$ girls out of $6$ girls is ${}_6{C}_2.$

Possible number of outcomes of choosing $4$ boys out of $8$ boys is ${}_8{C}_4.$

The required probability is given by,

  $\begin{array}{l}P\left(2girls,4boys\right)=\frac{{}_6{C}_2{\cdot }_8{C}_4}{{}_{14}{C}_6}\\ =\frac{\frac{6!}{2!\left(4!\right)}\cdot \frac{8!}{4!\left(4!\right)}}{3003}\\ =\frac{1050}{3003}\end{array}$

The probability that salespeople will first help $5$ boys is as follows,

That is, the probability that the salespeople will help $5$ boys and $1$ girl as,

Possible number of outcomes of choosing $1$ girl out of $6$ girls is ${}_6{C}_1.$

Possible number of outcomes of choosing $5$ boys out of $8$ boys is ${}_8{C}_5.$

The required probability is given by,

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

The probability that salespeople will first help $6$ boys is as follows,

Possible number of outcomes of choosing $6$ boys out of $8$ boys is ${}_8{C}_6.$

The required probability is given by,

  $\begin{array}{l}P\left(6boys\right)=\frac{{}_8{C}_6}{{}_{14}{C}_6}\\ =\frac{\frac{8!}{6!\left(2!\right)}}{3003}\\ =\frac{28}{3003}\end{array}$

Now, the probability that the salespeople will help at least $4$ boys is,

  $\begin{array}{l}P\left(atleast4boys\right)=P\left(4boys\right)+P\left(5boys\right)+P\left(6boys\right)\\ =\frac{1050}{3003}+\frac{336}{3003}+\frac{28}{3003}\\ =\frac{202}{429}\\ =0.47086\end{array}$

Hence, the probability $P\left(atleast4boys\right)is\mathrm{0.47086.}$