Chapter 12, Problem 23SGR

a.

To determine

Create a frequency, relative frequency table of the data.

Answer to Problem 23SGR

  $\begin{array}{ccccccccccccc}marbles& BB& BG& BR& BW& GG& GR& GW& RR& RW& WW& Total& \\ Frequency& 28& 32& 80& 24& 6& 40& 12& 45& 30& 3& 300& \\ \mathrm{Re}lativefrequency& 0.0933& 0.1067& 0.2667& 0.08& 0.02& 0.1333& 0.04& 0.15& 0.1& 0.01& 1& \end{array}$

Explanation of Solution

Given information:

Sammy has a sack of $25$ marbles, $8$ black, $10$ red, $4$ green and the rest are white. He pulls two marbles out of the bag.

Calculation:

There are $8$ black marbles so the number of ways to selecting two marbles are,

  $\begin{array}{l}C\left(8,2\right)=\frac{8!}{2!6!}\\ =28\end{array}$

Since, there are $8black$ and $4green$ marbles, so number of ways of selecting $1$ black and $1$ green is,

  $\begin{array}{l}C\left(8,1\right)C\left(4,1\right)=\frac{8!}{1!7!}\times \frac{4!}{1!3!}\\ =32\end{array}$

A frequency table and relative frequency of the data.

  $\begin{array}{ccccccccccccc}marbles& BB& BG& BR& BW& GG& GR& GW& RR& RW& WW& Total& \\ Frequency& 28& 32& 80& 24& 6& 40& 12& 45& 30& 3& 300& \\ \mathrm{Re}lativefrequency& 0.0933& 0.1067& 0.2667& 0.08& 0.02& 0.1333& 0.04& 0.15& 0.1& 0.01& 1& \end{array}$

b.

To determine

Find most likely occurred outcome.

Answer to Problem 23SGR

Red, Red.

Explanation of Solution

Given information:

Sammy has a sack of $25$ marbles, $8$ black, $10$ red, $4$ green and the rest are white. He pulls two marbles out of the bag.

Calculation:

  $8$ Black, $10$ red, $4$ green and the rest are white. The most likely occur marble in first draw is red and in second draw it is also red.

Hence, most likely occurred outcome is Red, Red.

c.

To determine

Find $P\left(blackandgreen\right)$ .

Answer to Problem 23SGR

  $10.7\%$

Explanation of Solution

Given information:

Sammy has a sack of $25$ marbles, $8$ black, $10$ red, $4$ green and the rest are white. He pulls two marbles out of the bag.

Calculation:

  $\begin{array}{l}P\left(blackandgreen\right)=\frac{32}{300}\\ =0.1067\\ or\\ 10.7\%\end{array}$

Hence, $P\left(blackandgreen\right)$ is $10.7\%$ .