Chapter 13.4, Problem 23E

To determine

To find the probability of selecting three different colored crayons.

Answer to Problem 23E

Independent, $\frac{35}{1024}$

Explanation of Solution

Given:

A box contains $5$ red, $4$ black and $7$ bluecrayons.

Calculation:

The events are independent because outcome of one eventdoesn’t affect the outcome of another event.

Let $A$ be the event of selecting a red crayon.

Let $B$ be the event of selecting a black crayon.

Let $C$ be the event of selecting a blue crayon.

Total sample space $=5+4+7=16$

Therefore probability of event $A$ is $P(A)=\frac{5}{16}$

And probability of event $A_2$ is $P(B)=\frac{4}{16}$

And probability of event $C$ is $P(C)=\frac{7}{16}$

Hence,

  $\begin{array}{l}P(A\text{ and B and}\text{C})=P(A)\times P(B)\times P(C)\\ P(A\text{ and }B\text{ and C})=\frac{5}{16}\times \frac{4}{16}\times \frac{7}{16}=\frac{35}{1024}\end{array}$