(a) Suppose you have two quarters and a dime in your left pocket and two dimes and three quarters in your right pocket. You select a pocket at random and from it a coin at random. What is the probability that it is a dime? (b) Let x be the amount of money you select. Find E(x).

(c) Suppose you selected a dime in (a). What is the probability that it came from your right pocket?

(d) Suppose you do not replace the dime, but select another coin which is also a dime. What is the probability that this second coin came from your right pocket?

It has been given that there aretwo quarters and a dime in your left pocket and two dimes and three quarters in your right pocket.

Probability means the chances of any event to occur is called it probability.

a) The situation is shown below so the probability of getting dime " d " is given by the expression mentioned below.

$\begin{array}{c}p\left(d\right)=R\times d+L\times d\\ =\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{5}\right)\\ =\left(\frac{11}{30}\right)\end{array}$

The situation is shown below.

The Probability is shown below.

b) Let the value of coin to be x where x is given as mentioned below.

$x=\{0.1,0.25\}$

The probability of each value.

$\begin{array}{c}P(x=0.1)=\left(\frac{11}{30}\right)\\ P(x=0.25)=\left(\frac{19}{30}\right)\end{array}$

The average$\mu$ is given as shown below.

$\begin{array}{c}\mu =\sum xp\left(x\right)\\ =\left(0.1\right)\left(\frac{11}{30}\right)+\left(0.25\right)\left(\frac{19}{30}\right)\\ \text{=\$0.195}\\ =19.5\text{cent}\end{array}$

c) The probability of getting a dime from Right pocket is given as mentioned below.

$\begin{array}{c}p(R/d)=\frac{p\left(Rd\right)}{P\left(d\right)}\\ =\frac{0.5\times \frac{2}{5}}{11/30}\\ =\frac{6}{11}\end{array}$

d) The probability of getting a dime is given as mentioned below.

$\begin{array}{c}p\left(d\right)=0.5\times \frac{1}{3}+0.5\times \frac{1}{4}\\ =\frac{7}{24}\end{array}$

Choose a dime the probability of getting a dime from a right.

$\begin{array}{c}p(R/d)=\frac{p\left(Rd\right)}{p\left(d\right)}\\ =\frac{0.5\times 0.25}{7/24}\\ =\frac{3}{7}\end{array}$

The situation is shown below.

The Probability is shown below.

Hence, the solutions are as mentioned below.

(a)$p\left(d\right)=\frac{11}{30}$

(b)$E\left(x\right)=19.5\text{cents}$

(c)$p(R/d)=\frac{6}{11}$

(d)$p(R/d)=\frac{3}{7}$