Chapter 3.3, Problem 24E

a

To determine

Find the probability that the person wears 2 blue socks.

Answer to Problem 24E

The probability that the person wears 2 blue socks is 0.0909.

Explanation of Solution

There are 4 blue, 5 gray, and 3 black socks.

The total number of socks is 12, and there are 4 blue socks.

The probability of selecting a blue sock from the 12 socks is $\frac{4}{12}$. The probability of again selecting a blue sock from the remaining 11 socks is $\frac{3}{11}$.

Thus, the probability of wearing two blue socks is as follows:

$\begin{array}{c}P\left(\text{wearing two blue socks}\right)=\left[\begin{array}{l}P\left(\text{wearing first blue sock}\right)\times \\ P\left(\text{wearing second blue sock|first one is blue}\right)\end{array}\right]\\ =\frac{4}{12}\times \frac{3}{11}\\ =0.0909\end{array}$

The probability that theperson wears 2 blue socks is 0.0909.

b.

To determine

Find the probability that the person wears no gray socks.

Answer to Problem 24E

The probability that theperson wears no gray socks is 0.3182.

Explanation of Solution

There are 4 blue, 5 gray, and 3 black socks.

The total number of socks is 12, and there are 5 gray socks. The remaining 7 socks are not gray.

The probability of not selecting a gray sock from the 7 socks is $\frac{7}{12}$. The probability of again not selecting a gray sock from the remaining 11 socks is $\frac{6}{11}$.

Thus, the probability of not wearing gray socks is as follows:

$\begin{array}{c}P\left(\text{not wearing gray socks}\right)=\left[\begin{array}{l}P\left(\text{not wearing first gray sock}\right)\times \\ P\left(\text{not wearing second gray sock|first one is not gray}\right)\end{array}\right]\\ =\frac{7}{12}\times \frac{6}{11}\\ =0.3182\end{array}$

The probability that theperson wears no gray socks is 0.3182.

c.

To determine

Find the probability that the person wears at least one black sock.

Answer to Problem 24E

The probability that theperson wears at least one black sockis 0.4545.

Explanation of Solution

There are 4 blue, 5 gray, and 3 black socks.

The total number of socks is 12, and there are 3 black socks. The remaining 9 socks are not black.

The probability of not selecting a black sock from the 9 socks is $\frac{9}{12}$. The probability of again not selecting a black sock from the remaining 11 socks is $\frac{8}{11}$.

Thus, the probability of wearing at least one black sock is as follows:

$\begin{array}{c}P\left(\text{wearing at least one black sock}\right)=1-\left[\begin{array}{l}P\left(\text{not wearing first black sock}\right)\times \\ P\left(\begin{array}{l}\text{not wearing second black }\\ \text{sock|first socks is not black}\end{array}\right)\end{array}\right]\\ =1-\left[\frac{9}{12}\times \frac{8}{11}\right]\\ =1-0.5455\\ =0.4545\end{array}$

The probability that theperson wears at least one black sock is 0.4545.

d.

To determine

Find the probability that the person wears a green sock.

Answer to Problem 24E

The probability that theperson wears a green sock is 0.

Explanation of Solution

There are 4 blue, 5 gray, and 3 black socks.

Here, there are no green socks.

Thus, the probability that theperson wears a green sock is 0.

e.

To determine

Find the probability that the person wears matching socks.

Answer to Problem 24E

The probability that theperson wears matching socks is 0.2879.

Explanation of Solution

Matching blue socks:

There are 4 blue, 5 gray, and 3 black socks.

From Part (a), the probability of wearing matching blue socks is 0.0909.

Matching gray socks:

The total number of socks is 12, and there are 5 gray socks.

The probability of selecting a gray sock from the 12 socks is $\frac{5}{12}$. The probability of again selecting a blue sock from the remaining 11 socks is $\frac{4}{11}$.

Thus, the probability of wearing two gray socks is as follows:

$\begin{array}{c}P\left(\text{wearing two gray socks}\right)=\left[\begin{array}{l}P\left(\text{wearing first gray sock}\right)\times \\ P\left(\text{wearing second gray sock|first one is gray}\right)\end{array}\right]\\ =\frac{5}{12}\times \frac{4}{11}\\ =0.1515\end{array}$

The probability of wearing matching gray socks is 0.1515.

Matching black socks:

The total number of socks is 12, and there are 3 black socks.

The probability of selecting a black sock from the 12 socks is $\frac{3}{12}$. The probability of selecting again a black sock from the remaining 11 socks is $\frac{2}{11}$.

Thus, the probability of wearing two black socks is as follows:

$\begin{array}{c}P\left(\text{wearing two black socks}\right)=\left[\begin{array}{l}P\left(\text{wearing first black sock}\right)\times \\ P\left(\text{wearing second black sock|first one is black}\right)\end{array}\right]\\ =\frac{3}{12}\times \frac{2}{11}\\ =0.0455\end{array}$

The probability of wearing matching black socks is 0.0455.

The probability of obtaining matching socks is calculated using the addition rule of disjoint events as follows:

$\begin{array}{c}P\left(\text{matching socks}\right)=P\left(\text{wearing two blue sock or two gray socks or two black socks}\right)\\ =\left[\begin{array}{l}P\left(\text{wearing two blue sock}\right)+\\ P\left(\text{two gray socks}\right)+\\ P\left(\text{two black socks}\right)\end{array}\right]\\ =0.0909+0.1515+0.0455\\ =0.2879\end{array}$

The probability that theperson wears matching socks is 0.2879.