Question

uestion 10 (1 point)
There are 12 brown socks and 6 blue socks in a drawer. In the dark, Jason pulls out a sock and puts it on his right foot. Then he pulls out another sock and puts it on his left foot.

What is the probability that Jason has two blue socks on his feet?

Answer 1

To find the probability that Jason has two blue socks on his feet, we start by calculating the total number of socks in the drawer.

1. **Total socks** = 12 brown socks + 6 blue socks = 18 socks.

Next, we determine the number of ways to pick two blue socks out of the total socks.

2. The number of ways to choose the first blue sock is 6 (since there are 6 blue socks).
3. After picking the first blue sock, there are now 5 blue socks left, making the number of ways to choose the second blue sock 5.

The total combinations for picking two socks can be calculated as follows:

4. When Jason picks the first sock from 18, there's no restriction. The number of total socks is still 18.
5. When he picks the second sock, there are now 17 socks remaining.

Next, we compute the probabilities:

- The probability of Jason picking a blue sock first is \( \frac{6}{18} \).
- If he picks a blue sock first, the probability of picking a blue sock second is \( \frac{5}{17} \).

Now we calculate the combined probability of both events happening (i.e., picking two blue socks):

\[
P(\text{two blue socks}) = P(\text{first blue sock}) \times P(\text{second blue sock | first blue sock})
\]
\[
= \frac{6}{18} \times \frac{5}{17}
\]

Now, simplifying \( \frac{6}{18} \):

\[
\frac{6}{18} = \frac{1}{3}
\]

Substituting back into our equation:

\[
P(\text{two blue socks}) = \frac{1}{3} \times \frac{5}{17} = \frac{5}{51}
\]

Thus, the probability that Jason has two blue socks on his feet is:

\[
\frac{5}{51}
\]