The probability that Jason has blue socks is
![(5)/(51)](https://img.qammunity.org/2020/formulas/mathematics/high-school/z2j0guo9e5lfuuor8z8vvpo9d2u3c0kp6h.png)
Solution:
Given, there are 12 brown socks and 6 blue socks in a drawer.
So, total number socks = 12 + 6 = 18 socks
In the dark, Jason pulls out a sock and puts it on his right foot.
Now, let the probability of pulled sock to be blue sock
![=\frac{\text {number of blue socks}}{\text {total number of socks}}=(6)/(18)=(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/high-school/bjm2szpnekyg6negd3cvlgljquoabx43ud.png)
Then, available socks = 18 – 1 picked sock = 17 socks and number of blue socks = 6 – 1 = 5
Then he pulls out another sock and puts it on his left foot
Now, the probability that second picked sock will also be blue =
![\text { probability of } 1 \text { sock } * \frac{\text { blue socks count }}{\text { socks count }}](https://img.qammunity.org/2020/formulas/mathematics/high-school/vylf2mpd0txfgqudlq65b0awrgnd7ickvm.png)
![=(1)/(3) * (5)/(17)=(5)/(51)](https://img.qammunity.org/2020/formulas/mathematics/high-school/jxgfr3192y17upi9r1nyh6vu03ycik5ig6.png)
Hence, the probability that Jason has blue socks is
![(5)/(51)](https://img.qammunity.org/2020/formulas/mathematics/high-school/z2j0guo9e5lfuuor8z8vvpo9d2u3c0kp6h.png)