Question

a drawer contains 6 black socks, 4 blue socks, and 2 white socks. if two socks are drawn at random without replacement, what is the probability that both socks are black?
a. $\frac{1}{4}$
b. $\frac{5}{22}$
c. $\frac{1}{3}$
d. $\frac{25}{144}$

Explanation:

Step1: Calculate total number of socks

The total number of socks is \(6 + 4+2=12\).

Step2: Calculate probability of first - sock being black

The probability that the first sock drawn is black is \(\frac{6}{12}=\frac{1}{2}\), since there are 6 black socks out of 12 total socks.

Step3: Calculate probability of second - sock being black given first is black

After drawing one black sock, there are 5 black socks left and 11 total socks left. So the probability that the second sock is black given that the first one was black is \(\frac{5}{11}\).

Step4: Calculate probability of both socks being black

By the multiplication rule for dependent events, the probability that both socks are black is \(\frac{1}{2}\times\frac{5}{11}=\frac{5}{22}\).

Answer:

B. \(\frac{5}{22}\)