a drawer contains two pairs of red socks, three pairs of yellow socks, and one pair of black socks. one sock…

a drawer contains two pairs of red socks, three pairs of yellow socks, and one pair of black socks. one sock is taken from the drawer and put back after checking its color. a second sock is then taken out. what is the probability that the first and the second are yellow?\na $\frac{1}{144}$\nb $\frac{5}{6}$\nc $\frac{1}{4}$\nd $\frac{1}{6}$

a drawer contains two pairs of red socks, three pairs of yellow socks, and one pair of black socks. one sock is taken from the drawer and put back after checking its color. a second sock is then taken out. what is the probability that the first and the second are yellow?\na $\frac{1}{144}$\nb $\frac{5}{6}$\nc $\frac{1}{4}$\nd $\frac{1}{6}$

Answer

Explanation:

Step1: Calculate total number of socks

There are (2\times2 + 3\times2+1\times2=4 + 6+2 = 12) socks.

Step2: Calculate probability of first - yellow sock

The number of yellow socks is (3\times2 = 6). The probability of picking a yellow sock on the first draw, (P(Y_1)=\frac{6}{12}=\frac{1}{2}).

Step3: Calculate probability of second - yellow sock

Since the sock is replaced, the total number of socks and the number of yellow socks remain the same. So the probability of picking a yellow sock on the second draw, (P(Y_2)=\frac{6}{12}=\frac{1}{2}).

Step4: Calculate joint - probability

Since the two draws are independent events, the probability that the first and the second are yellow is (P(Y_1\cap Y_2)=P(Y_1)\times P(Y_2)). Substituting the values, we get (P(Y_1\cap Y_2)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}).

Answer:

C. (\frac{1}{4})