Question

37. Socks. In your sock drawer you have 4 blue socks, 5 grey socks, and 3 black ones. Half asleep one morning, you grab 2 socks at random and put them on. Find the probability you end up wearing a) 2 blue socks. b) no grey socks. c) at least 1 black sock. d) a green sock.

Answer 1

Answer:

a) $\frac{1}{22}$

b) $\frac{7}{22}$

c) $\frac{5}{11}$

d) $\frac{35}{66}$

Step-by-step explanation:

Given,

Number of blue socks = 4,

Grey socks = 5,

Black socks = 3,

Total socks = 4 + 5 + 3 = 12,

Ways of choosing any 2 socks = $^{12}C_2$

$=\frac{12!}{2!10!}$

$=\frac{12\times 11}{2}$

$=6\times 11$

= 66,

a) Ways of choosing 2 blue socks = $^3C_2$

= 3,

Since, $\text{Probability}=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

Thus, the probability of selecting 2 blue socks = $\frac{3}{66}=\frac{1}{22}$

b) Ways of choosing 2 shocks, non of them are grey socks = $^7C_2$

$=\frac{7!}{2!5!}$

$=7\times 3$

$=21$

Thus, the probability of selecting no grey socks = $\frac{21}{66}$

$=\frac{7}{22}$

c) ways of selecting atleast 1 black sock = ways of selecting 1 black sock + ways of selecting 2 black socks

$=^3C_1\times ^9C_1+^3C_2$

$=3\times 9 + 3$

$=27 + 3$

= 30,

Thus, the probability of selecting at least 1 black sock = $\frac{30}{66}$

$=\frac{5}{11}$

d) Ways of selecting a green sock = $^5C_1\times ^7C_1$

= 35,

Thus, the probability of selecting a green sock = $\frac{35}{66}$