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

a) 1/11

b) 7/22

c) 5/11

d) 0

e) 19/66

Explanation:

b=blue socks

g=grey socks

B=black socks

A)

P(2blue socks) = 4/12*3/11 = 1/11 . That is , there is a 4/12 chance the first sock drawn is blue . Once the first sock is drawn , there is a 3/11 chance the second sock is blue as well .

B)P(no grey socks)=7/12*6/11 = 7/22

c)1-P(no black socks) = 1- 9/12*8/11 = 5/11

d) There are no green socks in the pile :)

P(1 green sock) = 0

e)P(bb)+P(gg)+P(BB) = 4/12*3/11 + 5/12*4/11 + 3/12*2/11

Answer 2

a)
`(1)/(22)`

b)
`(7)/(22)`

c)
`(5)/(11)`

d)
`(35)/(66)`

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`

`=(12!)/(2!10!)`

`=(12* 11)/(2)`

`=6* 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 =
`(3)/(66)=(1)/(22)`

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

`=(7!)/(2!5!)`

`=7* 3`

`=21`

Thus, the probability of selecting no grey socks =
`(21)/(66)`

`=(7)/(22)`

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

`=^3C_1* ^9C_1+^3C_2`

`=3* 9 + 3`

`=27 + 3`

= 30,

Thus, the probability of selecting at least 1 black sock =
`(30)/(66)`

`=(5)/(11)`

d) Ways of selecting a green sock =
`^5C_1* ^7C_1`

= 35,

Thus, the probability of selecting a green sock =
`(35)/(66)`