[X] In a drawer, there are 33 socks. 21 of them are crew socks, and the rest are
ID: 3334100 • Letter: #
Question
[X] In a drawer, there are 33 socks. 21 of them are crew socks, and the rest are low-cut ones. Of the crew socks, 12 are black, 7 are white and 2 are gray. Of the low-cut socks, 5 are black, 4 are white, and 3 are gray. a) If two socks are randomly selected without replacement, what is the probability that the lst is a crew sock and the 2nd is either a white or a gray sock? A fair die is tossed. If1 or 6 shows up, two socks will be drawn at random from the drawer without replacement. If neither 1 nor 6 shows up, three socks will be drawn at random without replacement. What is the probability that, among the drawn socks, there will be at least one white sock? b)Explanation / Answer
Answer
Given, Crew socks=21
Low cut Socks=12
Black
white
Gray
Total
Crew Socks
12
7
2
21
Low cut socks
5
4
3
12
A.a) Probabiliy that 1st is a crew sock=P(X)=21/33=0.64
1st sock picked can be black or white or grey from crew socks
Probability that 2nd is either a white or grey sock=P(Y)
Here we have 3 scenarios
If 1st sock picked is black
P(Y1)=16/32
If 1st sock picked is white
P(Y2)=15/32
If 1st sock picked is grey
P(Y3)=15/32
P(Y)={P(Y1)+ P(Y2)+ P(Y3)}/3
P(Y)={16/32+15/32+15/32}/3
P(Y)=0.48
Fianl answer=P(X)*P(Y)
Fianl answer=0.64*0.48
Fianl answer=0.3054
A.b) A die is tossed
Outcome 1 having 1 or 6=2/6
Probability of having at least one white=1-probability of having no white
Probability of having at least one white =1-(22/33*21/32)
Probability of having at least one white =0.56
P(X)=0.56*2/6=0.1875
Outcome of having 2,3,4,5=4/6
Probability of having at least one white=1-probability of having no white
Probability of having at least one white=1-(22/33*21/32*20/31)
Probability of having at least one white=0.28
P(Y)=0.28*4/6=0.1882
Fianl probability=P(X)+P(Y)
=0.1875+0.1882=0.3757
Fianl probability=0.3757
Black
white
Gray
Total
Crew Socks
12
7
2
21
Low cut socks
5
4
3
12
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