Question

1. socks in a drawer a drawer contains 11 identical red socks and 8 identical black socks. suppose that you choose 2 socks at random in the dark.

a. what is the probability that you get a pair of red socks?
b. what is the probability that you get a pair of black socks?
c. what is the probability that you get 2 unmatched socks?
d. where did the other red sock go?

Explanation:

Step1: Calculate total number of socks

Total number of socks \(n = 11+8=19\)

Step2: Calculate probability for part a

Probability of first red sock \(P(R_1)=\frac{11}{19}\)
After taking one red sock, number of red socks left is \(10\), total socks left is \(18\)
Probability of second red sock \(P(R_2)=\frac{10}{18}\)
Probability of pair of red socks \(P(RR)=P(R_1)\times P(R_2)=\frac{11}{19}\times\frac{10}{18}=\frac{110}{342}=\frac{55}{171}\)

Step3: Calculate probability for part b

Probability of first black sock \(P(B_1)=\frac{8}{19}\)
After taking one black sock, number of black socks left is \(7\), total socks left is \(18\)
Probability of second black sock \(P(B_2)=\frac{7}{18}\)
Probability of pair of black socks \(P(BB)=P(B_1)\times P(B_2)=\frac{8}{19}\times\frac{7}{18}=\frac{56}{342}=\frac{28}{171}\)

Step4: Calculate probability for part c

Probability of unmatched socks \(P(\text{unmatched}) = 1-(P(RR)+P(BB))\)
\(P(RR)+P(BB)=\frac{55 + 28}{171}=\frac{83}{171}\)
\(P(\text{unmatched})=1-\frac{83}{171}=\frac{171-83}{171}=\frac{88}{171}\)

Answer:

a. \(\frac{55}{171}\)
b. \(\frac{28}{171}\)
c. \(\frac{88}{171}\)
d. This is a trick question, there is no information given in the problem to determine where the "other red sock" (assuming some prior context which isn't there) went. It could be a joke or a misdirection. If we consider the problem as given (11 red socks initially), there is no basis for this question within the probability - calculation framework of parts a - c.