AAPL wrote: ↑

Mon Nov 08, 2021 5:35 pm

Magoosh

A drawer contains 36 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?

1) The probability is 4/9 that the first sock is black
2) The number of white socks in the drawer is 4 more than the number of black socks

OA A

Given: A drawer contains 36 socks, and 2 socks are selected at random without replacement.

Target question: What is the probability that both socks are black?

Statement 1: The probability is 4/9 that the first sock is black.
P(1st sock is black) = (total number of black socks)/(total number of socks)
Substitute to get: 4/9 = (total number of black socks)/36
Solve to get: Total number of black socks = 16
Now that we know there are 16 black socks, we can answer the target question with certainty (although we wouldn't waste valuable time on test day calculating the probability)
Statement 1 is SUFFICIENT

Aside: P(both socks are black) = 16/36 x 15/35 = 4/21

Statement 2: The number of white socks in the drawer is 4 more than the number of black socks
Important: We don't know how many different colored socks there are.
Consider these two scenarios that satisfy statement 2:
Case a: There is 1 black sock, 5 white socks, and 30 green socks. In this case, the answer to the target question is P(both socks are black) = 0 [since we only have ONE black sock]
Case b: There are 2 black socks, 6 white socks, and 28 green socks. In this case, the answer to the target question is P(both socks are black) = something other than 0
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent