GMATPrepNow wrote:
A drawer contains 6 socks. If two socks are randomly selected without replacement, what is the probability that both socks will be black?
(1) The probability is less than 0.3 that the first sock selected will be black.
(2) The probability is greater than 0.4 that both socks will be white.
We are given that a drawer contains 6 socks. If we let b = the number of black socks, we see that the probability of a black sock on the first pick is b/6. Since there is no replacement, the probability of a black sock on the second pick is (b - 1)/5. We need to determine the product of (b/6) x (b-1)/5.
Statement One Alone:
The probability is less than 0.3 that the first sock selected will be black.
Using the information in statement one, we have:
b/6 < 3/10
10b < 18
b < 1.8
We see that there is at most 1 black sock. Thus, the probability of pulling 2 black socks in two picks is equal to 0. Statement one alone is sufficient to answer the question.
Statement Two Alone:
The probability is greater than 0.4 that both socks will be white.
Suppose that there were exactly 4 white socks in the drawer. In this case, the probability of drawing 2 white socks would be 4/6 x 3/5 = 12/30 = 2/5 = 0.4. Since the probability of getting 2 white socks is greater than 0.4, there must be more than 4 white socks in the drawer; therefore, there can be at most 1 black sock in the drawer. The probability of pulling 2 black socks is 0. Statement two alone is sufficient to answer the question.
Answer: D
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