A drawer contains 36 socks, and 2 socks are selected at

You are right, it's not mentioned and that is why (2) is not sufficient. But for (1): saying that the probability of drawing the FIRST black sock is 4/9 means that we have 4 chance out of 9 to have black OR as there are total of 36 socks, 16 chances out of 36. That's clearly gives us the NUMBER OF BLACK SOCKS in the drawer, even though we still don't know the number and colors of the other socks (well, there are total 36-16=20 others but we don't know their color). So (1) is still sufficient to determine the probability that both socks are black 4/9*15/35=4/21.

Answer: A.

I think it's D:
(1) is obvious
(2) The number of white socks in the drawer is 4 more than the number of black socks.
Let’s B be the number of black socks and W the number of white.
W = B +4 and we know 36 = B +W then W = 36 – B = B + 4
=> 2B = 32 => B =16 (Sufficient !)

But you are wrongfully assuming from the question that there are only B and W socks, which is not necessarily the case.

IMO : A

This was how I analyzed it! And if I'm doing what you're doing, I'm finally getting somewhere..

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Let the number of black socks = b

So the question can be rephrased as b/36 * (b-1)/35 = ?


Stmt 1 : The probability of the 1st sock being black = b/36 = 4/9

Cross multiply and you get b = 16

Hence we know that the 16 is the total number of black socks

This information helps us calculate the exact probability of getting 2 black socks. Hence stmt 1 is sufficient.



Stmt 2 : Let the number of white socks be w hence, w = b+4

This information does not help us arrive at the exact probability of getting 2 black socks because we are not told whether the 36 socks consist of only black and white socks or whether there are more colours as well.

Since we can not derive exact information,this statement is insufficient.


Answer : A

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.

From either statement, one can conclude that there are 16 black and 20 white socks. So the desired probability is 16/36 x 15/35

D.

Hi,
While evaluating both options you have assumed there are only white and black socks. Which is incorrect.
While in 1 we know the exact number of black socks and total number of socks. So we can conclude it is sufficient condition.
In option 2 we are given relative number of white socks as compared to black. But it is nowhere state black + white = total number of socks.
Hence it is insufficient.

Please give kudos if you like the explanation.

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

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