A drawer contains 8 socks, and 2 socks are selected at random without

I had initially thought answer as D but later changed to A(my mistake). My thought process was that with second choice I don't know if remaining sock is black or not. I failed to calculate that the answer of two draws would be zero in this case also

How about I tell you on what lines to think and perhaps you can arrive at the answer?
Let's say there are x black socks in the drawer. When I pick the first one, the probability of picking a black sock is x/8 which is less than 1/5. Any ideas?

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Hi Bunuel, Why is that in this problem the socks are assumed to be either white or black? No where in the question is it mentioned that the socks are either white or black.

So, shouldn't the answer be E?

Number of Socks HAS to be an integer. Therefore, (1) & (2) are saying the same thing because

\(\frac{(Favorable Cases)}{(Total Number of Cases)} = Probability\)

=> Favorable Cases = Probability * Total Number of Cases

=> Number of Ways to pick a Black Sock in 1st Attempt = Prob. * 8

(1) The probability is less than 0.2 that the first sock is black. => Prob of 1at Black sock < 0.2

(2) The probability is more than 0.8 that the first sock is white. => Prob of 1st white > 0.8
=> Prob of 1st sock of any color other than white < 0.2

Therefore, Prob of 1st Black Sock < 0.2

Hence, the answer cannot be anything but D or E. D if the statement is sufficient otherwise E.


Now, Number of Ways to pick a Black Sock in 1st Attempt < 0.2 * 8 < 1.6
Hence, we have 0 or 1 Black sock.

Answer: D

Bunuel, after calculating the range in each statements, how did you deduce the fact (highlighted in red) and concluded that black socks are 0 ( highlighted in blue)?
I still can't get what am I missing!

Solution:

Two socks out of 8 socks in a drawer are chosen. We need to calculate the probability that both socks are black.

Statement One Only:

(1) The probability is less than 0.2 that the first sock is black.
If the probability that the first sock is black were equal to 0.2 (instead of less than 0.2), then we would be able to say that out of 10 socks, there must be exactly 2 black socks, and the probability would be 0.2. However, there are only 8 socks in the drawer, and if the probability of a black sock were (again) equal to 0.2, we see that we could have at most 8 * 0.2 = 1.6 black socks in the drawer.

But we know that the probability of a black sock is actually less than 0.2, so we know that there must be fewer than 1.6 black socks in the drawer. So there are either no black socks or 1 black sock in the drawer. Thus, it is impossible to draw two black socks out of the drawer, and so the probability is zero.

Statement one is sufficient to answer the question.

Statement Two Only:

(2) The probability is more than 0.8 that the first sock is white.

Using similar logic to that used for statement one, we see that, if the probability that a white sock is more than 0.8, then the probability of a black sock must be less than 1 - 0.8 = 0.2. This is now an identical statement to statement one, and so we see that the question can be answered with the information in statement two.

Statement two is sufficient to answer the question.

Answer: D

In questions on probability, especially a question like the above, the best approach is to stick to the basics – use the fundamental equation used to calculate Probability.

Probability of an Event = \(\frac{Number of outcomes of the Experiment favourable to the event}{ Total Possible Outcomes of the Event (also known as Sample Space)}\).
You see how this simplifies problem solving. You just have to identify the experiment and the event. The experiment will give you the total number of outcomes and the event will show you the favourable outcomes.

The experiment that we are doing in this question is ‘to pick 2 socks at random from a drawer containing 8 socks’. Using counting methods, we know that the number of ways in which this experiment can be done is 8*7 ,since we are picking sequentially without replacement. Now, this is our denominator.

Of the 8 socks, let’s say ‘x’ are black; therefore (8-x) are not black. The event for which we are trying to calculate probability – both the socks that we picked should be black. This can happen in \(x_C_2\) ways i.e. \(\frac{x(x-1) }{ 2}\) ways. This is our numerator.

Therefore, we are trying to find the value of \(\frac{x(x-1) }{ 112}\) which should tell us that any data that gives us x will be sufficient data.

From statement I alone, the probability that the first sock is black is less than 0.2. This means,
\(\frac{x}{8}\) < \(\frac{1}{5}\) (remember, 0.2 = \(\frac{1}{5}\)). Simplifying, we have x < \(\frac{8}{5}\) which essentially means x<2. Since the number of socks needs to be a whole value, this only means that there is ONE black socks or NONE i.e. x = 0 or 1.

Substituting either of these values in the expression that we derived from the question data, we see that the probability turns out to be ZERO in both cases.
Thus, statement I alone is sufficient to say that the probability is ZERO. Answer options B, C and E can be eliminated. The possible answer options at this stage are A or D.

From statement II alone, the probability that the first sock is white is more than 0.8. If you look at this statement carefully, it’s actually telling what statement I said, albeit in a different way.
So, \(\frac{(8-x) }{ 8}\) > \(\frac{4 }{ 5}\) (remember, 0.8 = \(\frac{4 }{ 5}\)). Simplifying, we have (8-x) > \(\frac{32 }{ 5}\) which essentially means (8-x) > 6 which leads us to the same conclusion that x < 2. So, again x = 0 or 1. We know what answer we got previously with these values.

So, statement II alone is also sufficient. Answer option A can be eliminated.
The correct answer option is D.

Hope that helps!

This question is not clear. Is there white and black socks only? Even it is not mentioned what is the color of socks....

Please read the whole thread: https://gmatclub.com/forum/a-drawer-con ... l#p1387076

thank you...i understood it...