BTG DS

[r2kins]

A drawer contains 8 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?

(1) The probability is less than 0.2 that the first sock is black.

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

[Geva@EconomistGMAT]

A drawer contains 8 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?

(1) The probability is less than 0.2 that the first sock is black.

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

Probability is defined as number of wanted outcomes / total number of outcomes.

Probability of a single sock being black is number of blacks out of total number of socks. Since we know that there are 8 socks, the missing piece of the puzzle is the number of blacks. If a statement manages to limit the number of black socks to a single value, we can find the required probability.

The interesting part about stat. (1) is the low probability of the first sock being black - only 0.2. For a total of 8 socks, this means a very low number of blacks - too low:
If there's only 1 black sock, then the probability of pulling that sock out is 1/8 = 12.5.
If there're 2 black socks, the probability the first sock being black is 2/8 = 0.25 = already greater than 0.2

So stat. (1) basically limits the number of black socks to "one sock", in which case it would be impossible to pull out two black socks from the drawer without replacement - there's only one sock! Thus, P(two black socks)=0. Sufficient.

Stat. (2) says the same thing as stat. (1) - if there's more than 0.8 probability of a white sock, then there's less than 0.2 probability of a black sock - which again means that there is, at most, 1 sock in the drawer, and the probability of drawing two socks is zero, as it is an impossible event. Sufficient.
the answer is D.

To stem further queries: yes, the question, as phrased, allows the possibility of socks of other colors as well, so the number of black socks can be 1 sock (prob of 0.125), or zero socks (if there are red socks as well, for example)). But in both of these cases, the answer to the question "what is the probability of pulling two black socks in a row without replacement?" is "zero", as there are fewer than two blacks in the drawer in the first place - which means that the statements are sufficient for both of these cases.

[Frankenstein]

Hi,
From(1): probability that the first sock is black is less than 0.2
Let n be the number of black socks
So, probability that first sock is black is n/8
Now, n/8 < 0.2. So, n<1.6
So, n =0 or 1
So, there cannot be more than 1 black socks in the 8 socks.
So, probability that both socks are black is 0
Sufficient
From(2):
The probability is more than 0.8 that the first sock is white
Let p be the number of white socks
So, probability that first sock is black is p/8
Now, p/8 > 0.8
So, p > 6.4. i.e. p=7 or 8
So, there cannot be more than 1 black socks in those 8.
So, probability that both socks are black is 0
Sufficient

Hence, D

[r2kins]

A drawer contains 8 socks, and 2 socks are selected at random without replacement. What is the probability that both socks are black?

(1) The probability is less than 0.2 that the first sock is black.

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

Probability is defined as number of wanted outcomes / total number of outcomes.

Probability of a single sock being black is number of blacks out of total number of socks. Since we know that there are 8 socks, the missing piece of the puzzle is the number of blacks. If a statement manages to limit the number of black socks to a single value, we can find the required probability.

The interesting part about stat. (1) is the low probability of the first sock being black - only 0.2. For a total of 8 socks, this means a very low number of blacks - too low:
If there's only 1 black sock, then the probability of pulling that sock out is 1/8 = 12.5.
If there're 2 black socks, the probability the first sock being black is 2/8 = 0.25 = already greater than 0.2

So stat. (1) basically limits the number of black socks to "one sock", in which case it would be impossible to pull out two black socks from the drawer without replacement - there's only one sock! Thus, P(two black socks)=0. Sufficient.

Stat. (2) says the same thing as stat. (1) - if there's more than 0.8 probability of a white sock, then there's less than 0.2 probability of a black sock - which again means that there is, at most, 1 sock in the drawer, and the probability of drawing two socks is zero, as it is an impossible event. Sufficient.
the answer is D.

To stem further queries: yes, the question, as phrased, allows the possibility of socks of other colors as well, so the number of black socks can be 1 sock (prob of 0.125), or zero socks (if there are red socks as well, for example)). But in both of these cases, the answer to the question "what is the probability of pulling two black socks in a row without replacement?" is "zero", as there are fewer than two blacks in the drawer in the first place - which means that the statements are sufficient for both of these cases.

Thanks for the explanation Geva!