danjuma wrote: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 < 0.2 that the first sock is black.

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

Can one use the probability formula for this type of question?

Thank you

I have some genuine concerns about this questions.

First of all we do not know that drawer contains only Black and White socks. There can be Red also

1: P(B) < 0.2
Total choices 8
then P(B) < 1.6/8 so there can be only 1 or 0 black sock.
How can we say that both of the socks will be black as at best there will be 1 black sock.

2: P(W) > 0.8
then P(W) > 6.4/8 so there can be only 7 or 8 white socks.
How can we say that both of the socks will be black as at best there will be 1 black sock. (if any other color is not there)

Combining both also we are not getting two black socks in any case. If there is no possibility of getting two socks then how can we draw 2 black socks. I think both are individually sufficient ... totally confused

IMO D

How can we say that both of the socks will be black as at best there will be 1 black sock.

We can't. Thus, the probability of both socks being black is 0.

(That the question is asking for the probability of both socks being black does not imply that that probability is non-zero).

Choose D.

Both the statements, individually, implies that the number of black socks is 0 or 1.
so Black socks is at max 1.
so the probablity of drawing a second socks is always 0.
P(B1)=Probability of drawing first black socks = 1/8.
P(B2)=Probability of drawing second black socks = 0/7
P(B1)*P(B2) = 0

Testluv wrote:

How can we say that both of the socks will be black as at best there will be 1 black sock.

We can't. Thus, the probability of both socks being black is 0.

(That the question is asking for the probability of both socks being black does not imply that that probability is non-zero).

Choose D.

Yeah!! Thanks you sir... I thought Q is wrong and chose E then thought both the cases m sure not able to determine then chose back D.

So the zero probability is also a Valid option.

What's the source? This is a good tricky question, because the colors don't matter here.

chendawg wrote:What's the source? This is a good tricky question, because the colors don't matter here.

I ran into this question in BTG practice set too
https://practice.beatthegmat.com/
and here's my thread of this question
https://www.beatthegmat.com/ambiguous-ds ... tml#328917

shovan85 wrote:

danjuma wrote: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 < 0.2 that the first sock is black.

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

Can one use the probability formula for this type of question?

Thank you

I have some genuine concerns about this questions.

First of all we do not know that drawer contains only Black and White socks. There can be Red also

1: P(B) < 0.2
Total choices 8
then P(B) < 1.6/8 so there can be only 1 or 0 black sock.
How can we say that both of the socks will be black as at best there will be 1 black sock.

2: P(W) > 0.8
then P(W) > 6.4/8 so there can be only 7 or 8 white socks.
How can we say that both of the socks will be black as at best there will be 1 black sock. (if any other color is not there)

Combining both also we are not getting two black socks in any case. If there is no possibility of getting two socks then how can we draw 2 black socks. I think both are individually sufficient ... totally confused

IMO D

I agree with the explanations above

shovan85 wrote:

danjuma wrote: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 < 0.2 that the first sock is black.

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

Can one use the probability formula for this type of question?

Thank you

I have some genuine concerns about this questions.

First of all we do not know that drawer contains only Black and White socks. There can be Red also

1: P(B) < 0.2
Total choices 8
then P(B) < 1.6/8 so there can be only 1 or 0 black sock.
How can we say that both of the socks will be black as at best there will be 1 black sock.

2: P(W) > 0.8
then P(W) > 6.4/8 so there can be only 7 or 8 white socks.
How can we say that both of the socks will be black as at best there will be 1 black sock. (if any other color is not there)

Combining both also we are not getting two black socks in any case. If there is no possibility of getting two socks then how can we draw 2 black socks. I think both are individually sufficient ... totally confused

IMO D

shovan your guess is correct
But look at both the statements
first statement is about availability of getting 1 black sock or zero sock
second statement is about the availability os 7 white socks or 8 socks.
So from this statement you can infer that there are no other colour socks in it.
So there is no possiblilty of getting two black socks i.e., the probability of getting a two black sock is zero.
This what we have in the question.

So choosing an option using the both the statement would be wise .
I hope its correct to choose using both the statements.

cheers

IMO D

Stmt 1:

P (1st sock is black) is <0.2
0.2 x 8 socks = 1.8 therefore p < 1.8, therefore there must be either 1 or 0 black socks, which means that the probability of drawing two black socks will be zero (i.e., 1/8 x 0/7 OR 0/8 x 0/8)
SUFFICIENT

Stmt 2:
P (1st sock white) > 0.8
therefore probability is 0.8 x 8 > 6.4. This means that there is either 7 or 8 white socks, which means there is only 1 or 0 black socks. Same result as 1 in terms of probability of drawing two blk socks = 0.
SUFFICIENT

be careful on the actual exams as 0.8 and 0.2 can be changed to 0.7 and 0.2;respectively. The answer will be different.

What is the OA?????

jayavignesh wrote:

shovan85 wrote:

danjuma wrote: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 < 0.2 that the first sock is black.

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

Can one use the probability formula for this type of question?

Thank you

I have some genuine concerns about this questions.

First of all we do not know that drawer contains only Black and White socks. There can be Red also

1: P(B) < 0.2
Total choices 8
then P(B) < 1.6/8 so there can be only 1 or 0 black sock.
How can we say that both of the socks will be black as at best there will be 1 black sock.

2: P(W) > 0.8
then P(W) > 6.4/8 so there can be only 7 or 8 white socks.
How can we say that both of the socks will be black as at best there will be 1 black sock. (if any other color is not there)

Combining both also we are not getting two black socks in any case. If there is no possibility of getting two socks then how can we draw 2 black socks. I think both are individually sufficient ... totally confused

IMO D

shovan your guess is correct
But look at both the statements
first statement is about availability of getting 1 black sock or zero sock
second statement is about the availability os 7 white socks or 8 socks.
So from this statement you can infer that there are no other colour socks in it.
So there is no possiblilty of getting two black socks i.e., the probability of getting a two black sock is zero.
This what we have in the question.

So choosing an option using the both the statement would be wise .
I hope its correct to choose using both the statements.

cheers

How would it help us by choosing C? I did not understand. Thank you!

probablitiy is 0 and it can be determined by either of the statements.

a black socks < 1.6. hence black sock = 1. probability = 0. sufficient.

b whites socks = 7 or 8.thus probability for 2 black socks = 0. sufficient.

D it is.