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Re: A drawer contains 6 socks. Two socks are randomly selected without rep [#permalink]
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Paul121 wrote:
Total: 6 socks.
Get 2 white socks probability without replacement: [(x)/6]*[(x-1)/5]=0.4
Equation: x^2-x-12=0
(x-4)(x+3)=0
x must be positive so the answer it 4.

So there are 4 white socks and another 2 socks (assuming black, not mentioned in the question, so if not black, the answer will be D).
If there are 4 white and 2 black socks. The probability to get 2 black socks is: (1/6)*(1/5) = 1/30.

The answer is B.


Not really.
Your very first expression should have " > 0.4 " and not " = 0.4"
which will yield (x-4)(x+3) > 0
and, x>4

x = {5,6}

Since white balls are either 5 or 6 in numbers, black balls are 0 or 1 in number.
So, the probability of getting 2 black balls is 0.

Hence, B is the answer.
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Re: A drawer contains 6 socks. Two socks are randomly selected without rep [#permalink]
Paul121 wrote:
Total: 6 socks.
Get 2 white socks probability without replacement: [(x)/6]*[(x-1)/5]=0.4
Equation: x^2-x-12=0
(x-4)(x+3)=0
x must be positive so the answer it 4.

So there are 4 white socks and another 2 socks (assuming black, not mentioned in the question, so if not black, the answer will be D).
If there are 4 white and 2 black socks. The probability to get 2 black socks is: (1/6)*(1/5) = 1/30.

The answer is B.


What about 30 in the denominator in the first equation? Can you expand on the first equation?
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Re: A drawer contains 6 socks. Two socks are randomly selected without rep [#permalink]
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Re: A drawer contains 6 socks. Two socks are randomly selected without rep [#permalink]
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