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Re: A drawer contains 6 brown socks and 4 black socks. [#permalink]
2
jaidevphadke wrote:
I'd like to know how this problem could be solved using combinations. Could anyone help me? (I know the probability method is easiest but this will improve my understanding of combinatorics).

Thank you!


I am not familiar with the forum math notation, but here we go:

Quantity of 6 brown pairs:

\(\frac{6!}{2!(6-2)!} \)= 15 stands for all the 2 socks pairs that you can do.

Quantity of 4 black pairs:

\(\frac{4!}{2!(4-2)!} \)= 6 stands for all the 2 socks pairs that you can do.

The number of possible outcomes is given by: 15+6 = 21

The number of total outcomes is all the pairs that you can do with both colors mixed:

\(\frac{10!}{2!(10-2)!}\) = 45


Finally, the result is given by possible outcomes divided by total outcomes:

\(\frac{21}{45}\) = \(\frac{7*3}{5*9}\) = \(\frac{7}{15}\)
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Re: A drawer contains 6 brown socks and 4 black socks. [#permalink]
Thank you Cote! Very helpful reply :-D
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Re: A drawer contains 6 brown socks and 4 black socks. [#permalink]
would like to know why the below mentioned approach is wrong? thanks
Probability = number of favorable outcomes/total number of outcomes

Total outcomes possible
1. Black,Black
2. Black,Brown
3. Brown,Black
4. Brown,Brown

favorable are 1 and 4 , so probability is 2/4 = 1/2
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Re: A drawer contains 6 brown socks and 4 black socks. [#permalink]
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tinku wrote:
would like to know why the below mentioned approach is wrong? thanks
Probability = number of favorable outcomes/total number of outcomes

Total outcomes possible
1. Black,Black
2. Black,Brown
3. Brown,Black
4. Brown,Brown

favorable are 1 and 4 , so probability is 2/4 = 1/2


In order for your approach to work, it must be the case that the four possible outcomes are all equally likely.

Since the number of brown socks is greater than the number of black socks, the outcome BROWN,BROWN is more likely than the outcome BLACK,BLACK
As such we can't use the this approach.

Here is an analogous question.
A box contains 1 red ball and 1,000,000 green balls. If a single ball is randomly drawn from the box, what is the probability that the ball is red?

Of course, we know the answer my here must be 1/1,000,001
However, let's see what happens if we list all possible outcomes:
1) The selected ball is RED
2) The selected ball is GREEN
Since 1 of the 2 possible outcomes is favorable, P(select red ball) = 1/2

As you can see, if the outcomes are equally likely, this approach of listing outcomes falls apart.
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Re: A drawer contains 6 brown socks and 4 black socks. [#permalink]
1
The probability that the first two socks pulled out will be the same color = The probability the first two socks pulled out will be brown + The probability that the first two socks pulled out will be black


\(= \frac{6}{10} \times \frac{5}{9} + \frac{4}{10} \times \frac{3}{9}\)

\(=\frac{1}{3} + \frac{2}{15}\)

\(=\frac{7}{15}\)

The answer is Choice B
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Re: A drawer contains 6 brown socks and 4 black socks. [#permalink]
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