Problem Solving

1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color?
A. 1/5
B. 2/5
C. 3/4
D. 4/5
E. 1

One way to solve this problem is to "think" about it. Probability is always 1!

Here's what I did :(long method)

10C2 {6C1 * 2C1 + 6C2 + 2C2} + 6C2*(2C2 + 10C1*2C1) + 2C2*6C1*10C1
--------------------------------------------------------------------
20C4

45*{12+15+1} + 15{1+20} + 60
= ----------------------------
4845

1260 + 315 + 60
= ---------------
4845

is not equal to 1! Why?

I did not go through your calculation but the correct answer is 1.

There are three colors and four picks and at least two socks with the same color in the bin. Even in the worst case scenario, you would pick the same color at least once:

Say you pick color 1 then color 2, then color 3; on your last pick you have to pick one of the previously picked colors.

You can read the Wikipeadia article about the "Pigeonhole Principle" for a detailed explanation

Pharo - I was specifically looking for an algebraic method. I knew the "Intuitive" method beforehand. Your post was not at all helpful.... Sorry for being blunt.

Being blunt is good, no worries. However, you could have easily converted what I told you to algebra (or have read on the pigeonhole principle.. that has more math ). You should have spent some time on it -- now it is your turn to forgive me for being blunt

Let's go with NOT(P(picking a pair));

NOT(P(picking a pair)) = P(picking a white) * P(picking a black) * P(picking a grey) * NOT(P(picking white OR black OR grey))

=(5/10)*(3/9)*(2/8)*0 = 0;
NOTE: since there are only three colors to pick from the chances of not picking one of them is 0

Since NOT(P(picking a pair)) + P(picking a pair) = 1; the total set of the outcomes

P(picking a pair) = 1 ; sub in P(not picking a pair) = 0;

The question is asking the probability of getting two socks of the same color and picking a pair (and picking two pairs but I included this probability in P(picking a pair) for ease of explanation) satisfies that condition. Hence answer is 1.

Hope this helps

I hoped did it right but the way I do my probability is easy for me.
I decided to do (5-4)(3-4)(2-4)/(4-2), the reason is because I took the chance of taking 4 from each possible set and then dividing it by the possibility of the total number of possibilities from the actual amount taken.