Her first pick is inconsequential. she can choose any shoe.

For her second pick, she now has to pick the 1 shoe that matches out of the 9 remaining shoes... so 1/9. C.

This should ideally take 20-30 seconds.

II wrote:Kim has 5 pairs of shoes; each pair is a different color. If Kim randomly selects 2 shoes without replacement from the 10 shoes, what is the probability that she will select 2 shoes of the same color ?

(A) 2/5
(B) 1/5
(C) 1/9
(D) 1/10
(E) 1/25

I know the answer to this one ... but am interested in learning the various approaches people are taking in answering this one.

Thanks.

I'll go with C also

You could also do :

5/10C2 = 5/45 = 1/9

but this isn't as fast as what i did before...

i go with 1/9 as well...

egybs wrote:You could also do :

5/10C2 = 5/45 = 1/9

but this isn't as fast as what i did before...

Thanks egybs.
Sorry for the basic question, buy can you please clarify what you mean by the C2 in "5/10C2"

My approach for this was:
There will be 5 pairs of shoes ... so there will be 5 colours represented. The probability of picking one colour is 2/10.
So once the first shoe is picked we have 1 more shoe which matches the one already picked ... and 9 shoes remaining ... so the probability is 1/9.
Since we are picking one shoe AND another shoe ... we multiply the two probabilities: 2/10 * 1/9 = 2/90 = 1/45

So 1/45 is the probability of picking a pair of shoes of the same colour.

Since the question doesnt specify a specific colour, and there are 5 colours ... we could pick the 1st colour OR the 2nd colour OR the 3rd colour OR the 4th colour OR the 5th colour ... (note with OR scenarios in probability, we add the probabilities).
We know the probability of picking 1 pair of shoes with the same colour is 1/45 ... so we have 1/45 + 1/45 + 1/45 + 1/45 + 1/45 = 5/45 = 1/9.

Instead of the actual answer itself, I am more interested in reviewing this problem in depth: finding out alternative approaches, key take-aways/lessons from this question, traps/tricks to be aware of etc.

Thanks.

Sure,

we know that there are 5 possible ways for Kim pick the same color shoes. But how many different ways can she select them?

10C2 ("10 choose 2") is the number of ways that you can choose 2 things out of 10 things.

The operation works as follows:

10!/(2!*(10-2)!)

This simplifies to:
10*9/2 = 45.

So Kim could choose the two shoes 45 different ways. 5/45 = 1/9.

Your method, while correct, is unnecessarily long. It doesn't matter what colour shoe she chooses first. Regardless of what color shoe she chooses first, she'll need to pick that same color shoe again out of the nine remaining shoes. which is 1/9.

This is very similar to the 4 letters in envelopes question. Where it doesn't matter which envelope is correctly placed:
https://www.beatthegmat.com/probability- ... 14363.html

In that question you could either do 4* 1/4 * 2/3 * 1/3 or you could just do 2/3 *1/3

Hope this helps...

II wrote:

egybs wrote:You could also do :

5/10C2 = 5/45 = 1/9

but this isn't as fast as what i did before...

Thanks egybs.
Sorry for the basic question, buy can you please clarify what you mean by the C2 in "5/10C2"

My approach for this was:
There will be 5 pairs of shoes ... so there will be 5 colours represented. The probability of picking one colour is 2/10.
So once the first shoe is picked we have 1 more shoe which matches the one already picked ... and 9 shoes remaining ... so the probability is 1/9.
Since we are picking one shoe AND another shoe ... we multiply the two probabilities: 2/10 * 1/9 = 2/90 = 1/45

So 1/45 is the probability of picking a pair of shoes of the same colour.

Since the question doesnt specify a specific colour, and there are 5 colours ... we could pick the 1st colour OR the 2nd colour OR the 3rd colour OR the 4th colour OR the 5th colour ... (note with OR scenarios in probability, we add the probabilities).
We know the probability of picking 1 pair of shoes with the same colour is 1/45 ... so we have 1/45 + 1/45 + 1/45 + 1/45 + 1/45 = 5/45 = 1/9.

Instead of the actual answer itself, I am more interested in reviewing this problem in depth: finding out alternative approaches, key take-aways/lessons from this question, traps/tricks to be aware of etc.

Thanks.