Problem Solving

GMATGuruNY wrote:

tester_twelve wrote:

Ian Stewart wrote:

jason49 wrote: I think that C is correct, trying to figure out what tgmat calculated, because it obviously includes more cases than necessary.

Can somebody comment on this?

Nice find. In probability questions, when you see the words 'at least', as in this question, that often means that you would need to consider a few different cases if you did the problem directly. So this is often a clue that it will be faster to do the opposite problem: determine the probability that the event does not happen, then subtract from 1. That's exactly what sudhir did above, and that's exactly how I'd approach the problem as well.

That said, it is possible to do the problem directly, just as long as we break down the problem correctly. This solution is not quite right:

torontogmat.com wrote:It does not matter which card is turned first.

There is a 1/11 chance that the next card turned will match. If it doesn't,
there is a 2/10 chance that the following card will match. If it still doesn't,
there is a 3/9 chance that the final card will match.

1/11 + 1/5 + 1/3 = 103 / 165.

The probability, for example, that the last card matches one of the first three is not equal to 1/3, which is what the above solution assumes. It is only equal to 1/3 when we know that none of the first three cards match. We can adjust this solution to get the correct answer, however:

We get at least one pair IF:

[1+2 match] OR [(1+2 do not match)AND(1/2 matches 3)] OR [(1,2,3 do not match)AND(4 matches 1,2 or 3)]

The above solution missed the AND conditions. Solving:

1/11 + (10/11)(1/5) + [(10/11)(8/10)](1/3) = 15/165 + 30/165 + 40/165 = 85/165 = 17/33

Still, that's more difficult than the approach sudhir took above, which is much better.

Hi Ian/others,

Is there a way to solve this problem by first starting out with the total number of ways 4 cards can be drawn as the denominator (i.e. 12C4) and then determine what we're looking for in the numerator?

Yes, although it's not the approach that I would recommend.

Total number of ways to choose a combination of 4 cards from 12 choices = 12C4 = 495.

To get at least 1 pair of the same value, we need either to combine 1 pair of the same value with 1 pair of different values or to combine 2 pairs, each of which is of the same value.

Total number of ways to choose exactly 1 pair of the same value:
Number of ways to choose a pair of the same value = 6

Number of ways to choose a pair of different values from the 10 remaining cards:
1st card = 10 choices
2nd card = 8 choices (9 cards left, but we can't choose the mate of the card just chosen, leaving us 9-1= 8 choices)
Order of the non-matching cards doesn't matter, so we divide by 2!: 10*8/2! = 40.

Multiplying, we have 6*40=240 ways to combine a pair of the same value with a pair of different values.

Total number of ways to choose 2 pairs, each of which is of the same value:
We have 6 pairs to choose from, and the order of the pairs doesn't matter, so 6C2 = 15.

So total number of ways to get at least 1 pair of the same value = 240+15 = 255.

P(at least 1 pair of the same value) = 255/495 = 51/99 = 17/33.