Probability of drawing a 2 and a 3 in a two-dices shot.

[LVVL1000]

Hi,

I have two different answers for this probability problem. Please help me to choose the right solution.

Question: What is the probability of drawing a 2 and a 3 in a two-dices shot?

First approach:

Considering all possible combinations between the two dices you have 6 * 6 = 36 combinations. To get a 2 and a 3 involve two possible outcomes: a "2 and 3" and a "3 and a 2". Thus, the answer would be 2/36 = 1/18.

However, there is a second approach:

Considering all possible combinations without repeating outcomes. This is, we will consider a 2 a 3 just 1 time, not two times (it is not relevant which dice shows up the 2 and which dice shows up the 3). Under this non-duplicative scheme we would have 21 combinations (all duplicated outcomes were excluded). Thus, the probability of drawing a 2 and a 3 would be 1/21.

I just simply find both approaches correct. However, I suspect that GMAT tests apply some common sense rule in these situations. Which is the correct approach?

Best regards,

Luis R. Villegas H.
Mexico.

[nitin_prakash_khanna]

Approach 1 is the correct solution.
Not sure what do you mean by duplciates, a 2 and a 3 is a different outcome from a 3 and a 2 on Dice 1 and Dice 2 respectively.

How did you come to 21 outcomes?

[LVVL1000]

Hi, Nitin Prakash Khanna.

The duplicates are, for example, a 3-2 and a 2-3. When throwing the dices, you actually do not care which dice is Dice 1 or Dice 2; you just want to get a 2 and 3 regardless of which dice shows up each number. So, the 21 possible outcomes, without duplicates, would be:

1-1
1-2 ______ 2-2
1-3 ______ 2-3 ______ 3-3
1-4 ______ 2-4 ______ 3-4 ______ 4-4
1-5 ______ 2-5 ______ 3-5 ______ 4-5 ______ 5-5
1-6 ______ 2-6 ______ 3-6 ______ 4-6 ______ 5-6 ______ 6-6

As you can see, in this table the outcome 2-3 is present, but the outcome 3-2 is not present. Another example would be drawing a 4 and a 5. The outcome 4-5 is counted but the outcome 5-4 is not. The nature of this reasoning is that, as I said before, when you throw a pair of dices you do not actually know which dice is Dice 1 and which dice is Dice 2.

Best regards,

Luis R. Villegas H.
Mexico.

[RonPurewal]

hi -

your "second approach" is flawed.

However, there is a second approach:

Considering all possible combinations without repeating outcomes. This is, we will consider a 2 a 3 just 1 time, not two times (it is not relevant which dice shows up the 2 and which dice shows up the 3).

it is VERY relevant which die* shows 2 and which shows 3.

here's an analogy:
instead of rolling dice, imagine 2 friends writing down random integers from 1-6 on pieces of paper.
there is only ONE way you could get a pair of 3's: each friend writes a 3.
there are TWO ways you could get one 3 and one 2, since either friend could write the 3 and the other could write the 2.

therefore, the probability of one 3 and one 2 is twice the probability of a pair of 3's.

same with the two dice.

--

*note that "dice" is a plural. the singular is "die".

--

what troubled me about your post, though, was the following:

I just simply find both approaches correct.

that's impossible, since the same event can't have two different probabilities.

you should have a more critical mindset toward these sorts of things. when you are in the market to find THE probability of AN event, there can only be one such probability!
...so if you encounter two conflicting accounts, then one of them must be wrong. period.

[LVVL1000]

Thank you, ManhattanGMAT Staff.

[Ben Ku]

I'm glad it helped!