In a simultaneous throw of a pair of dice, find the probability of getting atleast one 6 IF THE TWO FACES ARE DIFFERENT.

Question

I'm so confused about this. I can find the probability of getting atleast one 6 but what exactly am I supposed to do in case of this question? It clearly mentions that find the probability of getting atleast one 6 IF THE TWO FACES ARE DIFFERENT. But I have no clue what that really means.

Answer 1

I think the question is not well defined. If you exclude same faces you also exclude (6,6) so you don't have "at least one six" but exactly one six. I give answers for both cases. First we consider the case where (6,6) is allowed. You have 36 possible outcomes. Take out the cases where the faces are the same, without a 6. That is 5 cases. So we are left with 31 cases. And the outcomes which have at least one six are, (6,6) and ten more with one six. So we get 11 outcomes. And the probability is 11/31.

Now assume that (6,6) is not allowed. Then we are left with 30 outcomes. And the outcomes with one six are 10, so the answer in this case is 10/30.

Answer 2

Hint:

You can rephrase the question as:

"If 2 distinct faces are selected randomly from a die then what is the probability that one of them carries number 6?"

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Edit:

The selection takes place by throwing two dice simultaneously and eventually more than once. The first throw that provides two distinct outcomes provides the selection.

Answer 3

You can answer many questions about roling two dice by having the following picture in your head (or sketching it on paper if necessary)

If the two faces need to be different then you can ignore the main diagonal - leaving 30 combinations rather than the usual 36. 10 of these have a 6 so your probability is 10 out of 30, or one third.