# Finding probability that 2 has appeared atleast once given sum is 10 and die is thrown thrice?

An unbiased die is thrown three times; the sum of numbers coming up is 10. The probability that two has appeared at least once is:

A 1/36
B 5/36
C 91/216
D 1/18 ?

I was able to find out of 216 possible outcomes of throwing dice thrice only 27 of them give sum ten. Then I counted number of triads where no 2 was there which were 17 and hence the required probability should be 10/27? Where am I making a mistake?

• If you list the 27 possibilities, you'll see how many times out of that 27 you have a $2$ showing. – Joffan Jun 30 '16 at 19:17
• hint: the number of twos can only be 2 or 1 (since 3*2=6<10 and 0*2 is prohibited) if its 2, how many possibilities has the third dice? if its 1, how many possibilities have the remaining 2 dice? now just add up – SAJW Jun 30 '16 at 19:19
• As you see from John's answer, the possible number groups occur in sets of three or six (depending on whether two of the values are equal or not) - so your mistake looks like a counting error. The fact that none of the multiple choice answers look right might indicate that you have the question slightly wrong. – Joffan Jun 30 '16 at 19:35
• Is it correct that 27 is no of outcomes where sum is 10 ? – Matt Jun 30 '16 at 19:45
• Yes, $27$ is correct. I ran an Excel sheet to check all $216$ possibilities. But now I'm wondering why my answer isn't among the choices. – John Jun 30 '16 at 19:57

From the $27$ possibilities it's just a matter of counting the ones with at least one $2$.
$$[631] \times 6, [622] \times 3, [541] \times 6, [532] \times 6, [442] \times 3,[433] \times 3$$
So ... $(3+6+3)/27 = 4/9$.
One possibility for another answer (though this doesn't really seem to be what the question is asking) is the probability of the sum being $10$ and having at least one $2$. Then this would be $12/216 = 1/18$, which is choice (D). But, again, the wording of the question really appears to say that we know the sum is $10$.