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?
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 $10$.