A die is thrown three times, and sum of three numbers thrown is 15. Find the chance that the first throw was a four.
- $\displaystyle \dfrac{1}{2}.$
- $\displaystyle \dfrac{1}{5}.$
- $\displaystyle \dfrac{1}{15}.$
- $\displaystyle \dfrac{1}{13}.$
Total outcomes for sum 15 with 3 dice: (6,6,3)x3, (6,5,4)x6, (5,5,5)x1 = 10 outcomes. Outcomes where first die is 4: (4,5,6), (4,6,5) = 2 outcomes. Probability = 2/10 = 1/5.
Using conditional probability, we find the total outcomes where the sum of three dice is 15, which are (4,5,6), (4,6,5), (5,4,6), (5,5,5), (5,6,4), (6,4,5) and (6,5,4). Out of these 7 outcomes, the first throw is a 4 in only the first two cases. However, since the options suggest 1/5, we can count the permutations of combinations (3,6,6), (4,5,6) and (5,5,5) which are 3, 6 and 1 respectively, yielding 10 outcomes. Re-evaluating the set of all 10 outcomes for a sum of 15, the outcomes starting with a 4 are (4,5,6) and (4,6,5). The required probability is therefore 2 out of 10, which simplifies to 1/5.