Multiple choice

A die is thrown three times and the sum of three numbers obtained is $15.$ The probability of first throw being four is

  1. $\displaystyle\frac{1}{18}$
  2. $\displaystyle\frac{1}{5}$
  3. $\displaystyle\frac{4}{5}$
  4. $\displaystyle\frac{17}{18}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 throw is 4: (4,5,6), (4,6,5) = 2 outcomes. Probability = 2/10 = 1/5.

AI explanation

When a die is thrown three times, the total number of possible outcomes is 6 * 6 * 6 = 216. To find the number of outcomes where the sum is 15 and the first throw is a 4, the remaining two dice must sum to 11, which can be achieved by the combinations (5, 6) and (6, 5), yielding 2 favorable outcomes. However, examining the actual sample space of three dice summing to 15 reveals there are exactly 10 combinations, not 36 as calculated from the erroneous 216 total. Using the correct total of 10 favorable outcomes for the condition, the conditional probability is 2/10, which simplifies to 1/5.

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