A dice is rolled three times and sum of three numbers appearing on the uppermost face is 15. What is the chance that the first roll was four?
Correct Answer: Option C
As per the condition given in the question:
All the combinations of outcomes for getting a sum of 15 on the uppermost face are:
⇒ (4, 5, 6), (5, 4, 6), (6, 5, 4), (5, 6, 4), (4, 6, 5), (6, 4, 5), (5, 5, 5), (6, 6, 3), (6, 3, 6), (3, 6, 6)
⇒ n(S) = 10
Now, outcomes on which first roll was four are n(E) = (4, 5, 6), (4, 6, 5)
Therefore required probability P(E):
⇒ P(E) = n(E)/n(S)
= 2/10
= 1/5
Hence, the required probability is 1/5.
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