A fair die is rolled six times. The probability of rolling more than 4 at least five times is
(a) 2/729 b) 3/729 c) 12/729 d) 13/729 (e) none of the above
Hello sri340The die shows more than 4 if it lands showing 5 or 6. We'll call this a 'success'. So the probability of a success on a single roll is $\displaystyle \frac26=\frac13$. The probability of a 'failure' on a single roll is $\displaystyle \frac23$.
We want the probability that when the die is rolled six times we get 5 sucesses, $\displaystyle p(5)$, or 6 successes, $\displaystyle p(6)$.
For $\displaystyle p(5)$ we want 5 successes and 1 failure. The failure can occur on any one of the 6 rolls. So $\displaystyle p(5) = 6 \times \left(\frac13\right)^5\times \left(\frac23\right) = \frac{12}{729}$.
For $\displaystyle p(6)$ we must have all 6 successes. So $\displaystyle p(6) = \left(\frac13\right)^6 = \frac{1}{729}$.
Therefore the probability that one or other of these events occurs is $\displaystyle \frac{12}{729}+\frac{1}{729}=\frac{13}{729}$
Grandad