Thus, the probability that exactly two of the four rolls result in a number greater than 7 is:
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- P(k) = \binom{n}{k} p^k (1 - p)^{n - k}
- Substitute the values:
- P(2) = \binom{4}{2} (0.3)^2 (0.7)^2 = 6 \cdot (0.09) \cdot (0.49) = 6 \cdot 0.0441 = 0.2646
- \boxed{0.2646}
- Question: A science administrator is reviewing grant applications and notices that each letter of a 7-character activity code is either 'E', 'X', or 'Y'. How many such codes contain exactly three 'E's and at least one pair of consecutive identical letters?
- We are to count the number of 7-character strings using only the letters 'E', 'X', 'Y' such that:
