Question

You roll a fair 6-sided die 8 times. What is the probability that you will roll at least four “4’s”?

Thanks in advance.

Answer 1

Answer on Question #42959, Math, Statistics and Probability

Conditions: You roll a fair 6-sided die 8 times. What is the probability that you will roll at least four "4's"? $P_{>=4}$-?.

Solution:

Probability to get n "4's":

$$P_n = \frac{1}{6} \sum_{8}^{n} C_n^n; \quad \text{Where} \quad C_n^n = \frac{8!}{(n!)(8-n)!};$$ $$\begin{aligned} P_{>=4} &= P_4 + P_5 + P_6 + P_7 + P_8 \\ &= \frac{1}{6} \sum_{8}^{4} C_n^4 + \frac{1}{6} \sum_{8}^{5} C_n^5 + \frac{1}{6} \sum_{8}^{6} C_n^6 + \frac{1}{6} \sum_{8}^{7} C_n^7 + \frac{1}{6} \sum_{8}^{8} C_n^{8} = \\ &\simeq 0.0618 \end{aligned}$$

Answer: $P_{>=4} = 0.0618$.

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