10. A 6 -sided die is blased in such a way that the probability of a "six" appearing on top is \( 20 \% \) and all other possibilities have an equal chance of appearing on top. If the die is thrown 5 times, what is the probability that a "six" will appear at most 3 times? 0.0026 0.0080

To find the probability of rolling a "six" at most 3 times when throwing a biased die 5 times, we can use the binomial probability formula. Given that the probability of rolling a six \( p = 0.20 \) and the number of trials \( n = 5 \), we can calculate the probabilities for getting 0, 1, 2, and 3 sixes and sum them up. For each \( k \) (number of sixes), the probability is calculated as: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Calculating for \( k = 0, 1, 2, 3 \): 1. For \( k = 0 \): \[ P(X=0) = \binom{5}{0} (0.20)^0 (0.80)^5 \approx 0.32768 \] 2. For \( k = 1 \): \[ P(X=1) = \binom{5}{1} (0.20)^1 (0.80)^4 \approx 0.40960 \] 3. For \( k = 2 \): \[ P(X=2) = \binom{5}{2} (0.20)^2 (0.80)^3 \approx 0.20480 \] 4. For \( k = 3 \): \[ P(X=3) = \binom{5}{3} (0.20)^3 (0.80)^2 \approx 0.05120 \] Now, summing all the probabilities: \[ P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \approx 0.32768 + 0.40960 + 0.20480 + 0.05120 \approx 0.99328 \] Thus, the probability that a "six" will appear at most 3 times in 5 rolls of this die is approximately \( 0.9933 \). So the answer to the problem is neither of the options provided (0.0026 or 0.0080). 0.9933