12 A six-sided die is biased. When the die is thrown the number 5 is twice as likely to appear as any other number. All the other faces are equally likely to appear. The die is thrown repeatedly. Find the probability that: a the first 5 will occur on the sixth throw b in the first eight throws there will be exactly three 5 s .

Let’s define the probabilities first: let p be the probability of rolling any number other than 5. Since the number 5 is twice as likely to appear, we have \( P(5) = 2p \) and the total probability must equal 1 for all sides of the die. The six faces of the die give: \[ P(5) + 5P(\text{other}) = 1 \] \[ 2p + 5p = 1 \] \[ 7p = 1 \] Thus, \( p = \frac{1}{7} \) and \( P(5) = 2p = \frac{2}{7} \). **a. Probability the first 5 will occur on the sixth throw:** This follows a geometric distribution where the probability of failure (not rolling a 5) is \( \frac{5}{7} \). The probability of the first 5 occurring on the sixth throw is: \[ P(\text{1st 5 on 6th}) = \left( \frac{5}{7} \right)^5 \times \left( \frac{2}{7} \right) \] Calculating it: \[ P(\text{1st 5 on 6th}) = \left( \frac{5}{7} \right)^5 \times \left( \frac{2}{7} \right) = \frac{3125}{16807} \times \frac{2}{7} = \frac{6250}{117649} \approx 0.0532 \] **b. Probability of exactly three 5’s in the first eight throws:** This follows a binomial distribution where \( n = 8 \) and \( k = 3 \) with the probability of success being \( \frac{2}{7} \). The formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where \( n = 8 \), \( k = 3 \), \( p = \frac{2}{7} \), and \( 1-p = \frac{5}{7} \). Calculating it: \[ P(X = 3) = \binom{8}{3} \left( \frac{2}{7} \right)^3 \left( \frac{5}{7} \right)^{5} \] \[ = 56 \times \frac{8}{343} \times \frac{3125}{16807} \] The binomial coefficient \( \binom{8}{3} = 56 \), Thus: \[ = 56 \times \frac{8 \times 3125}{343 \times 16807} = 56 \times \frac{25000}{574703} \approx 0.1907 \] So, the answers are: a. \( P(\text{1st 5 on 6th}) = \frac{6250}{117649} \) b. \( P(\text{exactly 3 fives in 8 throws}) \approx 0.1907 \)