<span>a) How long will it take to roll "snake-eyes" (a pair of ones)? P(a pair of ones) = 1/36 Therefore, it takes 36 rolls to roll "snake-eyes".
b) What is the probability of rolling a sum of 7 on the first roll? P(a sum of 7) = 6/36 = 1/6 P(a sum of 7 on the first roll) = 1/6
c) What is the probability of rolling a sum of 7 on the fourth roll? P(a sum of 7 on the fourth roll) = </span><span><span>(5/6)^3 * 1/6 = 125/1296</span>
d) What is the probability of rolling a sum of 7 by the fourth roll? P(a sum of 7 by the fourth roll) = 1/6 + (5/6) * 1/6 + (5/6)^2 * 1/6 + (5/6)^3 * 1/6 = 1/6 + 5/36 + 25/216 + 125/1296 = 671/1296
e) What is the probability of it takin</span>g more than 10 rolls to roll the sum of 7? P(more than 10 rolls to roll the sum of 7) = 1 - P(no sum of 7 in the first 10 rolls) = 1 - (5/6)^10 = 1 - 0.1615 = 0.8385
