Question

Roll a fair die three (3) times. a) What is the probability of rolling at least one 6 dotted side (ㅁ. ) in 3 rolls of a fair die? That is, calculate P (at least one □ in 3 rolls). Use the complement rule. (0.5pr ) b) Now reconsider the same question - what is the chance of rolling at least one 6 dotted side ([)) in 3 rolls of a fair die? But, for part b), do NOT use the complement rule. Instead, add the probability that you roll exactly 1 ixs to the probability that you roll exactly 2 sixes to the probability that you roll exactly 3 sixes. ( 1pr )

Answer 1

To calculate the probability of rolling at least one 6 dotted side in 3 rolls of a fair die, you can use the complement rule or consider the three cases of rolling one, two, or three sixes. The probability using the complement rule is 91/216, and the probability without using the complement rule can be found by adding the probabilities of each case.

Step-by-step explanation:

To calculate the probability of rolling at least one 6 dotted side in 3 rolls of a fair die, we can use the complement rule. The complement of rolling at least one 6 is rolling no 6's. In three rolls, there are 6 possible outcomes for each roll, so the total number of possible outcomes is 6^3 = 216. The number of outcomes where no 6's appear is 5^3 = 125. Therefore, the probability of rolling at least one 6 in 3 rolls is 1 - (125/216) = 91/216, which is approximately 0.4213.

To calculate the probability without using the complement rule, we need to consider the three cases: rolling exactly 1 six, rolling exactly 2 sixes, and rolling exactly 3 sixes. The probability of rolling exactly 1 six is (1/6) * (5/6)^2. The probability of rolling exactly 2 sixes is (1/6)^2 * (5/6). The probability of rolling exactly 3 sixes is (1/6)^3. Adding these probabilities together gives us the probability of rolling at least one 6 in 3 rolls.