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aksik
14 days ago
5

Problem 5 (4+4+4=12) We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. 1) Fin

d the probability that doubles (i.e., having an equal number on the two dice) were rolled. 2) Given that the roll resulted in a sum of 4 or less, find the conditional probability that doubles were rolled. 3) Given that the two dice land on different numbers, find the conditional probability that at least one die is a 1.
Mathematics
1 answer:
tester [8.8K]14 days ago
3 0
1 2 3 Step-by-step explanation: Generally, during the roll of two fair 6-sided dice, the doubles result in (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). Therefore, the total for doubles is N = 6. The outcome of rolling two fair 6-sided dice yields n = 36. Thus, the probability of rolling doubles (matching numbers on both dice) is calculated mathematically. When rolling two fair dice, outcomes that sum to 4 or less are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1). Observing this, we see two doubles present. Consequently, the conditional probability of rolling doubles is represented mathematically. Lastly, when rolling the two fair dice, outcomes that show different numbers result in L = 30, while outcomes where at least one die shows a 1 give W = 10. Hence, the conditional probability of having at least one die show a 1 is presented mathematically.
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Response:

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