ExplanationA There are a total of 62 = 36 possibilities for each toss. There are a total of 8 ways we can get a total of 7 or 11 on the first toss: 6 ways to get a total of 7—(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), or (6, 1)—plus 2 ways to get a total of 11—(5, 6) or (6, 5).

Therefore, the probability of getting a total of either 7 or 11 on the first toss is \(\frac{8}{36}=\frac{2}{9}\).

The probability of getting a total of 7 on the second toss is \(\frac{6}{36}=\frac{1}{6}\) so the probability that both of these independent events occur is the product \(\frac{2}{9}\times \frac{1}{6}=\frac{1}{27}\),

choice A.
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Sandy

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