The probability of throwing either 7 or 11 with two dice is
Correct Answer: |
B) \( \Large \frac{2}{9} \) |
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Description for Correct answer:
\( \Large n \left(S\right) \) = 36
Events to get 7 = \( \Large \{ \left(1, 6\right) \left(2, 5\right) \left(3, 4\right) \left(4, 3\right) \left(5, 2\right) \left(6, 1\right) \} \)
\( \Large n \left(E_{1}\right) = 6 \)
\( \Large p \left(E_{1}\right) = \frac{n \left(E_{1}\right) }{n \left(S\right) }=\frac{6}{36} \)
Events to get 11 = \( \Large \{ \left(5, 6\right), \left(6, 5\right) \} \)
\( \Large n \left(E_{2}\right)=\frac{n \left(E_{2}\right) }{n \left(S\right) }=\frac{2}{36} \)
\( \Large P \left(E_{1}E_{2}\right) = P \left(E_{1}\right)+P \left(E_{2}\right) \)
= \( \Large \frac{6}{36}+\frac{2}{36}=\frac{8}{36}=\frac{2}{9} \)
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