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Two dice are thrown together. The probability of getting a total of 3 or 8 is
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A) \( \Large \frac{24}{36} \)\ |
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B) \( \Large \frac{8}{36} \) |
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C) \( \Large \frac{11}{36} \) |
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D) \( \Large \frac{7}{36} \) |
Correct Answer:
| D) \( \Large \frac{7}{36} \) |
Description for Correct answer:
The events to get total of 3 is \( \Large E_{1}=\{ \left(1, 2\right) \left(2, 1\right) \} \) Therefore, \( \Large n \left(E_{1}\right)=2 \)
The events to get total of 8 is \( \Large E_{2}=\{ \left(2, 6\right) \left(3, 5\right) \left(4, 4\right) \left(5, 3\right) \left(6, 2\right) \} \)
\( \Large n \left(E_{2}\right) = 5 \)
\( \Large P \left(E_{1}\right)=\frac{n \left(E_{1}\right) }{n \left(S\right) }=\frac{2}{36} \)
\( \Large P \left(E_{2}\right)=\frac{n \left(E_{2}\right) }{n \left(S\right) }=\frac{5}{36} \)
\( \Large P \left(E_{1}\right)P \left(E_{2}\right) = \{ \} \)
\( \Large P \left(E_{1}E_{2}\right)=P \left(E_{1}\right)+P \left(E_{2}\right) \)
=\( \Large \frac{2}{36}+\frac{5}{36}=\frac{7}{36} \)
The events to get total of 3 is \( \Large E_{1}=\{ \left(1, 2\right) \left(2, 1\right) \} \) Therefore, \( \Large n \left(E_{1}\right)=2 \)
The events to get total of 8 is \( \Large E_{2}=\{ \left(2, 6\right) \left(3, 5\right) \left(4, 4\right) \left(5, 3\right) \left(6, 2\right) \} \)
\( \Large n \left(E_{2}\right) = 5 \)
\( \Large P \left(E_{1}\right)=\frac{n \left(E_{1}\right) }{n \left(S\right) }=\frac{2}{36} \)
\( \Large P \left(E_{2}\right)=\frac{n \left(E_{2}\right) }{n \left(S\right) }=\frac{5}{36} \)
\( \Large P \left(E_{1}\right)P \left(E_{2}\right) = \{ \} \)
\( \Large P \left(E_{1}E_{2}\right)=P \left(E_{1}\right)+P \left(E_{2}\right) \)
=\( \Large \frac{2}{36}+\frac{5}{36}=\frac{7}{36} \)
Part of solved Statistics questions and answers : >> Elementary Mathematics >> Statistics
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