In a single throw of two dice, what is the probability of getting a to

1.	In a single throw of two dice, what is the probability of getting a total of 11.(a) \(\frac{1}{9}\)(b) \(\frac{1}{18}\)(c) \(\frac{1}{12}\)(d) \(\frac{35}{36}\)
Answer» (b) \(\frac{1}{18}\) Total number of exhaustive cases = 6 × 6 = 36 A total of 11 may be obtained in 2 ways as (5, 6), (6, 5). \(\therefore\) P (total of 11) = \(\frac{2}{36}\)=\(\frac{1}{18}\).

In a single throw of two dice, what is the probability of getting a total of 11.(a) \(\frac{1}{9}\)(b) \(\frac{1}{18}\)(c) \(\frac{1}{12}\)(d) \(\frac{35}{36}\)

Answer»

(b) \(\frac{1}{18}\)

Total number of exhaustive cases = 6 × 6 = 36 A total of 11 may be obtained in 2 ways as (5, 6), (6, 5).

\(\therefore\) P (total of 11) = \(\frac{2}{36}\)=\(\frac{1}{18}\).

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In a single throw of two dice, what is the probability of getting a total of 11.(a) \(\frac{1}{9}\)(b) \(\frac{1}{18}\)(c) \(\frac{1}{12}\)(d) \(\frac{35}{36}\)

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