question_answer

In a single throw of a pair of different dice, what is the probability of getting

(i) a prime number on each dice?

(ii) a total of 9 or 11?

Answer:

Total outcomes = {(1, 1), (1, 2) (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

No. of outcomes \[=36\]

(i) Let \[{{E}_{1}}\] be the event of getting a prime number on each dice.

Favourable outcomes = {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}

\[\Rightarrow \] No. of favourable outcomes \[=9\]

\[\therefore P({{E}_{1}})=\frac{9}{36}=\frac{1}{4}\]

(ii) Let \[{{E}_{2}}\] be the event of getting a total of 9 or 11.

Favourable outcomes = {(3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5)}

\[\Rightarrow \]No. of favourable outcomes \[=6\]

\[\therefore P({{E}_{2}})=\frac{6}{36}=\frac{1}{6}\]