ProbabilityHard
Question
In a throw of a pair of dice, the probability of ′A total of 8 but not 11′.
Options
A.5/36
B.1/26
C.2/36
D.1/9
Solution
A → sum is 8 , B → sum is 11
If A occurs naturally B is not allowed so ′A total of 8 but not 11′ is equivalent to sum of ′8′ is obtained now n(S) = 6 × 6
n(E) = {(2, 6), (6, 2), (4, 4), (3, 5), (5, 3)}
⇒ P = 5/36
If A occurs naturally B is not allowed so ′A total of 8 but not 11′ is equivalent to sum of ′8′ is obtained now n(S) = 6 × 6
n(E) = {(2, 6), (6, 2), (4, 4), (3, 5), (5, 3)}
⇒ P = 5/36
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