Correct Answer - Option 3 :
\(\dfrac{1}{12}\)
Concept:
Let S be sample space. The probability of an event is P (E)
⇒ P (E) = \(\rm \dfrac {n(E)}{n(S)}\)
Calculations:
Given, S = a simultaneous throw of a pair of dice.
{(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) }
⇒ n(S ) = 36
E = event of getting a total more than 10
E = {(5, 6), (6, 5), (6, 6)}
⇒ n(E ) = 3
The probability of getting a total more than 7 is P (E)
⇒ P (E) = \(\rm \dfrac {n(E)}{n(S)}\)
⇒ P (E) = \(\rm \dfrac {3}{36}\)
⇒ P (E) = \(\dfrac{1}{12}\)
