A pair of dice is rolled.
So
Sample space
S = {(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
1) Let A be the event of sum not more than 9
A = {(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)
(5,1) (5,2) (5,3) (5,4)
(6,1) (6,2) (6,3) }
n(A) = 30
P(A) = n(A) /n(S) = 30/36 = 1/6
2)
Let B be the event of sum not more than 6
B = {(1,1) (1,2) (1,3) (1,4) (1,5)
(2,1) (2,2) (2,3) (2,4)
(3,1) (3,2) (3,3)
(4,1) (4,2)
(5,1)}
n(B) = 15
P(B) = n(B) /n(S) = 15/36 = 5/12
3)
Let C be the event of sum between 6 and 10
C = {(1,6)
(2,5) (2,6)
(3,4) (3,5) (3,6)
(4,3) (4,4) (4,5)
(5,2) (5,3) (5,4)
(6,1) (6,2) (6,3) }
n(C) = 15
P(C) = n(C) /n(S) = 15/36 = 5/12
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SOLUTION :
Pair of dice rolled.
Only part(a) is required to be answered.
P(sum is not greater than 9)
= P( sum ≤ 9)
= 1 - P(sum > 9)
Total space = 6 * 6 = 36 events
Events, (sum > 9 ) are :
(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6,6) = 6 events
So, P( sum > 9) = 6/36 = 1/6
Hence, P (sum ≤ 9) = 1 - 1/6 = 5/6 (ANSWER).
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