Question

two dice are rolled, find the probability of - Event A: the sum of digit on upper face is at least 9​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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

i. Let A be the event that the sum of the digits on the upper faces is at least 10.

∴ A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}

∴ n(A) = 6

∴ P(A) =

n

(

A

)

n

(

S

)

n(A)n(S) = 6/36

∴ P(A) = 1/6

ii. Let B be the event that the sum of the digits on the upper faces is 33.

The sum of the digits on the upper faces can be maximum 12.

∴ Event B is an impossible event.

∴ B = { }

∴ n(B) = 0

∴ P(B) =

n

(

B

)

n

(

S

)

n(B)n(S)= 0/36

∴ P(B) = 0

iii. Let C be the event that the digit on the first die is greater than the digit on the second die.

C = {(2, 1), (3, 1), (3,2), (4,1), (4,2), (4, 3), (5, 1), (5,2), (5,3), (5,4), (6,1), (6,2), (6, 3), (6, 4), (6, 5),

∴ n(C) = 15

∴ P(C) =

n

(

C

)

n

(

S

)

n(C)n(S)= 15/36

∴ P(C) = 5/12

∴ P(A) = 1/6 ; P(B) = 0; P(C) = 5/12