In A Single Throw Of Two Dice , Find The Probability That Neither A D

1.	In A Single Throw Of Two Dice , Find The Probability That Neither A Doublet Nor A Total Of 8 Will Appear?
Answer» n(S) = 36 A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) } n(A)=6, n(B)=5, n(ANB)=1 Required probability = P(A?B) = P(A)+P(B)-P(AnB) = 6/36+5/36-1/36 = 5/18. n(S) = 36 A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) } n(A)=6, n(B)=5, n(AnB)=1 Required probability = P(A?B) = P(A)+P(B)-P(AnB) = 6/36+5/36-1/36 = 5/18.

In A Single Throw Of Two Dice , Find The Probability That Neither A Doublet Nor A Total Of 8 Will Appear?

Answer»

n(S) = 36

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

n(A)=6, n(B)=5, n(ANB)=1

Required probability = P(A?B)

= P(A)+P(B)-P(AnB)

= 6/36+5/36-1/36 = 5/18.

n(S) = 36

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

n(A)=6, n(B)=5, n(AnB)=1

Required probability = P(A?B)

= P(A)+P(B)-P(AnB)

= 6/36+5/36-1/36 = 5/18.