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A die is tossed twice. A is the event 'the sumof the two numbers uppermost is greater than 5', B the event 'the sum of the two numbers uppermost is less than 8'. Find the probability of these events: (a) A (b) B (c) AB (d) A U B Hence show that P(A U B)=P(A)+P(B)-P(A ^ B). Are A and B mutually exclusive?

A die is tossed twice. A is the event 'the sumof the two numbers uppermost is greater than 5', B the event 'the sum of the two numbers uppermost is less than 8'. Find the probability of these events:


(a) A


(b) B


(c) AB


(d) A U B


Hence show that P(A U B)=P(A)+P(B)-P(A ^  B).


Are A and B mutually exclusive?

Grade:

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello Student,
Please find answer to your question below
Let x & y be the random number occuring on die 1st & 2nd time respectively.
1\leq x\leq 6
1\leq y\leq 6
Then for the 1stevent
x + y > 5
Number of possiblity
(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (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)
Total possibility = 36
P(A) = \frac{26}{36} = \frac{13}{18}
For the 2ndevent:
x + y > 8

(3,6), (4,5), (4,6), (5,4),(5,5), (5,6), (6,3),(6,4), (6,5), (6,6)
P(B) = \frac{10}{36} = \frac{5}{18}

No they are not mutually exclusive. As you can see there is an intersection b/w events of A & B.

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