Flag Algebra> Probablity and statistics...
question mark

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?

Burhan Tayyab , 15 Years ago
Grade
anser 1 Answers
Jitender Singh
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.
Last Activity: 11 Years ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments