Question

(B) Two events A and B in the sample space of a random
experiment are mutually exclusive. If 3 P (A)=4 P(B) = 1
then find P(AUB).​

Answer 1

Answer:

7/12

Explanation:

If 2 events are mutually exclusive then it means they cannot occur together, like head and tail of a coin. And their intersection is zero i.e. P(A∩B)=0

In this case, we can write

P(AUB) = P(A) + P(B)

given: 3P(A) =1 and 4P(B)=1

therefore we can say, P(A) = 1/3 and P(B) = 1/4

Finally, P(AUB)= P(A)+P(B) = 1/3 + 1/4 =7/12

Hope this helps.