213k views
1 vote
The lifetime of a device behaves according to the probability law for [(t, [infinity])]=1/t for t>1. let a be the event "lifetime is greater than 4," and b the event "lifetime is greater than 8.

Find the probability of A intersect B, and A union B.

1 Answer

4 votes

Final answer:

The probability of A intersect B, where A is the event 'lifetime greater than 4' and B is the event 'lifetime greater than 8' for a device, is 1/8. The probability of A union B is 1/4, which is the same as the probability of event A alone since A entirely includes B.

Step-by-step explanation:

Let's analyze the probability of events A (lifetime greater than 4) and B (lifetime greater than 8) for a device with lifetime probability P(t, ∞) = 1/t for t > 1. Since both events are about the lifetime exceeding certain values, they are not independent; instead, they are sequentially dependent. Event B is essentially a subset of event A, since any lifetime greater than 8 is also greater than 4. Thus P(A ∩ B) = P(B), as every occurrence of B is inherently an occurrence of A.

For event A (t > 4), the probability of occurrence is:

P(A) = 1/4

For event B (t > 8), the probability of occurrence is:

P(B) = 1/8

By definition of intersection of events:

P(A ∩ B) = P(B) = 1/8

The addition rule of probability tells us the probability of the union of A and B is the sum of the probabilities of A and B minus the probability of their intersection (which is just the probability of B in this case):

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

P(A ∪ B) = 1/4 + 1/8 − 1/8

P(A ∪ B) = 1/4

Therefore, the probability of A union B is 1/4, the same as the probability of just event A, since event A completely encompasses event B.

User Kartik Bhiwapurkar
by
9.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories