Question

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.

Answer 1

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.