The time between failures of a machine has an exponential distribution with parameter l. Suppose that the prior distribution for
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Answer:
Step-by-step explanation:
Hello!
Given the probabilities:
P(A₁)= 0.35
P(A₂)= 0.50
P(A₁∩A₂)= 0
P(BIA₁)= 0.20
P(BIA₂)= 0.05
a)
Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)
Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.
b)
Considering that you can clear the intersection from the formula
and apply it for the given events:
c)
The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:
P(B)= (A₁∩B) + P(A₂∩B)= 0.07 + 0.025= 0.095
d)
The Bayes' theorem states that:
Then:
I hope it helps!