For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator
had an icemaker, and C be the event that the customer purchased an extended warranty. Relevant probabilities are:
$P(A)=.75$, $P(B| A)=.9$, $P(B|A')=.8$, $P(C|A\cap B)=.8$, $P(C|A\cap B')=.6$, $P(C|A'\cap B)=.7$, $P(C|A'\cap B')=.3$
a. Construct a tree diagram consisting of first-, second-, and third generation branches, and place an event label and appropriate probability next to each branch. Ignore this (a.) part, I am not looking for this answer. Take a look at rest of the questions.
b. Compute $P(A\cap B\cap C).$
c. Compute $P(B\cap C).$
d. Compute $P(C).$
e. Compute $P(A|B\cap C)$, the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased.
Here are the solutions:
I really do not understand why there are 6 given answers, when there are only 5 questions. I double checked everything. The number of problem is correct and indeed, these are the given solutions for it. If someone can explain this to me, please do.
My attempt for b.:
From $P(B|A)={P(B\cap A)\over P(A)}$ I conclude $P(B\cap A)=P(B|A)*P(A)=0.675.$
And then from $P(C|A\cap B)={P(A\cap B \cap C) \over P(A \cap B)}$ I conclude $P(A\cap B\cap C)=P(C|A\cap B) * P(A\cap B) = 0.54.$
But solution for this should be .05, as it is stated above. Please, show me my mistake.
Best Answer