Question

The probabilities that John will choose Physics, Chemistry and Biology in the summer
semester are 0.4, 0.7 and 0.6, respectively. If John has decided to choose Physics, the
probabilities of choosing Chemistry and Biology are increased to 0.8 and 0.75,
respectively. On the other hand, if he has decided not to choose Biology, the probability
of choosing Physics only becomes 0.05. The probability that John will choose both
Chemistry and Biology but not Physics is 0.15.
(a) What is the probability that John will choose Physics only?
(b) What is the probability that John will choose all three courses?
(c) What is the probability that John will not choose any course in the summer
semester?

Answer 1

Define events:

John will choose Physics = $P$

John will choose Chemistry = $C$

John will choose Biology = $B$

Let probability of event A = $P(A)$, and $A^c$ is complement of set A.

Conditional probability

$P(B|P)=P(P\cap B)/P(P)=0.75,\\ \ P(P\cap B)=0.75\cdot0.4=0.3$

$P(P|B^c)=P(P\cap B^c)/P(B^c)=0.05, \\ P(P\cap B^c)=0.05\cdot(1-0.6)=0.05\cdot0.4=0.02$

But

$(P\cap B)\cup (P\cap B^c)=P\cap(B\cup B^c)=P\ and \ (P\cap B)\cap(P\cap B^c)=\empty$

Therefore $P(P)=P(P\cap B)+P(P\cap B^c)=0.3+0.02=0.32\ne 0.4$

and problem conditions contradict each other, hence problem has no solution.