Answer to Question #99516 in Statistics and Probability for Sromona Roy

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 Chemistry becomes 0.6. If John has decided to choose Biology, the probability of choosing Chemistry but not Physics becomes 0.35.
(a) What is the probability that John will choose all three courses?
(b) What is the probability that John will not choose any course in the summer semester?
(c) If John has decided to choose Chemistry, what is the probability that John will choose exactly two courses in the summer semester?

Answer 1

Let $P=$ the event ''John decides to choose Physics'', $C=$ the event ''John decides to choose Chemistry'', $B=$ the event ''John decides to choose Biology.''

Given that

$$P(P)=0.4,P(C)=0.7,P(B)=0.6$$

$$P(C|P)=0.8, P(B|P)=0.75$$

$$P(C|B^C)=0.6, P(C\cap P^C|B)=0.35$$

Conditional probability

If John has decided to choose Physics, the probability of choosing Biology

$$P(B|P)={P(B\cap P) \over P(P)}=>P(B\cap P)=P(B| P)P(P)$$ $$P(B\cap P)=0.75\cdot0.4=0.3$$

If John has decided to choose Physics, the probability of choosing Chemistry

$$P(C|P)={P(C\cap P) \over P(P)}=>P(C\cap P)=P(C| P)P(P)$$ $$P(C\cap P)=0.8\cdot0.4=0.32$$

If John has decided not to choose Biology, the probability of choosing Chemistry 

$$P(C|B^C)={P(C\cap B^C) \over P(B^C)}=>P(C\cap B^C)=P(C| B^C)(1-P(B))$$ $$P(C\cap B^C)=0.6(1-0.6)=0.24$$

If John has decided to choose Biology, the probability of choosing Chemistry but not Physics 

$$P(C\cap P^C|B)={P(C\cap P^C\cap B) \over P(B)}=>$$

$$=>P(C\cap P^C\cap B)=P(C\cap P^C|B)P(B)$$

$$P(C\cap P^C\cap B)=0.35\cdot0.6=0.21$$

(a) What is the probability that John will choose all three courses? 

$$P(P\cap C\cap B)=P(C)-P(C\cap B^C)-P(C\cap P^C\cap B)$$ $$P(P\cap C\cap B)=0.7-0.24-0.21=0.25$$

(b) What is the probability that John will not choose any course in the summer semester? 

$$P(P\cap C\cap B)=0.25$$

$$P(P\cap C^C\cap B)=P(P\cap B)-P(P\cap C\cap B)=$$ $$=0.3-0.25=0.05$$

$$P(P\cap C\cap B^C)=P(P\cap C)-P(P\cap C\cap B)=$$ $$=0.32-0.25=0.07$$

$$P(C\cap P^C\cap B)=0.21$$

$$P(P \cup B\cup C)=P(A)+P(B)+P(C)-$$ $$-P(P\cap C^C\cap B)-P(P\cap C\cap B^C)-$$ $$-P(C\cap P^C\cap B)-2P(P\cap C\cap B)$$

$$P(P \cup B\cup C)=0.4+0.6+0.7-$$ $$-0.05-0.07-0.21-2\cdot0.25=0.87$$

$$P(no \ course)=1-P(P\cup B \cup C)=$$ $$=1-0.87=0.13$$

The probability that John will not choose any course in the summer semester is $0.13$

(c) If John has decided to choose Chemistry, what is the probability that John will choose exactly two courses in the summer semester?

$$P(exactly\ 2|C)={P(P\cap C\cap B^C)+P(C\cap P^C\cap B) \over P(C)}=$$

$$={0.07+0.21\over 0.7}=0.4$$