**Question 1: **For a hypothetical magazine, the probability that the reader is male
given that the reader is at least 35 years old is 0.30. The probability
that a reader is male, given that the reader is under 35, is 0.65. If
75% of the reader are under thirty five what is the probability that a
randomly chosen reader is

(a) male

(b) female

(c) under 35 if it is given the reader is a female

** Solution: **

Let us put the information given in the form of a table.

(a) Let A_{1} be the event of the reader being at least 35 years and A_{2} the event of the reader being under 35 years old

and B the event of the reader being a male.

P(A_{2}) = 0.75 and P(A_{1}) = 1 - P(A_{2}) = 1- 0.75 = 0.25

P(B | A_{1}) = 0.30 and P(B | A_{2}) = 0.65

B = (A_{1} ∩ B) U (A_{2} ∩ B) as the male reader can either under 35 or at least 35 years old.

P(B) = P(A_{1} ∩ B) + P(A_{2} ∩ B)

= P(A_{1}) . P(B | A_{1}) + P(A_{2}) . P(B | A_{2})

= (0.25)(0.30) + (0.75)(0.65) = 0.5625

The probability the reader is male = 0.5625

(b) P(female) = 1 - P(male) = 1 - 0.5625 = 0.4375

The probability the reader is female = 0.4375.

(c) Let E_{1} be the event of the reader is under 35 and E_{2} the event of the reader is at least 35 and the F the event of the

reader being a female.

We need to find P( E_{1} | F) and we have the probabilities

P(F | E_{1}) = 1 - 0.65 = 0.35 , P(F | E_{2}) = 1 - 0.30 = 0.70, P(E_{1}) = 0.75 and P(E_{2}) = 0.25

P( E_{1} | F) = $\frac{P(E_{1})P(F|E_{1})}{P(E_{1})P(F|E_{1})+P(E_{2})P(F|E_{2})}$

= $\frac{(0.75)(0.35)}{(0.75)(0.35)+(0.25)(0.70)}$

= 0.6

Probability the reader is under 35 given that the reader is a female = 0.6.

**Question 2: **The probability the event A will occur is 0.4 The conditional
probability of another event B occurring given A is 0.6 and the
conditional probability of B occurring given A has not occurred is 0.3.
Find the probability of

(a) the event A occurring given B

(b) the event of A not occurring given B

** Solution: **

Expressing the given probabilities using notation

P(A) = 0.4. Hence P(A^{c}) = 1 - 0.4 = 0.6

P(B | A) = 0.6 and P(B | A^{c}) = 0.3

(a) We need to compute P(A | B)

Using Baye's Theorem

P(A | B) = $\frac{P(A)P(B|A)}{P(A)P(B|A)+P(A^{c})P(B|A^{c})}$

= $\frac{(0.4)(0.6)}{(0.4)(0.6)+(0.6)(0.3)}$

= 0.571

Probability of event A occurring given B = 0.571

(b) To find P(A^{c} | B)

P(A^{c} | B) = 1 - P(A | B)

= 1 - 0.571 = 0.429

Probability of event A not occurring given B + 0.429.