Pls help to solve this probability problem.. Thanks...?

Question

Consider families with two children, assuming equal probability for both genders. Let A be the event that the chosen family has at most one boy, B be the event that the chosen family has a boy and a girl and C be the event that either the chosen family has two boys or two girls.

a) Is P(AB) = P(A)P(B)? Are A and B independent? Are they mutually exclusive?

b) Is P(BC) = P(B)(C)? Are B and C independent? Are they mutually exclusive?

However, prove that if A and B are independent events and P(A)>0 and P(B)>0, then A and must be not mutually exclusive. (A very simple proof consists of one line only)

Answer 1

try...OK
The families can be classed into: bb, bg, gb, gg where b is boy and g is girl. The probability of each occurring is ½*½=¼

Event A consists of {bg, gb, gg}. Thus P(A)=¾.
Event B consists of {bg, gb}. Thus P(B)=½.
Event C consists of {bb, gg}. Thus P(C)=½.

Event AB means the common in A & B. Event BC means the common in B & C.

To answer:
(a) The equation P(AB) = P(A)P(B) is not true. P(AB)=½.
Then A & B are not independent.
A and B are not mutually exclusive since they have some common elements. Particularly, AB=B.

(b)The equation P(BC) = P(B)P(C) is not true. P(BC)=0.
Then B & C are not independent.
B and C are mutually exclusive since BC is empty.

Last proof: If A & B are independent, then
P(AB)=P(A)P(B)>0. Thus AB is not empty.

Answer 2

If C and D are mutually exclusive events, then p(C or D) = p(C) + p(D).

Two events C and D are independent if and only if p(C and D) = p(C)p(D).

The experiment is to randomly choose a family that has two children.

The sample space for this experiment has 4 members: {BB, BG, GB, GG}

There are 3 ways event A can occur: BG, GB, GG

There are 2 ways event B can occur: BG, GB

There are 2 ways event C can occur: BB, GG

probability of an event = number of ways event occurs / number of members in sample space

Part A

A and B = {BG, GB}
p(A and B) = 2 / 4 = 1 / 2
p(A) = 3 / 4
p(B) = 2 / 4 = 1 / 2

p(A and B) = p(A)p(B)?
1 / 2 = (3 / 4)(1 / 2) is false.
So, A and B are NOT independent.

A or B = {BG, GB, GG}
p(A or B) = 3 / 4
p(A or B) = p(A) + p(B)?
3/ 4 = 3 / 4 + 1 / 2 is false.
So, A and B are not mutually exclusive.

Part B

B and C = { }, the empty set
p(B and C) = 0 / 4 = 0
p(B) = 2 / 4 = 1 / 2
p(C) = 2 / 4 = 1 / 2

p(B and C) = p(B)p(C)?
0 = (1 / 2)(1 / 2) is false.
So, B and C are not independent.

B or C = {BB, BG, GB, GG}
p{B or C} = 4/4 = 1

p(B or C) = p(B) + p(C)?
1 = 1/2 + 1/2 is true.
So, B and C are mutually exclusive.

If p(A and B) = p(A)p(B) and p(A) > 0 and p(B) > 0, then p(A and B) is not equal to zero which A and B can occur together.

Answer 3

Possibilities are:
bb
bg
gb
gg
a) P(AB) = P(B), A contains B
b) P(BC) = 1, B & C are mutually exclusive
A and B cannot be independent. In order for B to occur A must occur.