Question

Discuss a probability model by 1. stating a sample space and assigning probabilities to outcomes in the space. 2. Is the model discrete or continuous? 3. Can you verify that it satisfies the rules of probability? 4. Can you identify any events that are independent, or any that are dependent? 5. Can you identify any events whose probability would require the inclusion-exclusion principle to calculate? 6. Can you identify any events whose probability would require conditional probability to calculate? Find these probabilities if possible.

Answer 1

Let us consider the random experiment of tossing a fair coin twice.

1. The sample space is given by

= {HH, HT, TH,TT} ,    H stands for occurrence of Head and T stands for Tail.

= {₁,₂,_3,4}, say where _i^{' s} are sample points.

P[{_i}] = 1/4 , i= 1(1)4

2. The probability model assumed here is discrete as it is concerned with the number of occurrences of Head and Tail which are isolated numbers.

3. Let us define three events as follows.

A₁= Event of getting of two Heads.

A₂= Event of getting one Head and one Tail.

A₃ = Event of getting two Tails.

P[A₁] = P[{HH}] = 1/4 , P[A₂] = P[{HT,TH}] =1/2 , P[A₃] = P[{TT}] = 1/4

Rule1: P[A] 0, where A is an event.

Clearly, P[A_i] 0 , i= 1,2,3.

Non-negativity property is satisfied.

Rule 2:  P[]=1

P[] = P[{HH,HT,TH,TT}] = 1   

Normed property is satisfied.

Rule 3: P[ A_i] = P[A_i] for disjoint A_i^'s i.e. A_i A_j = i j .

Note that, the events A₁ , A₂ and A₃ are disjoint.

P[A₁ A₂ A₃] = P[A₁] + P[A₂] + P[A₃]  

Finite additivity property is satisfied

Rule 4 : P[] = 1 - P[A]

Note that, P[₁] = 1 - P[A₁] = 3/4

P[₂] = 1- P[A₂] = 1/2

P[₃] =  1- P[A₃] = 3/4

Complementation property is satisfied.

4. Two events A and B are statistically independent if P[AB]=P[A}P[B]

Here, P[A₁A₂] = 0 P[A₁]P[A₂] = 1/2 * 1/4 = 1/8

So the events A₁ and A₂ are not independent .

P[A₁A₃] = 0   P[A₁] P[A₃] = 1/4

So the events A₁ and A₃ are not independent.

Similarly, for the events A₂ and A₃ , P[A₃A₂] = 0 P[A₃]P[A₂] = 1/2 * 1/4 = 1/8

So the events A₂ and A₃  are dependent.