Page 102 - DMTH404_STATISTICS

Statistics

Notes Solution.

The sample space of the experiment is S = {(H, H), (H, T), (T, H), (T, T)}
(i) A = {(T, H), (T, T)} and A = {(H, H), (T, H)}
1 2
1 b
A
Also A  g = {(T, H)}, Since A 1  A 2   , A and A are not mutually exclusive.
2
1
2
Further, the coins are given to be unbiased, therefore, all the elementary events are equally
likely.
2 1 2 1 P A b 1
 P A b g   , P A b g   , 1  A g 
2
2
1
4 2 4 2 4
1 b 1 1 1 3
Thus, P A  g  +   .
A
2
2 2 4 4
(ii) When both the coins show heads; A = {(H, H)}
1
When both the coins show tails; A = {(T, T)}
2
Here A  A  ,  A and A are mutually exclusive.
1
2
2
1
1 b 1 1 1
Thus, P A  g  +  .
A
2
4 4 2
Alternatively, the problem can also be attempted by making the following nine-square
tables for the two cases :
ii
i
( ) A 2 A 2 Total ( ) A 2 A 2 Total
A 1 1 1 2 0 1 1
A 1 1 1 2 1 2 3
Total 2 2 4 1 3 4
Theorem 5. Multiplication or Compound Probability Theorem
A compound event is the result of the simultaneous occurrence of two or more events. For
convenience, we assume that there are two events, however, the results can be easily generalised.
The probability of the compound event would depend upon whether the events are independent
or not. Thus, we shall discuss two theorems; (a) Conditional Probability Theorem, and (b)
Multiplicative Theorem for Independent Events.
(a) Conditional Probability Theorem
For any two events A and B in a sample space S, the probability of their simultaneous occurrence,
is given by
 b P A P B Ag

P A Bg b g b /
P B P A Bg
or equivalently b g b /

Here, P(B/A) is the conditional probability of B given that A has already occurred. Similar
interpretation can be given to the term P(A/B).

Proof.
Let all the outcomes of the random experiment be equally likely. Therefore,

( n A  ) B no. of elements in (A  B )
P (A  ) B  
n ( ) S no. of elements in sample space

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