Problem 1

1. The probability of a certain event is
2. The probability of an impossible event is
3. If an event cannot happen, the probability of the event will be
4. The probability of an event lies between the limits
5. If P(A) = 1, then A is called

Solution

1. 1
2. 0
3. 0 [an impossible event]
4. 0 and 1
5. A certain event

Problem 2

Which of the following are incorrect statements?

1. P(J) = 0.0
2. P(K) = 0.1
3. P(L) = 1.0
4. P(M) = 1.8
5. P(N) = − 0.2

Solution

1. TRUE

P(J) represents the probability of an impossible event.

2. TRUE

The probability of an event lies between 0 and 1.

3. TRUE

P(L) represents the probability of a certain event.

4. FALSE

0 ≤ P(EVENT) ≤ 1. The probability of an event should lie between 0 and 1.

5. FALSE

0 ≤ P(EVENT) ≤ 1. The probability of an event should lie between 0 and 1.

Problem 3

The probability of any event is always a proper fraction

Solution

A proper fraction is a fraction where the numerator is less than the denominator.
More specifically, a Proper fraction is a fraction of the form
 P Q
, where P, Q > 0 and
 P Q
< 1

Probability of occurrence of Event A

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(Event) =
 m n
Since m ≤ n always,
 m n
is always a proper fraction

Exceptions

The events which are either impossible or certain form an exception to this rule.
1. Probability of an impossible event

The probability of an impossible event is 0, which is possible only when the numerator is equal to zero, in which case the fraction is not an improper fraction.
2. Probability of a certain event

The probability of a certain event is 1, which is possible only when the numerator is equal to the denominator, in which case the fraction is not an improper fraction.

Problem 4

1. Events are said to be equally likely if
2. By exhaustive events we mean

1. Equally likely cases
2. Not equally likely cases
3. Containing all possible cases
4. None of these

Solution

1. Two or more events are said to be equally likely if their probability of occurrence is equal.
2. 3 - Exhaustive events mean events which cover all the possible cases within them.

If P(A) =
 1 4
then P(A) is

Solution

P(A) = 1 − P(A)
= 1 −
 1 4
=
 4 − 1 4
=
 3 4

Problem 6

If E is any event and E, its complement, then P(E) + P(E) =

Solution

 P(E) + P(E) = P(E) + [1 − P(E)] = P(E) + 1 − P(E) = 1

Problem 7

If P(A) = 0.05, P(B) = 0.15 then P(A) + P(B) is equal to ...

Solution

 P(A) + P(B) = [1 − P(A)] + [1 − P(B)] = [1 − 0.05] + [1 − 0.15] = 0.95 + 0.85 = 1.8

If P(A) =
 1 3
, P(A) is