Classical/Mathematical Definition of Probability : Basic Concepts : Problems Solutions

No. Problems & Solutions
01.
(a) The probability of a certain event is

(b) The probability of an impossible event is

(c) If an event cannot happen, the probability of the event will be

(d) The probability of an event lies between the limits

(e) If P(A) = 1, then A is called

Solution  
 
(a) 1 ;

(b) 0 ;

(c) 0 [an impossible event] ;

(d) 0 and 1

(e) A certain event

No. Problems & Solutions
02.

Which of the following are incorrect statements?

(a) P (J) = 0.0

(b) P (K) = 0.1

(c) P (L) = 1. 0

(d) P (M) = 1.8

(e) P (N) = - 0.2

Solution  
 
(a) TRUE → This represents the probability of an impossible event.

(b) TRUE → The probability of an event lies between 0 and 1.

(c) TRUE → This represents the probability of a certain event.

(d) FALSE → The probability of an event should lie between 0 and 1.

(e) FALSE → The probability of an event lies between 0 and 1.

No. Problems & Solutions
03.
The probability of any event is always a proper fraction

Solution  
 
A Proper fraction is a fraction of the form
P
Q
, where P, Q > 0 and
P
Q
< 1

In simple terms, a proper fraction is a fraction where the numerator is less than the denominator.

Probability of occurrence of Event "A" =
Number of Favourable/Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(Event) =
m
n

Since m is always ≤ n,
m
n
is always a proper fraction

• Exceptions

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

No. Problems & Solutions
04.
(a) Events are said to be equally likely if
(b) By exhaustive events we mean
1) Equally likely cases 2) Not equally likely cases 3) Containing all possible cases 4) None of these

Solution  
 
(a) → Two or more events are said to be equally likely if their probability of occurrence is equal.

(b) → (c) - Exhaustive events mean events which cover all the possible cases within them.

No. Problems & Solutions
05.
If P(A) =
1
4
then P(A) is

Solution  
 

P(A) = 1 − P(A)
=
1 −
1
4
=
3
4

No. Problems & Solutions
06.

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

Solution  
 

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

No. Problems & Solutions
07.
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

No. Problems for Practice
01.
If P(A) = 1/3, P(A) is
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