Page 1 of 40
Percentile Classes
Probability
Table of Content
Theory……………………………………..….02
Exercise 01 : Probability…….....…….06
Exercise 02 : Probability..……..……..16
Exercise 03 : Probability……..……….26
Exercise 04 : OLD is GOLD 15………33
No Substitute to Hardwork
, Page 2 of 40
Probability Theory:
Mutually Exclusive Events: Let S be the sample space associated with a random
experiment and let E1 and 𝐸2 be the two events. Then 𝐸1 and 𝐸2 are mutually exclusive
events if 𝐸1 ∩ 𝐸2 ≠ ∅,
Mutually Exclusive and Exhaustive System of Events: Let S be the sample space
associated with a random experiment, Let 𝐸1 , 𝐸2 , …. 𝐸𝑛 be the subsets of S such that
(i) 𝐸1 ∩ 𝐸𝑗 = ϕ for i ≠ j and
(ii) 𝐸1 ∪ 𝐸2 ∪ 𝐸3 ∪ …. ∪ 𝐸𝑛 = S
When the set of events 𝐸1 , 𝐸2 , 𝐸3 …..𝐸𝑛 is said to form a mutually exclusive and
exhaustive system of events.
Definition of Probability: In a random experiment, let S be the sample space and let
E ⊆ S.
Where E is a an event.
The probability of occurrence of the event E is defined as
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
P(E) =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝐸 𝑛(𝐸)
= =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑆 𝑛(𝑆)
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠 𝑖𝑛 𝐸
=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠 𝑖𝑛 𝑆
From the above definitions it is clear that
(i) 0 ≤ P(E) ≤ 1
(ii) P (∅) = 0
(iii) P(S) = 1
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠 𝑖𝑛 𝐸̅
Also, P(𝐸̅ ) =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠 𝑖𝑛 𝑆
𝑛(𝑆)−𝑛(𝐸)
=
𝑛(𝑆)
𝑛 (𝐸)
= 1-
𝑛(𝑆)
= 1- P(E)
➔ P(𝐸̅ ) = 1- P (E)
∴ P(E) = + P (𝐸̅ ) = 1
Odd in favour of An event and odds against an event
No Substitute to Hardwork
, Page 3 of 40
In m be the number of ways in which an event occurs and n be the number of
ways in which it does not occur, then
𝑚
(i) odds in favour of the events = (or m:n)
𝑛
𝑛
(ii) odds against the event = (or n:m)
𝑚
Some important results:
(A) If, A, B and C are three events, then
(i) P [Exactly one of A, B, C occurs]
= P(A) + P(B) + P(C) – 2[A∩B] + (B∩C) + (A∩C)] +3P (A∩B∩C)
(ii) P (Atleast two of A, B, C occur)
= P (A∩B) + P(B∩C) + P(A∩C) – 2P (A∩B∩C)
(B) If A and B are two events, then P (exactly one of A, B occurs)
= P(A) + P(B) + 2P(A∪B)
= P(A∪B) – P (A∩B)
Conditional Probability: Let A and B be two events associated with a random
experiment, then, the probability of occurrence of A under the condition that B has
already occurred and P(B)≠ 0 is called the conditional probability and it is denoted by
𝐴
P( )
𝐵
𝐴
Thus, P( ) = Probability of occurrence of A given that B has already occurred.
𝐵
𝐵
Similarly, P ( ) = Probability of occurrence of B given that A has already occurred.
𝐴
𝐴
NOTE: (i) Sometimes P( ) is used to denote the probability of occurrence of A when B
𝐵
occurs.
𝐵
(ii) Similarly P( ) is used to denoted the probability of occurrence of B when A
𝐴
occurs.
The above two cases happens due to the simultaneous occurrence of two events
since the two events are the subsets of the same sample space.
Multiplication Theorem:
Let A and B be two events associated with the same random experiment then
𝐵
P(A∩B) = P(A)P( ) if P(A) ≠ 0 …(i)
𝐴
No Substitute to Hardwork
, Page 4 of 40
𝐴
Or P(A∩B) = P(B) P( ), P(B) ≠ … (ii)
𝐵
𝐵 𝑃(𝐴∩𝐵)
NOTE: P( ) = from (i)
𝐴 𝑃(𝐴)
𝐴 𝑃(𝐴∩𝐁)
And P( ) = from (ii)
𝐵 𝑃(𝐵)
In general, if 𝐴1 , 𝑎2 , 𝑎3 ….𝑎𝑛 are events associated with the same random
experiment, then
P(𝐴1 ∩ 𝐴2 ∩ 𝐴3 ∩……∩𝐴𝑛 )
𝐴 𝐴3 𝐴𝑛
= P(𝐴1 ) P ( 3) P (
𝐴 𝐴
) …P (𝐴 )
1 1 ∩𝐴2 1∩𝐴2 ∩…∩𝐴𝑛−1
Independent Events: Events are said to be independent, if the occurrence of one
does not depend upon the occurrence of the other
Suppose an urn contains m red balls and n green balls. Two balls are drawn from
the urn one after the other.
If the ball drawn in the first draw is not replaced back in the bag, then two events of
drawing the ball are dependent because first draw of the ball determine the
probability of drawing the second ball.
If the ball drawn in the first draw is replaced back in the bag, then two events are
independent because first draw of a ball has no effect on the second draw.
Theorem I: Two events A and B associated with the same sample space of a
random experiment are independent if and only if
P (A∩B) = P (A) . P(B)
Theorem 2. If 𝐴1 , 𝐴2 , 𝐴3 …. 𝐴𝑛 are independent events associated with a random
experiment, then
P(𝐴1 ∩ 𝐴2 ∩ 𝐴3 ….∩𝐴𝑛 ) = P(𝐴1 ) P (𝐴2 )…P(𝐴𝑛 )
Theorem 3. If 𝐴1 , 𝐴2 , ….𝐴𝑛 are n independent events associated with a random
experiment, then
P(𝐴1 ∪ 𝐴2 ∪…..∪ 𝐴𝑛 ) = 1-P(𝐴 ̅̅̅1 ) P (𝐴
̅̅̅2 ) ….P(𝐴
̅̅̅̅
𝑛)
Important results:
If A and B are independent events then the following events are also independent.
(i) A ∩ 𝐵̅ (ii) 𝐴̅ ∩ B (iii) 𝐴̅ ∩ 𝐵̅
No Substitute to Hardwork
Percentile Classes
Probability
Table of Content
Theory……………………………………..….02
Exercise 01 : Probability…….....…….06
Exercise 02 : Probability..……..……..16
Exercise 03 : Probability……..……….26
Exercise 04 : OLD is GOLD 15………33
No Substitute to Hardwork
, Page 2 of 40
Probability Theory:
Mutually Exclusive Events: Let S be the sample space associated with a random
experiment and let E1 and 𝐸2 be the two events. Then 𝐸1 and 𝐸2 are mutually exclusive
events if 𝐸1 ∩ 𝐸2 ≠ ∅,
Mutually Exclusive and Exhaustive System of Events: Let S be the sample space
associated with a random experiment, Let 𝐸1 , 𝐸2 , …. 𝐸𝑛 be the subsets of S such that
(i) 𝐸1 ∩ 𝐸𝑗 = ϕ for i ≠ j and
(ii) 𝐸1 ∪ 𝐸2 ∪ 𝐸3 ∪ …. ∪ 𝐸𝑛 = S
When the set of events 𝐸1 , 𝐸2 , 𝐸3 …..𝐸𝑛 is said to form a mutually exclusive and
exhaustive system of events.
Definition of Probability: In a random experiment, let S be the sample space and let
E ⊆ S.
Where E is a an event.
The probability of occurrence of the event E is defined as
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
P(E) =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝐸 𝑛(𝐸)
= =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑆 𝑛(𝑆)
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠 𝑖𝑛 𝐸
=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠 𝑖𝑛 𝑆
From the above definitions it is clear that
(i) 0 ≤ P(E) ≤ 1
(ii) P (∅) = 0
(iii) P(S) = 1
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠 𝑖𝑛 𝐸̅
Also, P(𝐸̅ ) =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠 𝑖𝑛 𝑆
𝑛(𝑆)−𝑛(𝐸)
=
𝑛(𝑆)
𝑛 (𝐸)
= 1-
𝑛(𝑆)
= 1- P(E)
➔ P(𝐸̅ ) = 1- P (E)
∴ P(E) = + P (𝐸̅ ) = 1
Odd in favour of An event and odds against an event
No Substitute to Hardwork
, Page 3 of 40
In m be the number of ways in which an event occurs and n be the number of
ways in which it does not occur, then
𝑚
(i) odds in favour of the events = (or m:n)
𝑛
𝑛
(ii) odds against the event = (or n:m)
𝑚
Some important results:
(A) If, A, B and C are three events, then
(i) P [Exactly one of A, B, C occurs]
= P(A) + P(B) + P(C) – 2[A∩B] + (B∩C) + (A∩C)] +3P (A∩B∩C)
(ii) P (Atleast two of A, B, C occur)
= P (A∩B) + P(B∩C) + P(A∩C) – 2P (A∩B∩C)
(B) If A and B are two events, then P (exactly one of A, B occurs)
= P(A) + P(B) + 2P(A∪B)
= P(A∪B) – P (A∩B)
Conditional Probability: Let A and B be two events associated with a random
experiment, then, the probability of occurrence of A under the condition that B has
already occurred and P(B)≠ 0 is called the conditional probability and it is denoted by
𝐴
P( )
𝐵
𝐴
Thus, P( ) = Probability of occurrence of A given that B has already occurred.
𝐵
𝐵
Similarly, P ( ) = Probability of occurrence of B given that A has already occurred.
𝐴
𝐴
NOTE: (i) Sometimes P( ) is used to denote the probability of occurrence of A when B
𝐵
occurs.
𝐵
(ii) Similarly P( ) is used to denoted the probability of occurrence of B when A
𝐴
occurs.
The above two cases happens due to the simultaneous occurrence of two events
since the two events are the subsets of the same sample space.
Multiplication Theorem:
Let A and B be two events associated with the same random experiment then
𝐵
P(A∩B) = P(A)P( ) if P(A) ≠ 0 …(i)
𝐴
No Substitute to Hardwork
, Page 4 of 40
𝐴
Or P(A∩B) = P(B) P( ), P(B) ≠ … (ii)
𝐵
𝐵 𝑃(𝐴∩𝐵)
NOTE: P( ) = from (i)
𝐴 𝑃(𝐴)
𝐴 𝑃(𝐴∩𝐁)
And P( ) = from (ii)
𝐵 𝑃(𝐵)
In general, if 𝐴1 , 𝑎2 , 𝑎3 ….𝑎𝑛 are events associated with the same random
experiment, then
P(𝐴1 ∩ 𝐴2 ∩ 𝐴3 ∩……∩𝐴𝑛 )
𝐴 𝐴3 𝐴𝑛
= P(𝐴1 ) P ( 3) P (
𝐴 𝐴
) …P (𝐴 )
1 1 ∩𝐴2 1∩𝐴2 ∩…∩𝐴𝑛−1
Independent Events: Events are said to be independent, if the occurrence of one
does not depend upon the occurrence of the other
Suppose an urn contains m red balls and n green balls. Two balls are drawn from
the urn one after the other.
If the ball drawn in the first draw is not replaced back in the bag, then two events of
drawing the ball are dependent because first draw of the ball determine the
probability of drawing the second ball.
If the ball drawn in the first draw is replaced back in the bag, then two events are
independent because first draw of a ball has no effect on the second draw.
Theorem I: Two events A and B associated with the same sample space of a
random experiment are independent if and only if
P (A∩B) = P (A) . P(B)
Theorem 2. If 𝐴1 , 𝐴2 , 𝐴3 …. 𝐴𝑛 are independent events associated with a random
experiment, then
P(𝐴1 ∩ 𝐴2 ∩ 𝐴3 ….∩𝐴𝑛 ) = P(𝐴1 ) P (𝐴2 )…P(𝐴𝑛 )
Theorem 3. If 𝐴1 , 𝐴2 , ….𝐴𝑛 are n independent events associated with a random
experiment, then
P(𝐴1 ∪ 𝐴2 ∪…..∪ 𝐴𝑛 ) = 1-P(𝐴 ̅̅̅1 ) P (𝐴
̅̅̅2 ) ….P(𝐴
̅̅̅̅
𝑛)
Important results:
If A and B are independent events then the following events are also independent.
(i) A ∩ 𝐵̅ (ii) 𝐴̅ ∩ B (iii) 𝐴̅ ∩ 𝐵̅
No Substitute to Hardwork
