Mathematics

Session

Probability -2

Session Objectives

Addition Theorem on Probability

Conditional Probability

Multiplication Theorem on Probability

Independent Events

Class Exercise

Addition Theorem For 2 Events

If A and B be are events in a sample space S. Then the probability

of occurrence of at least one of the events A and B is given by

P A B =P A +P B - P A B

I f A and B are two mutually exclusive events i.e. A B=

Then, P A B =P A +P B

Addition Theorem For 3 Events

If A, B and C are three events in the sample space S. Then,

P A B C =P A +P B +P C - P A B - P B C

- P C A +P A B C

If A, B and C are three mutually exclusive events, then

P A B C =P A +P B +P C

Addition Theorem For n Events

then1 2 3 nI f A , A , A ,... A are mutually exclusive events,

1 2 3 n 1 2 nP A A A ... A =P A +P A +...+P A

Event A or not A

The events ‘A’ and ‘not A’ are mutually exclusive events as each

outcome of the experiment is either favourable to ‘A’ or ‘not A’.

Therefore, the event ‘A or not A’ is a certain (or sure) event whose

probability is 1.

P A or not A =P A A' =1

P A +P A' =1

or P A =1- P A' P A' =1- P A

Example-1

Solution: A and B are two mutually exclusive events.

Given two mutually exclusive events A and B such and

, find P(A or B).

1P A =

2

1P B =

3

P A B =P A +P B Addition theorem

1 1 5= + =

2 3 6

Example-2

An integer is chosen at random from the first 200 positive integers.

Find the probability that the integer is divisible by 6 or 8.

Solution: Let S be sample space. Then,

S = {1, 2, 3, …200}, n(S) = 200

Let A : event that the number is divisible by 6.

A = {6, 12, 18 ... 198}, n(A) = 33

Let B : event that number is divisible by 8.

B = {8, 16, 24 ... 200}, n(B) = 25

Solution Cont.

(A B) : event that the number is divisible by 6 and 8.

A B = {24, 48, ... 192}, n(A B) = 8

A B : event that the number is divisible either by 6 or 8.

P A B =P A +P B - P A B

33 25 8= + -

200 200 200

50 1= =

200 4

Example-3

Find the probability of drawing a black king and probability of drawing

a black card or a king from a well-shuffled pack of 52 cards.

Solution: Let A and B be two events such that

Total number of cards n(S) = 52

A : drawing a black card, B : drawing a king

Number of black cards n(A) = 26

Number of kings n(B) = 4

Solution (Cont.)

Number of black kings n A B =2

n A 26 1P A = = =

n S 52 2

n B 4 1P B = = =

n S 52 13

n A B 2 1

P drawing a black king =P A B = = =n S 52 26

Solution (Cont.)

P A B =P A +P B - P A B

and P(drawing a black card or a king)

1 1 1 7= + - =

2 13 26 13

Example-4

Find the probability of getting an even number on the first die or a

total of 8 in a single through of two dice. (CBSE 2004)

Solution: Let S be sample, then n(S) = 36

Let A and B be two events such that

A = getting an even number on first die, B = getting a total of 8

A B=getting an even number on first die and a total of 8

Solution Cont.

A ={(2, 1), (2, 2), …, (2, 6), (4, 1), (4, 2), …, (4, 6),

(6, 1), (6, 2), …, (6, 6)},

B = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}

and A B= 2, 6 , 6, 2 , 4, 4

18 5 3

P A = , P B = and P A B =36 36 36

Solution (Cont.)

Required probability =P A B =P A +P B - P A B

18 5 3

+36 36 36

20 5

36 9

Example-5

Two dice are tossed together. Find the probability that the sum of

the numbers obtained on the two dice is neither a multiple of 3 nor

a multiple of 4. (CBSE 1996)

Solution: Let S be sample, then n(S) = 36

Let A and B be two events such that

A = the sum of the numbers obtained on the two dice is a multiple of 3.,

B = the sum of the numbers obtained on the two dice is a multiple of 4

Solution (Cont.)

Then, A = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2),

(4, 5), (5, 1), (5, 4), (6, 3), (6, 6)},

B = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}

n A =12, n B =9 and n A B =1

and A B= 6, 6

12 9 1

P A = , P B = and P A B =36 36 36

Solution (Cont.)

P sum is multiple of 3 or 4 =

P A B =P A +P B - P A B

12 9 1 20= + - =

36 36 36 36

P sum is neither multiple of 3 nor 4

20 16 4

=1-36 36 9

Conditional Probability

In a random experiment, if A and B are two events, then the

probability of occurrence of event A when event B has already

occurred and , is called the conditional probability and it

is denoted by

P B 0

.P A/B

Cont.

P A B

Similarly, P B/A = , P A 0P A

Number of outcomes favourable to A which are also favourable to BP A/B =

Number of outcomes favourable to B

P A B

P A/B = , P B 0P B

Multiplication Theorem on Probability

If A and B are two events associated with a random experiment, then

P A B =P A/B ×P B , if P B 0

or P A B =P A/B ×P A , if P A 0

Example-6

I f A and B are two events such that P A =0.5, P B =0.6

and P A B =0.8. Find P A / B and P B / A .

Solution: Given that P(A) = 0.5, P(B) = 0.6 and P A B =0.8

P A B =P A +P B - P A B

By addition theorem

0.8=0.5+0.6 - P A B

P A B =1.1- 0.8=0.3

Solution Cont.

P A B 0.3 1

P A/B = = =P B 0.6 2

By multiplication theorem

P A B 0.3 3

and P B/A = = =P A 0.5 5

Example –7

A die is rolled twice and the sum of the numbers appearing on them

is observed to be 7. What is the conditional probability that the number

2 has appeared at least once. (CBSE 1990, 2003)

Solution: Let A and B be two events such that

A = getting number 2 at least once

B = getting 7 as the sum of the numbers on two dice

A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}

Solution (Cont.)

B = {(2, 5), (5, 2), (6, 1), (1, 6), (3, 4), (4, 3)}

11 6 2

P A = , P B = and P A B =36 36 36

P A B

Required probability =P A / B =P B

2136= =

6 336

Independent Events

Two events A and B are said to be independent if occurrence of one

does not affect the occurrence of other; then

P A/B =P A and P B/A =P B

are independent events. P A B =P A ×P B , where A and B

Example –8

The probability of student A passing an examination is

and the probability of student B passing is . Find the probability of

(i) only A passing the examination

(ii) only one of them passing the examination

29

59

Solution: (i) Let A : event of A passing the examination 2

=9

P(A') = 1 – P(A)2 7

=1- =9 9

Solution Cont.

Let B : event of B passing the examination5

9

P(B’) = 1 – P(B)5 4

=1- =9 9

A P B' i P only A passing the examination =P A B' =P

[As A and B' are independent events]

2 4 8= × =

9 9 81

Solution Cont.

(ii) P(only one passing the examination)

=P A B' A' B

=P A B' A' B

=P A ×P B' +P A' ×P B

[As A and B' are independent events and A’ and B are also independent events]

8 7 5 8 35 43= + × = + =

81 9 9 81 81 81

Example –9

Three bags contain 7 white 7 red, 5 white 8 red, and 8 white 6 red

balls respectively. One ball is drawn at random from each bag. Find

the probability that the three balls drawn are of the same colour.

Solution: W1 : White ball from bag I W2 : White ball from bag II

W3 : White ball from bag III

R1 : Red ball from bag I

R2 : Red ball from bag II

R3 : Red ball from bag III

Solution (Cont.)

P(Three balls are of same colour)

1 2 3 1 2 3=P W W W R R R

1 2 3 1 2 3=P W W W R R R

1 2 3 1 2 3=P W ×P W ×P W +P R ×P R ×P R

[As W1, W2 and W3 are independent events and R1, R2 and R3 are also

independent events]

7 5 8 7 8 6 22= × × + × × =

14 13 14 14 13 14 91

Example –10

A speaks truth in 60% of the cases and B in 90% of the cases. In what

percentage of the cases they are likely to contradict each other in

stating the same fact? (CBSE 2003)

Solution: Let A : event of A speaking the truth.

B : event of B speaking the truth.

60 40P A = , P A' =

100 100

90 10P B = , P B' =

100 100

Solution (Cont.)

P(A and B contradicting each other)

'=P A B A' B

'=P A B +P A' B

=P A ×P B' ++P A' ×P B

[As A and B' are independent of each other and A' and B are

also independent of each other.

60 10 40 90 21= × + × =

100 100 100 100 50

Example –11

The probabilities of A, B and C solving a problem are

respectively. If the problem is attempted by all simultaneously,

find the probability of exactly one of them solving it. (CBSE 2004)

1 1 1, and

2 3 4

1 1 1

Solution: P A = P A' =1- =2 2 2

1 1 2

P B = P B' =1- =3 3 3

1 1 3

P C = P C' =1- =4 4 4

Solution (Cont.)

Required probability

=P A B' C' A' B C' A' B' C

P P =P A B' C' A' B C' A' B' C

P B' P C' P B P C' P B' P C =P A +P A' P A'

[As A, B’ and C’ are independent events; A’, B and C’ are independent

events; A’, B’ and C are also independent events]

1 2 3 1 1 3 1 2 1= × × + × × + × ×

2 3 4 2 3 4 2 3 4

6+3+2 11= =

24 24

Thank you