8/10/2019 Conditional & Independent Events

1/15

8/10/2019 Conditional & Independent Events

2/15

Conditional events

We often wish to consider the probability of anevent B amongst occurrences of another event A.Consider a class of 30 students, consisting of 20girls and 10 boys. There are 15 girls and 5 boyswho wear spectacles.A student is chosen at random from the class.Let A be the event that the student wears spectaclesand B be the event that the student is a girl.An example of a conditional event :B|A the student chosen is a girl, knowing that the

student chosen wears spectacles.

8/10/2019 Conditional & Independent Events

3/15

Conditional probability

In a sense, B| A restricts the sample space to the A.Thus,

)()(

)|( An

A Bn A B P

)()()(

)(

S n

An

S n A Bn

)()(

A P A B P

In the above example,n (A) = 20, andn (B A) = 15.Thus, P( B |A) = 15 / 20 =

P( B A) = 15 / 30 =

P( A) = 20 / 30 = 2 / 3

P( B |A) = 32

21

8/10/2019 Conditional & Independent Events

4/15

Conditional probability

We have )()()|(

A P

A B P A B P

Note that)(

)()|( B P

B A P B A P

)()()|( A B P A P A B P So

So )()()|( B A P B P B A P

But P( A B) = P( B A)Thus P( B| A)P( A) = P( A| B)P( B)

8/10/2019 Conditional & Independent Events

5/15

Example : A number is selected randomly from the set{1, 2, 3, , 30}. If this number chosen is odd, what is

the probability that it is a perfect square?Given n(S ) = 30.

Let A be the event getting an odd number. Then n( A)= 15So, P( A) = 15 / 30.

Let B be the event getting a perfect square.

B = {1, 4, 9, 16, 25}. So, n( B) = 5P( B) = 5 / 30But n( B A) = 3P( B A) = 3 / 30

Conditional probability

8/10/2019 Conditional & Independent Events

6/15

P(number is a perfect square | number is odd}= P( B| A)= P(number is a perfect square number is odd) / P(number is odd)= P( B A) / P( A)

=3 / 30

/ 15 / 30 =3

/ 15 =1

/ 5P(number is odd | number is a perfect square}= P( A| B)= P(number is odd number is perfect square ) / P(number is a perfect square)

=P( A B)

/ P( B)= 3 / 30 / 5 / 30 = 3 / 5So, P( B| A)P( A) = 1 / 5 1 / 2 = 1 / 10

and P( A| B)P( B) = 3 / 5 5 / 30 = 1 / 10

Conditional probability

S A B

1

925

35

7111315

1719212327 29

4

16

2 6 8 10 12 14 18

20 22 24 26 28 30

8/10/2019 Conditional & Independent Events

7/15

Note that B = ( B A) ( B A ).( B A) and ( B A ) are exclusive events, soP( B) = P( B A) + P( B A ).From the rule of conditional probability,

P( B A) = P( B| A)P( A)and P( B A ) = P( B| A )P( A ).Thus,

P( B) = P( B| A)P( A) + P( B| A )P( A ).

Conditional probability complementary events

S

A

B

B A B A

8/10/2019 Conditional & Independent Events

8/15

Independent events

When an event A is independent of B, the probability of A happening does not depend on whether B happens or not.That is, P( A| B) = P( A).

From the relationship P( A B)=P( A| B)P( B), when we putP( A| B) = P( A)

We have, P( A B) = P( A)P( B).

This is the multipli cation rule .

8/10/2019 Conditional & Independent Events

9/15

Example 1 : Tossing two fair diceP(getting 6 on both dice)

= P( getting 6 on the first die and 6 on the second die )= P( getting 6 on the first die ) P( getting 6 on the second die )= 1 / 6 1 / 6

= 1 / 36

Independent events

1

12

2 3

34

4 5

56

6 No on first die

N o on

s e c on

d d i e

Just look atthe sample

space!

8/10/2019 Conditional & Independent Events

10/15

8/10/2019 Conditional & Independent Events

11/15

Three or more events

For two events, A and B, in the sample space S ,P( A B) = P( A) + P( B) P( A B).

This can be extended to three events, A, B andC , in the sample space S :

P( A B C )= P( A) + P( B) + P( C ) P( A B) P( B C ) P(C A) + P( A B C )

A

BC

S

8/10/2019 Conditional & Independent Events

12/15

Example : Three candidates, A, B and C , contest in threedifferent constituencies during an election. The probabilities ofthem winning are 0.5, 0.3 and 0.8 respectively.Let A represent the event A wins the election, B represent theevent B wins the election and C represent C wins the election.That is, P( A) = 0.5, P( B) = 0.3, P( C ) = 0.8

Three or more events

(a) P(only one of them wins the elections)= P( A B C ) P( A B C ) P( A B C )

= P( A B C ) + P( A B C ) + P( A B C )= P( A)P( B )P( C ) + P( A )P( B)P( C ) + P( A )P( B )P( C )= 0.5 0.7 0.2 + 0.5 0.3 0.2 + 0.5 0.7 0.8= 0.07 + 0.03 + 0.28

= 0.38

A, B, C areindependent

8/10/2019 Conditional & Independent Events

13/15

Example : Three candidates, A, B and C , contest in threedifferent constituencies during an election. The probabilities ofthem winning are 0.5, 0.3 and 0.8 respectively. That is, P( A) = 0.5, P( B) = 0.3, P( C ) = 0.8

Three or more events

(b) P(all of them win the election)= P( A B C )= P( A)P( B)P( C )= 0.5 0.3 0.8

= 0.12

(c) P(at least one of them win the election)= 1- P(none of them win)= 1 - P( A B C )= 1 - P( A )P( B )P( C )= 1 - 0.5 0.7 0.2= 1 0.07 = 0.93

8/10/2019 Conditional & Independent Events

14/15

Tree diagrams

Tree diagrams are useful in solving problemsin probability.Independent events are placed in stages and

probabilities are labelled at each branch sothat calculations can be easily done.

8/10/2019 Conditional & Independent Events

15/15

Example : A shop sells pots supplied bytwo factories, A and B, in the ratio 3 : 7.Given that 20% and 10% of the potssupplied by factories A and B respectively

have cracks, what is the probability that a pot selected at random from the shop hascracks?Let C represents the event a pot hascracks.P(C ) = P( C A) + P( C B)

= P( C | A)P( A) + P( C | B)P( B)= 0.2 0.3 + 0.1 0.7= 0.06 + 0.07 = 0.13

Tree diagrams

A

B

Cracks

Cracks

No cracks

No cracks

P( A) = 0.3

P( B) = 0.7

P(C|A) = 0.2

P(C|A ) = 0.8

P(C|B ) = 0.9

P(C|B ) = 0.1