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PROBABILITY

It is a measure of the likelihood of a random phenomenon or chance behavior. Probability describes the long-term proportion with which a certain outcome will occur in situations with short-term uncertainty.

EXAMPLE:

Simulate flipping a coin 100 times. Plot the proportion of heads against the number of flips. Repeat the simulation.

THEOREMS OF PROBABILITY

ADDITIONTHEOREM

MULTIPLICATIONTHEOREM

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Addition Rule for Mutually Exclusive Events

If E and F are mutually exclusive events, then

P(E or F) = P(E) + P(F)

In general, if E, F, G, … are mutually exclusive events, then

P(E or F or G or …) = P(E) + P(F) + P(G) + …

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:2An 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 Ç B) : event that the number is divisible by 6 and 8.

A Ç B = {24, 48, ... 192}, n(A Ç 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

When Events are not mutually Exclusive

For finding the probability of one or more of two events that are not mutually exclusive the modified addition theorem is used:

P(A or B) = P(A) + P(B) – P(A and B) Where P(A or B) = Probability of happening of A and B when A and B are

not mutually exclusive.

P(A) = Probability of happening of event A.

P(B) = Probability of happening of event B.

P(AB) = Probability of happening of events A and B together in case of three events

P(A or B or C) = P(A) + P(B) +P(C) – P(AB) – P(AC) – P(BC) + P( ABC)

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MULTIPLICATION THEOREM

INDEPENDENT EVENTS:

Two events E and F are independent if the occurrence of event E in a probability experiment does not affect the probability of event F. Two events are dependent if the occurrence of event E in a probability experiment affects the probability of event F.

DEFINITION OF INDEPENDENT EVENTS:

Two events E and F are independent if and only if P(F | E) = P(F) or P(E | F) = P(E)

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Example : 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.

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 46+3+2 11

= =24 24

MULTIPLICATION THEOREM IN CASE OF 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

PB0

. PA/B

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

P A B

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

Independent and Dependent Events

Two events are independent if the occurrence of one of the events does NOT affect the probability of the occurrence of the other event. Two events A and B are independent if:

P(B|A) = P(B) or if P(A|B) = P(A)

Events that are not independent are dependent

EXAMPLE :Q) Two cards are selected without replacement, from a

standard deck. Find the probability of selecting a king and then selecting a queen.

Solution:

Because the first card is not replaced, the events are dependent.

P(K and Q) = P(K) ● P(Q|K)

So the probability of selecting a king and then a queen is about .0006

006.02652

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4

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