PROBABILITY Presented by:Savi Arora

BASIC TERMS

Probability: Measure if likeliness than an event will occur.

Sample Space: Set of all possible outcomes.

Event: Any subset of any outcomes of any experiment.

Mutually exclusive event: Event which cannot occur together.

Dependent and Independent events.

PROBABILITY MANTRA

outcomepossibleofnoTotal

outcomefavourableofNoEP )(

TODAY WE WILL PLAY WITH

Coins

Dice

Cards

&

A bag full of colorful balls.

PRACTICE – 1

A dice is thrown, find probability

P(even no) =

P(no less than 3) =

P(not 3) =

In a deck of 52 cards

P(ace) =

P(red card) =

P(face card) =

CLARIFICATION: AND – OR

)()( BAPBorAP

)()( BAPBandAP A B

RULE OF ADDITION

Mutually exclusive event:

Not Mutually exclusive event:

)()()( BPAPAorBP

)()()()( BandAPBPAPAorBP

PRACTICE – 2

A Dice is thrown once: P( 2 or 3) =

P(multiple of 2 or multiple of 3) =

In Deck of 52 cards P(red or face) =

One ball drawn: Bag contains – 10 Red, 15 Yellow, 25 Green P( Yellow or Green) =

RULE OF MULTIPLICATION

Independent Events:

Dependent Event:

)()()( BPAPBandAP

)|()()( ABPAPBandAP )|()()( BAPBPBandAP Or

PRACTICE – 3

Bag contains – 10 Red, 15 Yellow, 25 Green

Two balls drawn (with replacement): P(1st Red, 2nd Yellow) = P(Both Red) = P(One Red, one Yellow) =

Two balls drawn (without replacement): P(1st Red, 2nd Yellow) = P(Both Red) = P(One Red, one Yellow) =

COIN

Two unbiased coins are tossed simultaneously:

P(one head) =

P(atleast one head) =

P(No head) =

P(one Head one Tail) =

COIN

Three unbiased coins are tossed simultaneously:

P(two head) =

P(at least two head) =

P(at most two head) =

P(All head) =

COIN

Four or More Coins are tossed??

You need Binomial for that.

PERMUTATION & COMBINATION Permutation: “one or more elements of a set where order does matter”

Combination: “one or more elements of a set where order does not matter”

For e.g.: Take 3 Letters – ABCPermutation of 2 of these letter: 3P2 = 6AB BAAC CABC CB

Combination of 2 of these letter: 3C2 = 3ABACBC

Now Take 4 : Letter – ABCDAnd write Permutation & Combination of 3 of these.

P&C

Permutation

Combination

3𝐶 2=3 !

(3−2 )! 2!=3n𝐶𝑟=

𝑛 !(𝑛−𝑟 )!𝑟 !

= 6n𝑃 𝑟=𝑛 !

(𝑛−𝑟 )!

*We will focus majorly on COMBINATION

PRACTICE – 4

Bag is Back : 10 Red, 15 Yellow, 25 Green

Three bolls drawn at random, find:

P(each different color) =

P(2 red 1 green) =

P(exactly 2 green) =

P(at least 2 green) =

P(at most 2 green) =

PRACTICE – 4

Bag is Back : 10 Red, 15 Yellow, 25 Green

Two draws of three balls each are drawn at random (with replacement)

P(1st three Red, 2nd three yellow) =

P(all different color, all different color) =

P(one draws three red, other draws three yellow) =*Wanna try without replacement??

BAYES’ THEOREM

What is the probability of outcome A given that outcome B has already occurred. i.e. P(A|B)

)'().'|()().|(

)().|()|(

APABPAPABP

APABPBAP

n. , . . . 2, 1, =k ,)B|P(A × )P(B

)B|P(A × )P(B = A)|P(B

1ii

ikk

n

i

Believe me! You don’t need to remember this, if you Still go with your Probability Mantra and your logics!

ILLUSTRATION

Late

BusCar Train

P(train)train)|P(lateP(bus)bus)|P(lateP(car)car)|P(late

P(car)car)|P(late late)|P(car

P(Car) = P(Bus) = P(Train) = 1/3

P(late|car) = 1/6P(late|bus) = 1/5P(late|train) = 1/4

Given that You are Late, what is the probability that you travelled by car??

PRACTICE – 5

A bulb is manufactured in one of the three machines X,Y,Z.

The machines produce in capacity: X – 30%, Y – 30%, Z – 40%.

Probability of bulb found defective in machine X – 10%, Y – 15%, Z – 20%.

A bulb is drawn at random and is found to be defective.

What is the probability that the bulb was produced in Machine Y??

THANK YOU

Queries are welcomed.