6-Probability- 13 Sept 11

7/31/2019 6-Probability- 13 Sept 11

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Probability

Prof. Dhananjay M.Apte

98231 90939

Rolling Dice.What is the possibility

(probability) of getting

6 when rolling a

standard Dice cube?

outcomespossible

outcomesfavorableyprobabilit

How many 6s are there?How many possible outcomes?

= 1/ 6


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Rolling Dice.

What is the P (getting 2 )

when rolling a standardDice cube?

outcomespossible

outcomesfavorableyprobabilit

How many 2s are there?

How many possible outcomes?

= 1/ 6

SpinnerWhat is the probability of the spinner landing on 1?

8

3


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Bunch of 52 playing cards. A card is drawn.

Probability of getting a King.

Ans. 4 /52

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P (any one Heart ) = 13 / 52 = 1/ 4 = 0.25 = 25 %


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Bunch of 52 playing cards. A card is drawn.

P (an Ace or a King )

4 /52 = 8 /524 /52+

Addition rule of Probability

Mutually Exclusive ( Do NOT occur together) Events

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P ( a king from 1st bunch & a Heart from 2nd bunch)

= 1/ 524 /52 13 /52X

Multiplication rule of Probability

Independent Events


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and

P (6 on Red Dice and 4 on black)

1/ 6 X 1/6 = 1/36

If two Dices Red & Black are rolled, Find

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P ( a king & a Queen)= 16/ (52 x 51)4 /52 4 / 51X

P ( two kings)

= 12/ (52 x 51)4 /52 3 / 51X

- 1 = 51

Drawing2 cards at random from a single bunch of cards

Dependent Events


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A throw of 2 Dices (one big and one small) is made. What is theProbability of getting 1 on a Dice and 4 on the other ?

anotherpossibility

& &

(1/6 X 1/6 ) (1/6 X 1/6)

=( 1/36 ) + (1/36) = 2/36 = 1/18

+

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A bag contains 6 white, 4 red, 10 black balls

2 balls are drawn at random.

Find P( both the 2 balls are black)

Can be solved using conventional approach.

Now using Permutations, Combinations Theory approach

Total 20 balls.. 2 balls are drawn at random..

2 Balls are drawn in 20 c 2 = 190 ways..(Total no of possibilities)

Possibilities of both 2 balls are black = 10 c 2 = 45..(Favorable possibilities)

Hence P( both the 2 balls are black) = 45/ 190 = 0.237


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A bag contains 8 white and 4 red balls

5 balls are drawn at random.

Find P (2 balls are red & 3 are white)

Its tedious to be solved using conventional approach. using Combinations approach



Favorable possibilities (2 red balls) = 4 c 2 = 6

Favorable possibilities (3 white balls) = 8 c 3 = 56

Favorable possibilities (2 red & 3 white) = 6 x 56 = 336

Hence P (2 balls are red & 3 are white) = 336/ 792 = 0.424

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A bag contains 5 white and 8 red balls3 balls are drawn at random.. for 2 times

Find P (3 balls are white in 1st drawing & 3 are red in the 2nd )



Favorable possibilities (3 white) = 5 c 3 = 10

Hence P (3 balls are white) = 10/ 286 = 0.0349

Now there are 10 balls3 balls are drawn

3 Balls are drawn in 10 c 3 = 120 ways..(Total no of possibilities) Favorable possibilities (3 red) = 8 c 3 = 56

Hence P (3 balls are red) = 56 / 120 = 0.4666

Hence P (3 balls are white in 1st drawing & 3 are red in the 2nd )= 0.0349 x 0.4666 = 0.0162


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Exercise

1) Drawing 6 cards at random from a single bunch of cards

Find P ( 3 kings & 3 Queens)

HintsNumber of ways 6 cards are drawn from 52 cards are 52 c 6 = 2,03,58,520

Number of ways 3 kings are drawn from 52 cards are

4 c 3.

2) With & Without out replacement

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NOT Approach

P (happening an event) = 1P ( NOT happening theevent)


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Bayes Theorem

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If a company buys 40% of

the cars for its customers

from agency I and 60%

from agency II. If6% of

the cars from agency I and

5% of the cars from

agency II break down,

what is the probability that

a car rented by thiscompany breaks down?

Agency I Agency II

40% = 0.4Prior Probability

60% = 0.6Prior Probability

6% = 0.06Conditional Probability

5% =0.05Conditional Probability

0.4 x 0.06 =0.024

Joint probability

0.6 x 0.05 =0.03

Joint probability

P (car breaks down)

= (.4) (.06) + (.6) (.05)

= .054 = 5.4%

P (break down car is from Agency 1)

= Posterior Probability

= 0.024/ (0.024 + 0.03)

Whats P of Agency 2?


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Probability Distribution

Binomial

Poisson

Normal

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Weighted Mean


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10 Possibilities

5 coin tosses. Find probability that you flip exactly 3 heads

= 0.3125The number of Possible combinations can also be determined by

using Combinations Theory.. n C r = 5 C 3 = 10

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UseBinomialProbabilityDistribution Theorem.


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Here Number of events (n) = 5

Number of times, the outcome (of getting heads) =3termed as r

Binomial considers two possibilities.

1) Success (Event happens)... Probability is termed p

2) Failure (Event do not happen)..Probability is termed q

In above problemgetting 3 heads is Success, Not getting 3 heads-Failure.

Probability of getting Head in a Coin (p) =

Probability ofNOT getting Head in a Coin (q) = 1-p =

P = (n c r )= 0.3125

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When to use the Binomial Distribution

A fixed number of trials

Only two outcomes (true, false; heads tails; girl, boy; six, not six ..)

Each trial is independent


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Problem n r p q Answer

A dice is rolled 5 times

What is the probability it will

show 6 exactly 3 times?

5 3 1/6 5/6

A coin is tossed 7 times.

Find the probability of

getting exactly 3 heads.

7 3 0.27

If I toss a coin 20 times,

whats the probability of

getting 2 or fewer heads?

20 2,

1,0

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Pascals Triangle

To determine n C r.we use

Pascals Triangle

5 C 0 = 1 , 5 C 1 = 5, 5 C 2 = 10,

7 C 0 = 1 , 7 C 1 = 7, 7 C 2 = 21,


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Pascals Triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

Edges are all 1s

Add the two

numbers in the

row above to

get the number

below, e.g.:

3+1=4; 5+10=15

To get the

coefficient for

expanding to

the 5th power,

use the row

that starts

with 5.

(p + q)5= 1p5 + 5p4q1 + 10p3q2 + 10p2q3+ 5p1q4+ 1q5

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Use of tables


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Poisson Distribution

Example-1 A rare disease has an incidence of1 in 1000 person per

year. Find the probability of 0 incidences (cases) (x = 0)Consider population = 10,000(Assume that members of the population are affected independently)

The probability is 1 in 1000, i.e. 1/ 1000 = 0.001

The expected value = 0.001*10,000 = 10 (Termed as mean, m )

10 new cases expected in this population per year

00227.!2

)10()2(

000454.!1

)10()1(

0000454.!0

)10()0(

)10(2

)10(1

)10(0

eXP

e

XP

eXP

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Example-2 If new cases of West Nile Virus in New

England are occurring at a rate of about 2 per month,

then find the probabilities that: 0,1, 2, 3, 4, 5, 6, to 1000

cases will occur in New England in the next month:


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Poisson Probability table

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Exercise

1a. If calls to your cell phone are a Poisson process with a

constant rate =2 calls per hour, whats the probability

that, if you forget to turn your phone off in a 1.5 hour

movie, your phone rings during that time?

1b. How many phone calls do you expect to get during the

movie?

Hint-Rate = 2 calls per hr..Expected Value = 2 x 1.5 hrs


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Answer

We have to find : P(X1) =1 P(X=0) =2 calls/hour..m = 2 x 1.5

P(X1)=1 .05 = 95% chance

1b. How many phone calls do you expect to get during the movie?

E(X) = t = 2(1.5) = 3

xxx

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6-Probability- 13 Sept 11