Lecture 5

Discrete Probability Distributions

Sample Space while rolling 2 dice.

P(sum of 8) = 5/36

Conditional Probability

Notation: P(B|A)

P(B|A) represents the probability of the event

B occurring after it is assumed that the event A has already occurred. (read P(B|A) as “B given A”)

Conditional Probability

andP A BP B A

P A

Example-1

Consider a well shuffled deck of 52 cards

Let A = drawing a face card and B= drawing a heart. Find P(B|A).

P(A) = 12/52

P(A and B) = 3/52

Conditional Probability

andP A BP B A

P A = 3/12

Example-2 A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results of the survey are shown below.

Find the probability that the respondent answered yes (Y), given that the respondent was a female (F).

P(Y|F) = 8/50 = 0.16

Factorial

Factorial is the product of all the positive

numbers from 1 to a number.

! 1 2 3 2 1

0! 1

n n n n

5 x 4 x 3 x 2 x 1 = 5!

10!

7!

10 9 8 7 .... 3 2 1

7 6 5 ... 3 2 1 10 9 8 = 720

Random Variables

A random variable x represents a numerical value

associated with each outcome of a probability distribution.

A random variable is discrete if it has a finite or countable

number of possible outcomes that can be listed.

x

2 10 6 0 4 8

A random variable is continuous if it has an uncountable

number or possible outcomes, represented by the intervals

on a number line.

x

2 10 6 0 4 8

Probability Histogram

The probability histogram is very similar to a relative frequency histogram, but the vertical scale shows probabilities.

0 P(x) 1 for every individual value of x.

P(x) = 1 where x assumes all possible values.

Requirements for

Probability Distribution

In words, The probability of each value of the discrete random variable is between 0 and 1, inclusive. And , the sum of all the probabilities is 1.

Example

Is the following probability distribution?

Answer:

0.16 + 0.18 + 0.22 + 0.10 + 0.3 + 0.01 = 0.97 <1 , Not a probability distribution.

What if P(X=5) = 0.04?

Example

The number of patients seen in the ENT in any given hour is a random variable represented by x. The probability distribution for x is:

x

10

11

12

13

14

P(x)

.4

.2

.2

.1

.1

Find the probability that in a given hour:

a. exactly 14 patients arrive

b. At least 12 patients arrive

c. At most 11 patients arrive

p(x=14)= .1

p(x12)= (.2 + .1 +.1) = .4

p(x≤11)= (.4 +.2) = .6

Rolling 2 Dice (Red/Green)

Red\Green 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

X = Sum of the up faces of the two die. Table gives value of y for all elements

in S.

Rolling 2 Dice – Probability Mass Function

x p(x)

2 1/36

3 2/36

4 3/36

5 4/36

6 5/36

7 6/36

8 5/36

9 4/36

10 3/36

11 2/36

12 1/36

inresult can die 2 waysof #

X tosumcan die 2 waysof #)( Xp

Rolling 2 Dice – Probability Mass Function

Dice Rolling Probability Function

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

2 3 4 5 6 7 8 9 10 11 12

y

p(y

)

Mean, Variance and Standard Deviation of a Probability Distribution

µ = [x • P(x)] Mean

2 = [(x – µ)

2 • P(x)] Variance

2

= [x2 • P(x)] – µ

2 Variance (shortcut)

= [x 2 • P(x)] – µ

2 Standard Deviation

Identifying Unusual Results

Probabilities

Using Probabilities to Determine When

Results Are Unusual

Unusually high: x successes among n

trials is an unusually high number of

successes if P(x or more) ≤ 0.05.

Unusually low: x successes among n

trials is an unusually low number of

successes if P(x or fewer) ≤ 0.05.

Expected Value

The expected value of a discrete random

variable is denoted by E, and it

represents the mean value of the

outcomes. It is obtained by finding the

value of [x • P(x)].

E = [x • P(x)]

Example Question: Construct a probability distribution of discrete random variable of x(number of heads) while tossing a coin 3 times.

Number of heads (X) Frequency, f Probability P(X) X.P(X)

0 1 1/8 = 0.125 0

1 3 3/8 = 0.375 0.375

2 3 3/8 = 0.375 0.750

3 1 1/8 = 0.125 0.375

Sum = 8 Sum = 1 Sum = 1.5

The mean or expected value is 1.5

Expected Value

Gain, x P (x)

At a raffle, 500 tickets are sold for $1 each for two prizes of

$100 and $50. What is the expected value of your gain?

Because the expected value is negative, you can expect to lose $0.70 for each ticket you buy.

Winning no prize

1500

1500

498500

$99

$49

–$1

E(x) = ΣxP(x).

1 1 498$99 $49 ( $1)

500 500 500

$0.70

Binomial

Distributions

Binomial Probability Distribution

A binomial probability distribution results from

a procedure that meets all the following

requirements

1. The procedure has a fixed number of trials.

2. The trials must be independent. (The outcome of any

individual trial doesn’t affect the probabilities in the other

trials.)

3. Each trial must have all outcomes classified into two

categories (commonly referred to as success and failure).

4. The probability of a success remains the same in all

trials.

Notation for Binomial Probability Distributions

S and F (success and failure) denote the two possible categories

of all outcomes; p and q will denote the probabilities of S and F,

respectively, so

P(S) = p (p = probability of success)

P(F) = 1 – p = q (q = probability of failure)

n denotes the fixed number of trials.

x denotes a specific number of successes in n

trials, so x can be any whole number between

0 and n, inclusive.

Probability Formula for Binomial Distribution

where

n = number of trials

x = number of successes among n trials

p = probability of success in any one trial

q = probability of failure in any one trial (q = 1 – p)

Example

Decide whether the experiment is a binomial experiment.

If it is, specify the values of n, p, and q, and list the possible

values of the random variable x. If it is not a binomial

experiment, explain why.

You roll a die 10 times and note the number the die

lands on.

This is not a binomial experiment. While each trial

(roll) is independent, there are more than two possible

outcomes: 1, 2, 3, 4, 5, and 6.

Example

Decide whether the experiment is a binomial experiment.

If it is, specify the values of n, p, and q, and list the possible

values of the random variable x. If it is not a binomial

experiment, explain why.

You randomly select a card from a deck of cards, and

note if the card is an Ace. You then put the card

back and repeat this process 8 times.

This is a binomial experiment. Each of the 8 selections

represent an independent trial because the card is

replaced before the next one is drawn. There are only

two possible outcomes: either the card is an Ace or not.

4 152 13

p 8n 1 12

113 13

q 0,1,2,3,4,5,6,7,8x

Example Question: A health research shows that there is roughly 80% chance that a person of age 20 years will be alive at age of 65 years. Suppose that 3 people of age 20 years are selected at random. Find the probability that the number alive at age 65 years will be:

(a) Exactly two (b) at most one (c) al least one

(d) Determine the probability distribution of the number of alive at age 65.

Solution: Here , Probability of Success, p=80% = 0.80, q = 0.20, n = 3 (# of trials)

(a) P(X=2) = 3!/(2!).(1!) (0.8)^2.(0.2)^1 = 0.384

(b) P(X<=1) = P(X=0) + P(X=1) = 0.104

(c) P(X>=1) = 1 – P(X=0) = 1-0.008 = 0.992

(d) P(X=3) = 1.(0.8)^3.(0.2)^0 = 0.512

# of alive at age 65 (X) Probability [ P(X)]

0 0.008

1 0.096

2 0.384

3 0.512

Example

Question: A six sided die is rolled 3 times. Find the probability of rolling exactly one 6.

Answer: Consider a binomial experiment; rolling a 6 is a success while rolling any other number is a failure. The values for n, p, q, and x are n = 3, p = 1/6, q = 5/6 and x = 1. The probability of rolling exactly one 6 is:

xnxqp

xxn

nxP

!)!(

!)(

347.072

25)

216

25(3)

36

25)(

6

1(3)

6

5)(

6

1(3

)6

5()

6

1(

!1)!13(

!3)1(

2

131

P

Binomial Distribution: Formulas

Std. Dev. = n • p • q

Mean µ = n • p

Variance 2 = n • p • q

Where

n = number of fixed trials

p = probability of success in one of the n trials

q = probability of failure in one of the n trials

Finding Mean, Variance and Standard Deviation

In Pittsburgh, 57% of the days in a year are cloudy. Find the mean,

variance, and standard deviation for the number of cloudy days

during the month of June. What can you conclude?

Solution: There are 30 days in June. Using n=30, p = 0.57, and q =

0.43, you can find the mean variance and standard deviation as

shown.

Mean: = np = 30(0.57) = 17.1

Variance: 2 = npq = 30(0.57)(0.43) = 7.353

Standard Deviation: = √npq = √7.353 ≈2.71

Example

Q) In a survey, out of 10 students 3 skip classes. Find the mean, variance and standard deviation if 50 students are randomly selected.

Answer:

Here, n = 50, p = 3/10 = 0.3 and q = 0.7

So =50(0.3)=15

= 50(0.3)(0.7) = 10.5

μ np

2σ npq

Binomial Probability Distribution Example:

A bag contains 10 chips. 3 of the chips are red, 5 of the chips are

white, and 2 of the chips are blue. Four chips are selected, with

replacement. Create a probability distribution for the number of red

chips selected. p = the probability of selecting a red chip 3

0.310

q = 1 – p = 0.7

n = 4

x = 0, 1, 2, 3, 4

x P (x) 0 0.240

1 0.412

2 0.265

3 0.076

4 0.008

The binomial probability formula is used to find each probability.

Finding Probabilities Example:

The following probability distribution represents the probability of

selecting 0, 1, 2, 3, or 4 red chips when 4 chips are selected.

a.) P (no more than 3) = P (x 3) = P (0) + P (1) + P (2) + P (3)

x P (x) 0 0.24

1 0.412

2 0.265

3 0.076

4 0.008

b.) Find the probability of selecting at

least 1 red chip.

a.) Find the probability of selecting no

more than 3 red chips.

= 0.24 + 0.412 + 0.265 + 0.076 = 0.993

b.) P (at least 1) = P (x 1) = 1 – P (0) = 1 – 0.24 = 0.76

Complement

Graphing Binomial Probabilities Example:

The following probability distribution represents the probability of

selecting 0, 1, 2, 3, or 4 red chips when 4 chips are selected. Graph

the distribution using a histogram.

x P (x) 0 0.24

1 0.412

2 0.265

3 0.076

4 0.008

Selecting Red Chips

0

0.4

0.3

0.2

x

Pro

bab

ility

0.1

0.5

0 3 1

Number of red chips 4 2

P (x)