Warm up

How many ways are there to spell SUCCESS? (moving only along diagonals)

S

S S

E E E

C C

C C C

U U

S

Warm up

How many ways are there to spell SUCCESS?

S18

S9 S9

E3 E6 E3

C3 C3

C1 C2 C1

U1 U1

S1

Binomial Distributions

Chapter 5.3 – Probability Distributions and PredictionsMathematics of Data Management (Nelson)MDM 4U

Our Problem… A batter either gets a hit or he doesn’t Suppose his lifetime batting average is 0.292

the probability that he will get a hit in any AB (at-bat) is 0.292

Why is lifetime average more reliable than season average?

As the leadoff hitter he gets 5 AB in a game What are the possible outcomes for the

number of hits he gets in 5 AB? Do they have different probabilities?

Binomial Experiments a binomial experiment is any experiment that

has the following properties: n identical trials Two possible outcomes for each trial, termed success and failure

The probability of success is p The probability of failure is 1 – p The probabilities remain constant The trials are independent

Bernoulli Trials - repeated independent trials with 2 possible outcomes (success and failure)

Bernoulli?

Jakob Bernoulli (Basel, December 27, 1654 - August 16, 1705)

Swiss Mathematician one of the great names in

probability theory

Binomial Distributions In a binomial experiment: The number of successes in n repeated

Bernoulli Trials is a discrete random variable (usually called X)

X is termed a binomial random variable and its probability distribution is called a binomial distribution

The following formula provides a method of solving highly complex situations involving probability

Binomial Probability Distribution Consider a binomial experiment in which there

are n Bernoulli trials, each with a probability of success p

The probability of k successes in the n trials is given by:

knk ppk

nkXP

1)(

Example 1a Consider a game where a coin is flipped 5 times.

You win the game if you get exactly 3 heads. What is the probability of winning?

We will let heads be a success n = 5 p = ½ k = 3

16

5

32

10

2

110

2

1

2

110

2

11

2

1

3

53

523

353

XP

Another solution

16

5

32

10

5.0105.05.010

5.05.03

53

523

23

XP

Example 1b Suppose the game is changed so that you win if

you get at least 3 heads what is the probability of winning now?

2

1

32

1

32

5

16

5

2

1

2

1

5

5

2

1

2

1

4

5

16

5

)5()4()3(3054

XPXPXPXP

The Batting Example the Expected Value of a binomial experiment

that consists of n Bernoulli trials with a probability of success, p, on each trial is E(X) = n(p)

Example: Consider a baseball player who has a lifetime batting average of 0.292

This means that his probability of getting a hit each time he is at bat is 0.292

Let a hit be a success where p = 0.292

a. What is the probability that he will get no hits in the next 5 at bats?

bat.at times5in hits noget willhe

y thatprobabilit 0.178 a is thereso

178.0178.011

708.0292.00

50

0,5,292.0

50

XP

knp

b. What is the probability of 2 hits in the next 8 at bats?

bat.at times8in hits 2get willhe

y thatprobabilit 0.300 a is thereso

300.0126.0085.028

708.0292.02

82

2,8,292.0

62

XP

knp

c. What is the probability of at least 1 hit in the next 10 at bats?

batat times10in hit 1least at be willthere

y thatprobabilit 0.968 a is thereso

968.0032.0111

708.0292.00

101

01111

10,292.0

100

XPXPXP

np

E(X) = np E(X) = (10)(0.292) = 2.92 ≈ 3 therefore the player is expected to get 3 hits

in the next 10 at bats

d. What is the expected number of hits in the next 10 at bats?

MSIP/ Homework

p. 299 #1, 3, 7, 8 – 12

Normal Approximation of the Binomial Distribution

Chapter 5.4 – Probability Distributions and PredictionsMathematics of Data Management (Nelson)MDM 4U

Recall… the probability of k successes in n trials (where p is

the probability of success) is

this formula can only be used if we have a binomial distribution: each trial is identical the outcomes are either success or failure

knk ppk

nkXP

1)(

This calculation is easy in simple cases… Find the probability of 30 heads in 50 trials

So there is about a 4.2% chance However, if we wanted to find the probability of

tossing between 20 and 30 heads in 50 trials, we would need to perform 11 of these calculations

But…there is an easier way

042.05.015.030

50

) trials50in heads 30(

305030

P

Graphing the Binomial Distribution If the distribution is normal, we can solve

complex problems in the same way we did in the last chapter

the question is: is the binomial distribution a normal one?

it turns out that if the number of trials is relatively large, the binomial distribution approximates a normal curve

What does it look like?

when graphed the distribution of probabilities of heads looks like this

what will the mean be?

what will the standard deviation be?

prob

abili

ty

0.00

0.02

0.04

0.06

0.08

0.10

0.12

heads0 5 10 15 20 25 30 35 40 45 50 55

Binomial Distribution Line Scatter Plot

So how do we work with all this it turns out that a binomial distribution can be

approximated by a normal distribution if: n x p > 5 and n x (1 – p) > 5

if this is the case, the distribution is approximated by the normal distribution

)1( and where),( 2 pnpnpxxN

But doesn’t a normal curve represent continuous data and a binomial distribution represent discrete data? Yes! so to use a normal approximation we have to

consider a range of values rather than specific discrete values

The interval for a value is from 0.5 below to 0.5 above, i.e., the interval for 10 goes from 9.5 to 10.5

Example 1

Tossing a coin 50 times, what is the probability that you will get tails less than 20 times

let success be tails, so n = 50 and p = 0.5 n x p = 50(0.5) = 25 > 5 n x (1 - p) = 50(1 - 0.5) = 25 > 5 now we can find the mean and the standard deviation

54.35.12)5.01)(5.0(50

25)5.0(50

x

Example 1 continued we will consider 0-19.5 times (values below

20 - the interval from 19.5-20.5), and use it to calculate a z-score

z = 19.5 – 25 = -1.55 3.54 therefore P(X < 19.5) = P(z < -1.55) = 0.0606 there is a 6% chance of less than 20 tails in

50 attempts

In terms of the normal curve, it looks like this all the values less than

19.5 are found in the shaded area

19.5 25.0

Example 2 Two dice are rolled and the sum recorded 40

times. What is the probability that a sum greater than 6 occurs in over half of the trials?

let p be the probability of getting a sum greater than 6

p = 6/36 + 5/36 + 4/36 + 3/36 + 2/36 + 1/36 p = 7/12 now we can do some calculations

Example 2 continued

the probability of getting a sum greater than 6 on at least half of the trials is 100 – 18 = 82%

7 540 23.3 5 (1 ) 40 16.6 5

12 12

( 20) ? x np 23.3

np(1-p) 9.72 3.118

20.5 23.30.91 0.1814 18.14%

3.118

np n p

P x

z or

Example 3 you have a drawer with one blue mitten, one

red mitten, one pink mitten and one green mitten

if you closed your eyes and picked a mitten at random 200 times (with replacement) what is the probability of choosing the pink mitten between 50 and 60 times (inclusive)?

so, success is considered to be drawing a pink mitten, with n = 200 and p = 0.25

Example 3 Continued check to see whether the normal

approximation can be used np = 200(0.25) = 50 n(1 – p) = 200(0.75) = 150 since both of these are greater than 5 the

binomial distribution can be approximated by the normal curve

now find the mean and standard deviation

Example 3 Continued

the probability of having between 50 and 60 pink mittens (inclusive) drawn is 0.9564 – 0.4681 = 0.4883 or about 49%

9564.0715.1124.6

505.60 Case

4681.0081.0124.6

505.49 Case

124.65.37)75.0)(25.0(2001

50)25.0(200

zSecond

zFirst

pnp

npx

MSIP/ Homework

Read the example on page 310 do p. 311 # 4-10