Download Dealing With Uncertainty

Minds on!






If a batter has a batting average of .300, how
many hits would you expect him to get in:
10 at-bats?
50 at-bats?
4 at-bats?
n at-bats?
3; 15; 1.2; 0.300n
Warm up











A roulette wheel contains the following 38 numbers:
 0, 00, 1-36 (18 red, 18 black)
What is P(red)?
18/38 = 0.47
What is P(red,red)?
(18/38)2 = 0.22
What is P(10 consecutive red)?
(18/38)10 = 0.00057
What is P(12 of 16 red)?
P(12R 4B) = (18/38)12(20/38)4 = 0.0000098
But…how many ways can this happen?
It’s complicated.
But…all those things happened one day
when I visited the 1000 Islands Casino!
5.3 Binomial Distributions
Chapter 5.3 – Probability Distributions and Predictions
Learning goal: Use Binomial Probabilities to calculate
the probability of k successes in n trials
Questions? p. 176 #1, 3b, 6, 8-10
MSIP/Home Learning: p. 299 #1, 3, 7, 8–12
Our Problem…


In any at-bat, a batter either gets a hit or he
doesn’t (2 outcomes: success or failure)
Suppose his lifetime batting average is 0.292





P(hit) = 0.292 in any at-bat
As the leadoff hitter he averages 5 AB per
game
What are the possible outcomes for the
number of hits he gets in 5 AB?
0, 1, 2, 3, 4, 5 hits are possible
Are they all equally likely?
Binomial Probability Distribution for a batter
Binomial Experiments

Any experiment that has the following
properties:







n identical trials
Two possible outcomes for each trial: 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,
Switzerland, 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, X
X is termed a binomial random variable
Its probability distribution is called a binomial
distribution
The following formula provides a method of
solving complex probability problems…
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:
n k
nk
P ( X  k )    p 1  p 
k 
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
3
5 3
p=½
 5  1  
1
P  X  3     1  
n=5
2
 3  2  
k=3
3
2
5
1 1
1
 10      10  
2 2
2
10
5


32 16
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?
P X  3  P ( X  3)  P ( X  4)  P ( X  5)
5  5  1 
    
16  4  2 
5 5
1
  
16 32 32
 1   5  1 
     
 2   5  2 
1

2
4
5
1
 
2
0
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) = np
Example: Consider a baseball player who
has a lifetime batting average of 0.292
Let success be a hit where p = 0.292
a. What is the probability that he will get
no hits in the next 5 at bats?
p  0.292, n  5, k  0
 5
0
5
P X  0   0.292 0.708
 0
 110.178  0.178
so there is a 0.178 probabilit y that
he will get no hits in 5 times at bat.
b. What is the probability that he gets 2
hits in the next 8 at bats?
p  0.292, n  8, k  2
8
2
6
P X  2   0.292 0.708
 2
 280.0850.126  0.3
so there is a 0.3 probabilit y that
he will get 2 hits in 8 times at bat.
c. What is the probability that he gets at
least 1 hit in the next 10 at bats?
p  0.292, n  10
P X  1  P X  1  P( X  2)  P( X  3)  ...  P( X  10)
OR
 1  P( X  0)
c. What is the probability that he gets at
least 1 hit in the next 10 at bats?
p  0.292, n  10
P X  1  1  P X  0 
10 
0
10
 1   0.292  0.708
0
 1  110.032   0.968
so there is a 0.968 probabilit y that
there will be at least 1 hit in 10 times at bat
d. What is the expected number of hits in
the next 10 at bats?




E(X) = np
E(X) = (10)(0.292)
= 2.92 ≈ 3
therefore the player is expected to get
approximately 3 hits in the next 10 at bats
MSIP/ Home Learning

p. 299 #1, 3, 7, 8 – 12
Warm up

Describe a situation that could be represented by the
following binomial probability:
 20 
15
P( X  5)   (0.17)5 0.83  0.135
5

It could be the probability that:

A 6 appears appears 5 times in 20 rolls of a
standard die
A player who scores on 17% of his shots scores
on 5 shots out of 20
In a deck with the aces removed, a K or Q is
drawn 5 times in 20 draws


5.4 Normal Approximation
of the Binomial Distribution
Chapter 5 – Probability Distributions and Predictions
Learning Goal: Use the Normal Distribution (z-scores)
to approximate Binomial Probabilities
Questions? p. 299 #1, 3, 7, 8 – 12
MSIP / Home Learning: p. 311 # 4-10
Quiz Wednesday
Recall…

the probability of k successes in n trials (where p is
the probability of success) is
n k
nk
P ( X  k )    p 1  p 
k 

this formula can only be used if we have a binomial
experiment:


each trial is identical
the outcomes are either success or failure
This calculation is easy in simple cases…

Find the probability of 30 heads in 50 trials
P(30 heads in 50 trials ) 
 50  30
 0.5 1  0.55030  0.042
 30 



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 this calculation for 20, 21,
22, …, 30
But…there is an easier way
Graphing the Binomial Distribution



If the shape of the distribution is Normal, we
can solve complex problems using z-Scores
the question is: is the binomial distribution
a Normal one?
if the number of trials is sufficiently 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?
Line Scatter Plot
0.12
0.10
probability

Binomial Distribution
0.08
0.06
0.04
0.02
0.00
0
5
10 15 20 25 30 35 40 45 50 55
heads
So how do we work with all this

It turns out that a binomial distribution can be
approximated by a Normal distribution if:


np > 5 and
n(1 – p) > 5
If this is the case, the distribution is approximated
by the Normal distribution
X ~ N ( x, 2 ) where x  np and   np(1  p)
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
Boundary 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
A binomial distribution





Suppose a batter with a
0.270 lifetime batting
average gets 9 at-bats in a
double-header (2 games).
What range of values
correspond to getting 4 or
more hits?
Greater than 3.5
What range of values
correspond to getting
between 4 and 6 hits?
3.5 – 6.5
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
Test:
 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
x  50(0.5)  25
  50(0.5)(1  0.5)  12.5  3.54
Example 1 continued

The boundary value is 19.5





20 is represented by the interval 19.5-20.5
Less than 20 is actually less than 19.5
z = 19.5 – 25 = -1.55  0.0606
3.54
There is about 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
= 7/12
now we can do some calculations
Example 2 continued
7
 5
np  40    23.3  5
n(1  p)  40    16.6  5
 12 
 12 
P( x  20)  ?
x  np  23.3
  np(1-p)  9.72  3.118
20.5  23.3
z
 0.91  0.1814 or 18.14%
3.118

the probability of getting a sum greater than 6 on
more than half of the trials is 100 – 18 = 82%
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

Test:




np = 200(0.25) = 50 > 5
n(1 – p) = 200(0.75) = 150 > 5
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
x  np  200 (0.25)  50
  np1  p   200 (0.25)( 0.75)  37 .5  6.124
49 .5  50
First Case z 
 0.081  0.4681
6.124
60 .5  50
Second Case z 
 1.715  0.9564
6.124

the probability of having between 50 and 60 pink
mittens (inclusive) drawn is 0.9564 – 0.4681 =
0.4883 or about 49%
MSIP/ Home Learning


Read the example on page 310
do p. 311 # 4-10