1 Section 7.5B Binomial Random Variables

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1 Section 7.5B Binomial Random Variables

Transcript Of 1 Section 7.5B Binomial Random Variables

+ Section 7.5B

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Binomial Random Variables

Learning Objectives
After this section, you should be able to…
 DETERMINE whether the conditions for a binomial setting are met
 COMPUTE and INTERPRET probabilities involving binomial random variables
 CALCULATE the mean and standard deviation of a binomial random variable and INTERPRET these values in context

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Binomial Random Variables

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 Binomial Settings
When the same chance process is repeated several times, we are often interested in whether a particular outcome does or doesn’t happen on each repetition. In some cases, the number of repeated trials is fixed in advance and we are interested in the number of times a particular event (called a “success”) occurs. If the trials in these cases are independent and each success has an equal chance of occurring, we have a binomial setting.
Definition:
A binomial setting arises when we perform several independent trials of the same chance process and record the number of times that a particular outcome occurs. The four conditions for a binomial setting are
B • Binary? The possible outcomes of each trial can be classified as “success” or “failure.”
I • Independent? Trials must be independent; that is, knowing the result of one trial must not have any effect on the result of any other trial.
N • Number? The number of trials n of the chance process must be fixed in advance.
S • Success? On each trial, the probability p of success must be the same.
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From Blood Type to Aces (continued)

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From Blood Type to Aces (continued)

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Binomial Random Variables

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 Binomial Random Variable
Consider tossing a coin n times. Each toss gives either heads or tails. Knowing the outcome of one toss does not change the probability of an outcome on any other toss. If we define heads as a success, then p is the probability of a head and is 0.5 on any toss.
The number of heads in n tosses is a binomial random variable X. The probability distribution of X is called a binomial distribution.
Definition:
The count X of successes in a binomial setting is a binomial random variable. The probability distribution of X is a binomial distribution with parameters n and p, where n is the number of trials of the chance process and p is the probability of a success on any one trial. The possible values of X are the whole numbers from 0 to n.
Note: When checking the Binomial condition, be sure to check the BINS and make sure you’re being asked to count the number of successes in a certain number of trials!

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Binomial Example

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 Binomial Probabilities Example 1
In a binomial setting, we can define a random variable (say, X) as the number of successes in n independent trials. We are interested in finding the probability distribution of X.

Type O Example

Each child of a particular pair of parents has probability 0.25 of having type O blood. Genetics says that children receive genes from each of their parents independently. These parents have 5 children. What is the probability that none of the children has type O blood?

•The count X of children with type O blood is a binomial random variable with n = 5 trials and probability p = 0.25 of a success on each trial. •It is reasonable to assume that each child’s blood type is independent of each other.
•In this setting, a child with type O blood is a “success” (S) and a child with another blood type is a “failure” (F).
•What’s P(X = 0)?

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Binomial Example

Example 1

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1) What is the probability that none of the children has type O blood?

P(Success) = (0.25) P(Failure) = (0.75)
P(X=0) = P(FFFFF) = (0.75)(0.75)(0.75)(0.75)(0.75) = (0.75)5 = .2373
There is about a 24% chance that none of the 5 children have type O blood.

Note: There is only one possible outcome in this scenario to have none of the children have type O blood

2) What is the probability that ONE of the children has 10 type O blood?

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2) What is the probability that EXACTLY ONE of the children has type O blood (continued)?

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3) What is the probability that EXACTLY TWO of the

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children have type O blood?

Probability of 2 with Type O blood
P(SSFFF) = (0.25)(0.25)(0.75)(0.75)(0.75) = (0.25)2(0.75)3 = 0.02637
However, there are a number of different arrangements in which 2 out of the 5 children have type O blood:
SSFFF SFSFF SFFSF SFFFS FSSFF FSFSF FSFFS FFSSF FFSFS FFFSS
Verify that in each arrangement, P(X = 2) = (0.25)2(0.75)3 = 0.02637 Therefore, P(X = 2) = 10(0.25)2(0.75)3 = 0.2637
There is about a 26% chance that 2 of the 5 children have type O blood.

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 Binomial Coefficient

Binomial Random Variables

Note, in the previous example, any one arrangement of 2 S’s and 3 F’s had the same probability. This is true because no matter what arrangement, we’d multiply together 0.25 twice and 0.75 three times.

We can generalize this for any setting in which we are interested in k successes in n trials. That is,
P(X  k)  P(exactly k successes in n trials) = number of arrangements  pk (1 p)nk

Definition:

The number of ways of arranging k successes among n observations is given by the binomial coefficient
n n!   k k!(n  k)!

for k = 0, 1, 2, …, n where

and 0! = 1.

n! = n(n – 1)(n – 2)•…•(3)(2)(1)

Binomial Random Variables

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 Binomial Probability
The binomial coefficient counts the number of different ways in which k successes can be arranged among n trials. The binomial probability P(X = k) is this count multiplied by the probability of any one specific arrangement of the k successes.

Binomial Probability – B(n,p)

If X has the binomial distribution with n trials and probability p of success on

each trial, the possible values of X are 0, 1, 2, …, n. If k is any one of

these values,

P(X



k)



n  pk (1

p) n  k

k

Number of arrangements of k successes

Probability of k successes

Probability of n-k failures

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Binomial Probabilities Example 3 (an easier way to

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find the number of possible outcomes)

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Now, Find These Binomial Coefficients

 5   0

 5   4

 5   1

 5   3

 5   5

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 Example 4 - Using the Binomial Probability Formula

 Type O Blood Example B(5, .25)

 Let X = the number of children with type O blood.

(a) Find the probability that exactly 3 of the children have type O blood.
5 P(X  3)   (0.25)3(0.75)2  10(0.25)3(0.75)2  0.08789
3

(b) Should the parents be surprised if more than 3 of their children have type

O blood?

To answer this, we need to find P(X > 3).

P(X  3)  P(X  4)  P(X  5)

5

5

  (0.25)4 (0.75)1   (0.25)5(0.75)0

4

5

 5(0.25)4 (0.75)1  1(0.25)5(0.75)0

 0.01465  0.00098  0.01563

Since there is only a 1.5% chance that more than 3 children out of 5 would have Type O blood, the parents should be surprised!

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Always State B(5, .25)

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Example 4 – Using the Calculator to find Binomial Probabilities
Answer Example 4 using the TI84

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 EXAMPLE 5: Expected Value and Expected Variance of a Binomial Distribution

 We describe the probability distribution of a binomial random variable just like any other distribution – by looking at the shape, center, and spread.

 Describe our probability distribution for:

X = number of children with type O blood in a family with 5 children.

xi

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pi 0.2373 0.3955 0.2637 0.0879 0.0147 0.00098

Shape: The probability distribution of X is skewed to the right. It is more likely to have 0, 1, or 2 children with type O blood than a larger value.

Center: The median number of children with type O blood is 1. Based on our formula for the mean (find the expected value):
X  xi pi  (0)(0.2373) 1(0.39551)  ... (5)(0.00098)
 1.25

Spread: The variance of X is X2  (xi X )2 pi  (0 1.25)2(0.2373)  (11.25)2(0.3955)  ...
(5 1.25)2(0.00098)  0.9375 The standard deviation of X is X  0.9375  0.968

Binomial Random Variables

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Mean and Standard Deviation of a Binomial Distribution Notice, the mean µX = 1.25 can be found another way. We can use the parameters n and p; and the method below:
Mean and Standard Deviation of a Binomial Random Variable
If a count X has the binomial distribution with number of trials n and probability of success p, the mean and standard deviation of X are
X  np  X  np(1 p)
Note: These formulas work ONLY for binomial distributions. They can’t be used for other distributions!

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TypeBloodBinomialProbabilityChildren