A **binomial** experiment is one that possesses the following properties:

The experiment consists of

*n*repeated trials;Each trial results in an outcome that may be classified as a

**success**or a**failure**(hence the name,**binomial**);The probability of a success, denoted by

*p*, remains constant from trial to trial and repeated trials are independent.

The number of successes *X* in *n* trials of a binomial experiment is called a **binomial random variable**.

The probability distribution of the random variable *X* is called a **binomial distribution.**

*n* = the number of trials

*x* = 0, 1, 2, ... *n [sample space]*

*p* = the probability of success in a single trial

*q* = the probability of failure in a single trial

*q*= 1 −*p*

## Mean and Variance of Binomial Distribution

If *p* is the probability of success and *q* is the probability of failure in a binomial trial, then the expected number of successes in *n* trials (i.e. the mean value of the binomial distribution) is

*E*(*X*) = *μ* = *np*

The **variance** of the binomial distribution is

*V*(*X*) = *σ*2 = *npq*

In a binomial distribution, only **2** parameters, namely *n* and *p*, are needed to determine the probability.

**EXAMPLES**

**A die is tossed 3 times. What is the probability of no fives turning up?**

**A die is tossed 3 times. What is the probability of 1 five turning up?**

**A die is tossed 3 times. What is the probability of 3 fives ?**

**Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?**

This is a **binomial **distribution because there are only 2 outcomes (the patient dies, or does not).

*X =*number who recover

n = 6 ; x=4

p= 0.25 probability of success

q=0.75 probability of failure

Let's make a table and graph first.

Insert a Scatter Chart, then change the siries type to Bar Chart

**Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover? **

P(x=4)=3.29%

**A Japanese archer finds that on the average he hits the middle of a target 4 times out of 5. If he fires 4 shots, what is the probability of**

**(A) more than 2 hits?**

n= 4 p=4/5

P(X>2)=P(x=3)+P(x=4)

**(B) at least 3 misses?**

3 misses =1 hit

4 misses = 0 hits.

**A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain**

**(A) no more than 2 rejects?**

In this case, "success" means rejection.

n=10 p=0.12

P(X<=2)=P(X=0)+P(X=1)+P(X=2)

**(B) at least 2 rejects?**

SOLUTION 1:

SOLUTION 2:

=1−*P*(*X*≤1)

=1−(*P*(*x*0)+*P*(*x*1))

=1−(0.2785+0.37977)

=0.34173

**The ratio of boys to girls at birth in Singapore is at 1.09 : 1.**

**What proportion of Singapore families with exactly 6 children will have at least 3 boys? (Ignore the probability of multiple births.)**

**A manufacture is making a product with 20% defect rate. If we select 5 items at the end of the assembly line, what is the probability of having 1 defected item? **

C(5,1)=5 WE GOT 5 DIFFRENCE WAYS HOW TO DO THAT.

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