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# 12. The Binomial Probability Distribution

### Notation

We use upper case variables (like X and Z) to denote random variables, and lower-case letters (like x and z) to denote specific values of those variables.

A binomial experiment is one that possesses the following properties:

1. The experiment consists of n repeated trials;

2. Each trial results in an outcome that may be classified as a success or a failure (hence the name, binomial);

3. 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, and is given by the formula:

P(X)=C_x^n p^x q^(n-x)

where

n = the number of trials

x = 0, 1, 2, ... n

p = the probability of success in a single trial

q = the probability of failure in a single trial

(i.e. q = 1 − p)

C_x^n is a combination

P(X) gives the probability of successes in n binomial trials.

## 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) = σ2 = npq

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

### Example 1 Image source

A die is tossed 3 times. What is the probability of

(a) No fives turning up?

(b) 1 five?

(c) 3 fives?

This is a binomial distribution because there are only 2 possible outcomes (we get a 5 or we don't).

Now, n = 3 for each part. Let X = number of fives appearing.

(a) Here, x = 0.

P(X=0) =C_x^np^xq^[n-x] =C_0^3 (1/6)^0 (5/6)^3 =125/216 =0.5787

(b) Here, x = 1.

P(X=1) =C_x^np^xq^[n-x] =C_1^3 (1/6)^1 (5/6)^2 =75/216 =0.34722

(c) Here, x = 3.

P(X=3)=C_x^np^xq^[n-x] =C_3^3 (1/6)^3 (5/6)^0 =1/216 =4.6296times10^-3

### Example 2

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).

Let X = number who recover.

Here, n = 6 and x = 4. Let p = 0.25 (success, that is, they live), q = 0.75 (failure, i.e. they die).

The probability that 4 will recover:

P(X) = C_x^np^xq^[n-x] =C_4^6(0.25)^4(0.75)^2 =15times 2.1973 times 10^-3 =0.0329595

#### Histogram of this distribution:

We could calculate all the probabilities involved and we would get:

 X text[Probability] 0 0.17798 1 0.35596 2 0.29663 3 0.13184 4 3.2959 times 10^-2 5 4.3945times10^-3 6 2.4414times10^-4

The histogram is as follows:

It means that out of the 6 patients chosen, the probability that:

• None of them will recover is 0.17798,
• One will recover is 0.35596, and
• All 6 will recover is extremely small.

### Example 3 Image source

In the old days, there was a probability of 0.8 of success in any attempt to make a telephone call. (This often depended on the importance of the person making the call, or the operator's curiosity!)

Calculate the probability of having 7 successes in 10 attempts.

Probability of success p = 0.8, so q = 0.2.

X = success in getting through.

Probability of 7 successes in 10 attempts:

text[Probability]=P(X=7)

=C_7^10(0.8)^7(0.2)^[10-7]

=0.20133

#### Histogram

We use the following function

C(10,x)(0.8)^x(0.2)^[10-x]

to obtain the probability histogram:

### Example 4

A (blindfolded) marksman finds that on the average he hits the target 4 times out of 5. If he fires 4 shots, what is the probability of

(a) more than 2 hits?

(b) at least 3 misses?

Here, n = 4, p = 0.8, q = 0.2.

Let X = number of hits.

Let x0 = no hits, x1 = 1 hit, x2 = 2 hits, etc.

(a) P(X)=P(x_3)+P(x_4)

=C_3^4(0.8)^3(0.2)^1+ C_4^4(0.8)^4(0.2)^0

=4(0.8)^3(0.2)+(0.8)^4

=0.8192

(b) 3 misses means 1 hit, and 4 misses means 0 hits.

P(X)=P(x_1)+P(x_0)

=C_1^4(0.8)^1(0.2)^3+ C_0^4(0.8)^0(0.2)^4

=4(0.8)^1(0.2)^3+(0.2)^4

=0.0272

### Example 5 Image source

The ratio of boys to girls at birth in Singapore is quite high 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.)

[Interesting and disturbing trivia: In most countries the ratio of boys to girls is about 1.04:1, but in China it is 1.15:1.]

The probability of getting a boy is 1.09/(1.09+1.00)=0.5215

Let X = number of boys in the family.

Here,

n = 6,
p = 0.5215,
q = 1 − 0.52153 = 0.4785

When x=3:

 P(X) =C_x^np^xq^(n-x) =C_3^6(0.5215)^3(0.4785)^3 =0.31077

When x=4:

 P(X) =C_4^6(0.5215)^4(0.4785)^2 =0.25402

When x=5:

P(X) =C_5^6(0.5215)^5(0.4785)^1 =0.11074

When x=6:

P(X) =C_6^6(0.5215)^6(0.4785)^0 =2.0115xx10^-2

So the probability of getting at least 3 boys is:

"Probability"=P(X>=3)

=0.31077+0.25402+ 0.11074+ 2.0115xx10^-2

=0.69565

NOTE: We could have calculated it like this:

P(X>=3) =1-(P(x_0)+P(x_1)+P(x_2))

### Example 6

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? (b) at least 2 rejects?

Let X = number of rejected pistons

(In this case, "success" means rejection!)

Here, n = 10, p = 0.12, q = 0.88.

(a)

No rejects. That is, when x=0:

P(X) =C_x^np^xq^(n-x) =C_0^10(0.12)^0(0.88)^10 =0.2785

One reject. That is, when x=1

P(X) =C_1^10(0.12)^1(0.88)^9 =0.37977

Two rejects. That is, when x=2:

P(X) =C_2^10(0.12)^2(0.88)^8 =0.23304

So the probability of getting no more than 2 rejects is:

"Probability"=P(X<=2)

=0.2785+ 0.37977+ 0.23304

=0.89131

(b) We could work out all the cases for X = 2, 3, 4, ..., 10, but it is much easier to proceed as follows:

"Probablity of at least 2 rejects"

=1-P(X<=1)

 =1-(P(x_0)+P(x_1))

 =1-(0.2785+0.37977)

=0.34173

#### Histogram

Using the function g(x)=C(10,x)(0.12)^x(0.88)^(10-x) and finding the values at 0, 1, 2, ..., gives us the histogram:

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