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**The Binomial Probability Distribution**

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, and is given by the formula:

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)

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) = σ^{2} = npq**Note:** In a binomial distribution, only 2 parameters, namely n and p, are needed to determine the probability.**Example 1.**

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

(a) No fives turning up?

(b) 1 five?

(c) 3 fives?**Solution.**

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.

(b) Here, x = 1.

(c) Here, x = 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?**Solution.**

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:

Histogram of this distribution:

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

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.**

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**Solution.**

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

X = success in getting through.

Probability of 7 successes in 10 attempts:

Probability = P(X = 7)

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?**Solution.**

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

Let X = number of hits.

Let x0 = no hits, x_{1} = 1 hit, x_{2} = 2 hits, etc.

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

**Example 5.**

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.]**Solution.**

The probability of getting a boy is

Let X = number of boys in the family.

Here,

When x = 3:

So the probability of getting at least 3 boys is:

Probability = P(X ≥ 3)**NOTE:** We could have calculated it like this:

**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?**Solution.**

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:

One reject. That is, when x = 1

Two rejects. That is, when x = 2:

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

(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

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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