# IB Class Notes

## Binomial Distribution

Aims: By the end of this note, you will be able to

1. calculate the expectation of a Binomial Distribution;
2. calculate the variance of a Binomial Distribution &
3. solve problems with Binomial Distribution.

Binomial Distribution

### What is a Binomial Distribution?

1. A Binomial distribution is a special type of discrete probability distribution that has many real life application.
2. A Binomial experiment is one in which we have a repeated n number of trials where each trial has only two mutually exclusive outcomes. These trials are called Bernoulli trials. The results are usually referred to as success (the event occurs) or failure (the event does not occur). We often denote P(success) and P(failure) by p and q respectively. So q = 1-p and p+q =1. Also notice that each probabiliy of success is the same because each trial is independent of previous results.
The random variable X is usually the total number of sucesses in n trials.
Example 1:
Suppose a student is blindfolded and asked to throw three darts one after another at a dart board. The probability that his dart hit the dart board at each throw is 1/5. Represent the probability distribution of X in (a) Tree diagram, (b) Tabular form, and (c) Functional form.
Thus, p=1/5 and q=4/5. We will let H and M to denote Hit and Miss respectively.
Tree diagram: Tabular form: From the tree diagram it is easy to calculate the following probability distribution.
 x 0 1 2 3 P(X=x) (4/5)3 3(1/5)(4/5)2 3(1/5)2(4/5) (1/5)3

Functional form:

 P(X=x) = ( 3 )(1/5)x (4/5)3-x x
where x=0,1,2,3
3. In general, given that a Binomial experiment with n trials with x successes and (n-x) failures then
 P(X=x) = ( n )px (1 - p)n-x x
where p=P(success) and x=0,1,2,3,...,n

P(X=x) is called the Binomial probability distribution function.
Mathematicians often express the Binomial distribution as
X B(n,p) or X Bin(n,p)read as "X is distributed binomially with n trials and p probability of success."
4. Thus, X B(6,0.3) will have
 P(X=x) = ( 6 )(0.3)x (0.7)6-x x
where x=0,1,2,3,4,5,6

Example 2:
If X B(8,0.67) then find the following:
(a) P(X=5)
(b) P(X ≤ 5)
(c) P(X ≥ 6)
Solutions:
(a) P(X=5) = 8C5(0.67)5 (0.33)3
≈ 0.272 (3s.f.)
 Use of GDC. [2nd][VARS] for DISTR, go down tobinompdf( [ENTER] binompdf(8,0.67,5) [ENTER] Note: binompdf(n trials, p probability of success, x of X=x) pdf represents probability distribution function in the case of discrete data.

(b) P(X ≤ 5) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= 8C5(0.67)5 (0.33)3 + 8C4(0.67)4 (0.33)4 + ... + 8C0(0.67)0 (0.33)8
≈ 0.524 (3.s.f.)
 Use of GDC. In this case, working out the answer for (b) can be very tedious. I advise you to write up the above two lines as part of your workings and then use GDC for the actual calculation. [2nd][VARS] for DISTR, go down to binomcdf( [ENTER] binomcdf(8,0.67,5) [ENTER] Note: binomcdf(n trials, p probability of success, x of X £x) cdf represents cumulative distribution function.

(c) P(X ≥ 6) = P(X=6) + P(X=7)) + P(X=8)
= 1 - P(X<6)
Since this is a discrete data then
P(X ≥ 6) = 1- P(X ≤ 5)
= 1 - 0.524
≈ 0.476 (3 s.f.)

Example 3:
A biased coin has P(tail)=0.7 at each toss. The coin is tossed 9 times.
(a) Find the probability that head occurs 6 times.
(b) Find the probability that tail occurs at least 3 times.

Solutions: Let H and T to represent the total number of Head and Tail respectively. Note that P(H)=0.3.
(a) P(H=6) = 9C6(0.3)6(0.7)3
P(H=6) ≈ 0.0210 (3 s.f.)

(b) P(T ≥ 3) = 1 - P(T<3)
= 1 - P(T ≤ 2)
= 1 - 0.004291
≈ 0.996 (3 s.f.)

5. Properties.
If the random variable X is such that X ~ B(n,p) then we have
(a) the expected value of X is m = E(X) = np.

(b) the mode of X is the value of x such that P(X=x) is largest.
In example 1 above the mode is x=2.

(c) the variance of X is Var(X) = np(1-p) = npq.

Note that the standard deviation of X is √(npq).

Example 4:
The random variable X ∼B(15,p) where p < 0.5. Given that Var(X)=3.4125, find the following:
(a) the value of p
(b) E(X)
(c) P(X=2)

Solutions:
(a) Var(X)=npq
15p(1-p) = 3.4125
p(1-p) = 0.2275
p2 - p + 0.2275 = 0
p=0.35 or p=0.65
Since p<0.5 then p=0.35.

(b)E(X)= np
E(X) = 15(0.35)
E(X) = 5.25

(c)P(X=2) = 15C2(0.35)2(0.65)13
P(X=2) ≈ 0.0476 (3 s.f.)

Exercise: Let X ∼ B(n,0.6) and E(X)=15. Find n and standard deviation of X.

Answer: n =25, s = 2.45 (3 s.f.)

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