## Binomial Sequences

A Bernoulli sequence is used to describe a sequence of trials that result in one of two outcomes (either a success or a failure). In addition, the probability of success in a single trial, p, must be constant. Hence, the probability of failure is (1-p). Note that the probability of success and failure must add up to one. This is why if the probability of success is p, the probability of failure must be (1-p). Lastly, a Bernoulli sequence is independent and the outcome of the trial is not affected by the outcome of any previous trial.

## Binomial Distribution

Now let’s look at binomial probability distribution. The number of successes in a Bernoulli sequence of n trial is called a binomial random variable and in general, we can find the probability of achieving x successes in n independent trials of a binomial experiment by using this formula.

$Pr(X=x)=\binom{n}{x}p^{x}(1-p)^{n-x}$ x=0,1,….n

where $\binom{n}{x}=\frac{n!}{x!(n-x)!}$

### Practice Question

The probability that a soccer player scoring a goal has ever scored a goal before is 0.72. Find the probability that of five soccer players chosen at random, at least three have scored a goal before.

Solution:

If X is the number of soccer player who have scored goals before, then

Pr(X=x) (0.72)x(0.28)5-x  x=0,1,2,3,4,5

Since the probability that they have scored a goal before is 0.72, according to the Bernoulli sequences, the probability that they have not scored a goal before is 1-0.72 which is 0.28

And Pr (X≥3)= Pr (X=3)+ Pr (X=4)+ Pr (X=5)

Since the question ask for at least three who have scored a goal before, this include 3,4 or 5 soccer players who have scored a goal before

$=\binom{5}{3}(0.72)^{3}(0.28)^{2}+\binom{5}{4}(0.72)^{4}(0.28)^{1}+\binom{5}{5}(0.72)^{5}(0.28)^{0} \newline =\frac{5!}{3!(5-3)!}(0.72)^{3}(0.28)^{2}+\frac{5!}{4!(5-4)!}(0.72)^{4}(0.28)^{1}+\frac{5!}{5!(5-0)!}(0.72)^{5}(0.28)^{0} \newline =0.8624$

Therefore, the probability that out of give soccer players, at least three have scored a goal before is 0.8624

## Expected Value and Variance

This is very similar to what we have learnt in the previous discrete random variable video. In order to find the expected value for a binomial random variable, we have to multiple the number of trials by the probability of success. It is given by E(X)= np, and the formula for variance is Var(X)=np(1-p), the number of trials multiplied by probability of success and probability of failure

### Practice Question

The probability of raining in this winter is known to be 0.2. Of 100 days, how many days would we expect to be raining? Find the standard deviation of the number of days it would be raining in this winter.

The number of rainy day is a binomial random variable, with parameter p=0.2 and n=100. We can expect:

np= 0.2 x 100 days= 20 days will be rainy.

Var(x)=np(1-p)

=100 x 0.2 x 0.8=16

And hence a standard deviation

$\sigma=\sqrt{var(x)}=\sqrt{16}=4$