Expected Value of Binomial Distribution

Average Successes across Repeated Sets of Trials

Suppose each free throw goes in independently with probability $p=0.8$ , and let $X$ count the baskets in $n=10$ attempts. Then $E(X)=8$ : across many sets of ten attempts, the average number of baskets approaches $8$ .

Expected value is the probability-weighted average of a random variable. It describes the mean over many repetitions. One set can give a different result. An expected value can also be noninteger even when the random variable itself only takes integer values.

Expected Value Formula for Binomial

A binomial random variable counts successes in a fixed number of independent trials, each with the same success probability. Its expected value gives the long-run mean number of successes. It is not necessarily the most likely count, which is described by the mode instead.

Binomial distribution $b_{n,p}(x)$ has an expected value:

E(X) = np

In this formula, $n$ is the number of trials and $p$ is the probability of success in each trial.

If the success probability in each trial is $p$ , then the expected number of successes in $n$ trials is:

E(X) = n \times p

Deriving the Expected Value

The formula follows by representing each trial with an indicator variable.

Let $I_k$ represent the $k$ -th trial. It equals $1$ for success and $0$ for failure.

The expected value for a single trial is:

E(I_k) = 1 \cdot p + 0 \cdot q = p

Since the total number of successes ( $X$ ) is the sum of all successes in each trial, then:

X = I_1 + I_2 + ... + I_n

By linearity of expectation, the expected values add:

E(X) = E(I_1) + E(I_2) + ... + E(I_n)

The expected number of successes is the number of trials multiplied by the success probability.

Dice Rolling Case Study

Problem:

A fair die is rolled $7$ times. What is the expected value for getting a five on the die from all the rolls?

Solution:

Verify that the experiment satisfies the binomial conditions:

There are only two possibilities: success (rolling a $5$ ) or failure (rolling a number other than $5$ )
Fixed number of trials: $7$ rolls
Each roll is independent: the result of previous rolls doesn't affect subsequent rolls
Same probability of success: $p = \frac{1}{6}$ on each roll

The parameters are:

S = \{1, 2, 3, 4, 5, 6\}

Apply the expected-value formula:

E(X) = np = 7 \times \frac{1}{6} = \frac{7}{6} \approx 1.167

So, what does this number $1.167$ mean?

Expected value $1.167$ doesn't mean you will get the number $5$ exactly $1.167$ times in one experiment (because that's impossible). It means that if you repeat the experiment with $7$ rolls hundreds or thousands of times, the number $5$ will appear about $1.167$ times per set on average.

In practice, in one set of $7$ rolls, you might get the number $5$ as many as $0$ , $1$ , $2$ , or even $3$ times. But when averaged over the long term, the result will approach $1.167$ .