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Data Analysis and Probability

Expected Value of Binomial Distribution

Understanding the Concept of Expected Value

Imagine you are a basketball player. On average, out of 1010 free throw attempts, you successfully make 88 shots. Well, this number 88 can be called your expected value or expectation of success.

Simply put, expected value is the average value we expect to occur from an experiment if that experiment is repeated many times under the same conditions. This doesn't mean you will definitely get that result every time, but it's a prediction of the long-term average.

Expected Value Formula for Binomial

This concept is very useful in binomial distribution. Remember, binomial distribution is used for experiments that have only two possible outcomes (success or failure) and are performed repeatedly. Binomial expected value helps us predict how many successes we're most likely to get.

Binomial distribution b(x;n,p)b(x;n,p) has an expected value:

E(X)=npE(X) = np

where nn is the number of trials and pp is the probability of success.

This formula is very intuitive. If the probability of success in one trial is pp, then in nn trials, the expected number of successes is:

E(X)=n×pE(X) = n \times p

Where Does the Formula Come From?

You might be curious, why is the formula so simple? Let's break down the logic.

Each trial in a binomial distribution can be thought of as a small random variable, let's call it IkI_k. This variable has a value of 11 if the kk-th trial is successful, and 00 if it fails.

The expected value for a single trial is:

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

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

X=I1+I2+...+InX = I_1 + I_2 + ... + I_n

Using the properties of expected value, we can sum all expected values from each trial:

E(X)=E(I1)+E(I2)+...+E(In)E(X) = E(I_1) + E(I_2) + ... + E(I_n)
E(X)=p+p+...+pn times=npE(X) = \underbrace{p + p + ... + p}_{n \text{ times}} = np

See? The total expected value is the product of the number of trials and the probability of success.

Dice Rolling Case Study

Let's apply this concept to an example.

Problem:

A fair die is rolled 77 times. What is the expected value for getting a five on the die from all the rolls?

Solution:

First, let's make sure this meets the requirements for binomial distribution:

  • There are only two possibilities: success (rolling a 55) or failure (rolling a number other than 55)
  • Fixed number of trials: 77 times
  • Each roll is independent: the result of previous rolls doesn't affect subsequent rolls
  • Same probability of success: p=16p = \frac{1}{6} on each roll

Now we identify the parameters:

S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}
p=16p = \frac{1}{6}
n=7n = 7

Now we can directly calculate the expected value using the formula:

E(X)=np=7×16=76=1.167E(X) = np = 7 \times \frac{1}{6} = \frac{7}{6} = 1.167

So, what does this number 1.1671.167 mean?

Expected value 1.1671.167 doesn't mean you will get "1.1671.167 times" the number 55 in one experiment (because that's impossible). What it means is: if you repeat this experiment of "rolling a die 77 times" hundreds or thousands of times, then on average you will get the number 55 about 1.1671.167 times per set of 77 rolls.

In practice, in one set of 77 rolls, you might get the number 55 as many as 00, 11, 22, or even 33 times. But when averaged over the long term, the result will approach 1.1671.167.