# Developing Discrete Probability Distributions Theoretically & Finding Expected Values Video

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• 0:03 Generating Discrete…
• 0:31 Tossing Two Coins
• 3:55 Expected Value
• 5:30 Roulette! A Game of Chance
• 8:30 Lesson Summary
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Lesson Transcript
Instructor: Rudranath Beharrysingh
In this lesson, we will look at generating a theoretical probability distribution for a discrete random variable and introduce the concept of expected value.

## Generating Discrete Probability Distribution

You may recall that a probability distribution summarizes the probability for a particular experiment.

Discrete probability distributions can be created from data or from theory based on what we know about a given experiment. Here we will look at theoretical discrete distributions like tossing coins, throwing dice, spinning a spinner or playing cards, but it could also be extended to other experiments where each outcome is known.

## Tossing Two Coins

Suppose you toss two coins, and we want to know the various possible outcomes and the probabilities associated with them.

First, we need to look at all the possible outcomes. Well, when we toss two coins, we can get no heads, one head or two heads. Similarly we can get no tails, one tail or two tails. But note: if we got no heads, then we actually got two tails, and one head is the same as one tail, and two heads is equivalent to no tails. So, we have to decide if we want to look at the number of heads or the number of tails before we proceed.

For this problem, we'll just look at the number of heads. So, note the number of heads could be 0, 1 or 2, so we define our random variable X to be the number of heads when we toss two coins.

And we want to fill out a table like this:

Number of Heads When You Toss Two Coins Probability of X: P(X)
0
1
2

But we need to calculate the probability for each value of X.

So, first of all, what is the probability that X is equal to 0? In shorthand math notation, we would write

P(0) = ?

Since X = 0 means no heads (or two tails), and the probability of no head on one coin is 1/2, then the probability of no heads on both coins (no heads and no heads) is equal to 1/2 * 1/2 = 1/4.

Note in probability, the word AND translates as MULTIPLY.

Thus, P(0) = 1/4 or 0.25

Number of Heads When You Toss Two Coins Probability of X: P(X)
0 0.25
1
2

Now let's calculate P(1): the probability of one head.

There are two possible outcomes. But, to look at this, let's label the coins as Coin One and Coin Two. So, the possible outcomes that have one head are: heads on Coin One and tails on Coin Two OR tails on Coin One and heads on Coin Two.

Let's look at the first outcome. The probability of heads on Coin One is 1/2, and then the probability of tails on Coin Two is 1/2. Multiplying them, we get 1/4. But we also have to look at the other outcome, which has exactly the same probability of 1/4. Since we are interested in only one head, we must take into consideration both of these outcomes, and so we add the possibilities (1/4 + 1/4) to get 1/2.

Note in probability, the word OR translates as ADD.

Thus the probability that X = 1, is 1/2 or in short hand math: P(1) = 1/2 or 0.5

Number of Heads When You Toss Two Coins Probability of X: P(X)
0 0.25
1 0.5
2

Finally, we want to find the probability of getting two heads or P(2).

The probability of heads on each coin is 1/2, and so the probability of heads on Coin One and heads on Coin Two is 1/2 * 1/2 = 1/4.

P(2) = 1/4.

We can summarize this in a table.

Number of Heads When You Toss Two Coins Probability of X: P(X)
0 0.25
1 0.5
2 0.25

This is a probability distribution for the number of heads when you toss two coins.

A probability distribution contains all the values of the random variable X and its probability.

Note in a probability distribution, the probabilities sum to one.

Number of Heads When You Toss Two Coins Probability of X: P(X)
0 0.25
1 0.5
2 0.25
Total = 1

## Expected Value

Suppose you toss two coins and you want to know how many heads to expect. Well, since there are only two coins and the probability of heads is 50% (or 0.5), we would expect that when tossing two coins we would get one head. Thus, the expected value of X is 1 for this experiment.

But how do we calculate this, especially for larger distributions? Formally, The Expected Value is the sum of (X times P(X)) for all values of a random variable X, or E(X) = The sum of X * P(X).

Applying this to the two coins example, we take each value of X and multiply by its probability and then sum them up. So we will have:

E(X) = (0 * 0.25) + (1 * 0.5) + (2 * 0.25)

Then, we sum these numbers up. So we have:

0 + 0.5 + 0.5 = 1

This is the expected value of X. In other words, we expect one head when we toss two coins. The calculation can be done using the probability distribution table:

Number of Heads When You Toss Two Coins Probability of X: P(X) X*P(X)
0 0.25 0 * 0.25 = 0
1 0.5 1 * 0.5 = 0.5
2 0.25 2 * 0.25 = 0.5
Sum: X * P(X) = 1

This does not mean that we will always get one head when we toss two coins. It means that, on average, when tossing two coins, you will get one head. Can you see where this information will be useful?

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