Random Variables

Example: Tossing a coin: we could get Heads or Tails.

Let's give them the values Heads=0 and Tails=1 and we have a Random Variable "X":

In short:

X = {0, 1}

Note: We could choose Heads=100 and Tails=150 or other values if we want! It is our choice.

Example: x + 2 = 6

In this case we can find that x=4

Example: X = {0, 1, 2, 3}

X could be 0, 1, 2, or 3 randomly.

And they might each have a different probability.

Example: Throw a die once

Random Variable X = "The score shown on the top face".

X could be 1, 2, 3, 4, 5 or 6

So the Sample Space is {1, 2, 3, 4, 5, 6}

Example (continued): Throw a die once

X = {1, 2, 3, 4, 5, 6}

In this case they are all equally likely, so the probability of any one is 1/6

• P(X = 1) = 1/6
• P(X = 2) = 1/6
• P(X = 3) = 1/6
• P(X = 4) = 1/6
• P(X = 5) = 1/6
• P(X = 6) = 1/6

Note that the sum of the probabilities = 1, as it should be.

Example: How many heads when we toss 3 coins?

X = "The number of Heads" is the Random Variable.

In this case, there could be 0 Heads (if all the coins land Tails up), 1 Head, 2 Heads or 3 Heads.

So the Sample Space = {0, 1, 2, 3}

But this time the outcomes are NOT all equally likely.

The three coins can land in eight possible ways:

			X = "Number of Heads"
HHH			3
HHT			2
HTH			2
HTT			1
THH			2
THT			1
TTH			1
TTT			0

Looking at the table we see just 1 case of Three Heads, but 3 cases of Two Heads, 3 cases of One Head, and 1 case of Zero Heads. So:

• P(X = 3) = 1/8
• P(X = 2) = 3/8
• P(X = 1) = 3/8
• P(X = 0) = 1/8

Example: Two dice are tossed.

The Random Variable is X = "The sum of the scores on the two dice".

Let's make a table of all possible values:

There are 6 × 6 = 36 possible outcomes, and the Sample Space (which is the sum of the scores on the two dice) is {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Let's count how often each value occurs, and work out the probabilities:

• 2 occurs just once, so P(X = 2) = 1/36
• 3 occurs twice, so P(X = 3) = 2/36 = 1/18
• 4 occurs three times, so P(X = 4) = 3/36 = 1/12
• 5 occurs four times, so P(X = 5) = 4/36 = 1/9
• 6 occurs five times, so P(X = 6) = 5/36
• 7 occurs six times, so P(X = 7) = 6/36 = 1/6
• 8 occurs five times, so P(X = 8) = 5/36
• 9 occurs four times, so P(X = 9) = 4/36 = 1/9
• 10 occurs three times, so P(X = 10) = 3/36 = 1/12
• 11 occurs twice, so P(X = 11) = 2/36 = 1/18
• 12 occurs just once, so P(X = 12) = 1/36

Example (continued) What is the probability that the sum of the scores is 5, 6, 7 or 8?

In other words: What is P(5 ≤ X ≤ 8)?

Example (continued) If P(X=x) = 1/12, what is the value of x?

Looking through the list above we find:

• P(X=4) = 1/12, and
• P(X=10) = 1/12

So there are two solutions: x = 4 or x = 10