First, let's think about what the probability of tossing a head or tail on one coin is. That would be 0.5 since the coin is "fair."
How about tossing three heads successively (HHH)? That would be 0.5 x 0.5 x 0.5, or (0.5)3
How about tossing two heads and then one tail in that order (HHT)? That would be 0.5 x 0.5 x 0.5, or (0.5)3
How about tossing two tails and then one head in that order (TTH)? That would again be (0.5)3.
Doing more of these examples, you can convince yourself that the probability of getting a particular outcome tossing three coins in a row is (0.5)3.
Let us now write out all the possible outcomes from tossing 3 coins in a row:
{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
In other words, there are 8 possible outcomes, and each outcome has a probability of (0.5)3
If X represents the number of heads that will result from the 3 tosses, X can be 0, 1, 2, or 3 based on the above outcome (that is, 0 heads can result, 1 head can result, 2 head can result or 3 head can result).
Now let's consider these questions:
What is the probability that X = 0? That would correspond to the outcome TTT, which would be (0.5)3
What is the probability that X = 1? That would correspond to the outcomes HTT, THT, or TTH, depending on when the head comes. That would be (0.5)3 + (0.5)3 + (0.5)3 or 3(0.5)3
A probability distribution is basically all the possible values of the random variable and their corresponding probabilities. So for our case, this would be the probability distribution:
X = 0, (0.5)3
X = 1, 3(0.5)3
X = 2, 3(0.5)3
X = 3, (0.5)3
The sum of all probabilities in a probability distribution should be 1, which is true in our case.
Graphing this would be simple. Since this is a discrete function, you should have to denote each value of x and its probability using vertical lines.
E(X) is basically the expectant value, or weighted average, of the probability distribution. To calculate the E(X), we need to sum x*p(x) for all values of x. So in this case, it would be 0*(0.5)3 + 1*3(0.5)3 + 2*3(0.5)3 + 3*(0.5)3, which is 1.5
V(X) is the variance and is a measure of how far each random variable is from the mean.
To calculate the variance we need to average the square of the difference of each random variable and the mean. So, in other words, V(X) = E[(X-E(X))2]
So, (0-1.5)2 (0.5)3 + (1-1.5)2 3(0.5)3 + (2-1.5)2 3(0.5)3 + (3 -1.5)2 (0.5)3 = 0.75
Then we divide 5 by the number of trials, which in this case was 3 (since we tossed the coin 3 times). So 5/3 is the variance
Note: this is an example of the binomial distribution! You can read about it further online. Knowing that it is a binomial distribution can provide many useful shortcuts, like E(X) = np, where n = 3 and p = 0.5, or V(X) = np(1-p)
