## Probability II – Multiple Events

Before you read on, have you read our article on Basic Probability?

A lot of the time you will be considering the probabilities of two or more events occurring. When dealing with these multiple probabilities you will need to consider whether these events are independent or dependent.

Independent Events – Multiple events are considered to be independent if the outcome of one event does not effect the outcomes of any others.

For example, if you roll three dice, the chances of getting a 6 on the third roll is not effected by what you have rolled on the first and second go.

Dependent Events – As the name suggests, multiple events are dependent if the outcome of 1 effects the outcome of the others.

For example, if you have a bag containing 5 black balls and 5 white balls. The probability of picking a white ball on the second pick depends on what you have picked first. If you picked a white ball first, then the chances of picking a white ball on the second go would be $\frac{4}{9}$. However if you picked a black ball first, then the chances of getting a white ball on your second go would be $\frac{5}{9}$.

Example 1

You roll a fair dice three times. What is the probability that on each roll you will roll a number greater than 4 on the first two rolls, and then a number less than 4 on the third roll?

The probability of rolling a a number greater than 4 is equal to the probability of rolling a 5 or a 6, which is $\frac{2}{6} = \frac{1}{3}$. Likewise, the probability of rolling a number less than 4 is $\frac{3}{6} = \frac{1}{2}$.

To find the total probability we multiply these individual probabilities together, resulting in: $\frac{1}{3} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{18} = 0.0556$.

We already know that if two events are independent, then the outcome of one does not effect the outcome of the other.

We now introduce a new term, Mutually Exclusive, (sometimes called disjoint). Two events are mutually exclusive when they can’t occur at the same time. For example, if you flip a coin once, the events ‘get a head’ and ‘get a tail’ are mutually exclusive, as you can only get one or the other.

The AND and OR Rules

When the events A and B are independent: $P(\mathrm{A \; and \; B}) = P(A) \times P(B)$.

When two events, A and B, are mutually exclusive: $P(\mathrm{A \; or \; B}) = P(A) + P(B)$

and $P(\mathrm{A \; and \; B}) = 0$.

Remember these only work for the specific examples where the events are independent or mutually exclusive. If you are not told that the events are either independent or mutually exclusive, don’t use these formulas!

Example 2

A bag contains 20 balls, 12 are black, and 8 are white. A ball is taken from the bag, it’s colour written down and then replaced. A second ball is then taken from the bag, what is the probability that the two balls are different colours?

We try to break the question down and find out exactly what it’s asking us. For the two balls to be different colours, we need to find; $P(\mathrm{1st \; ball \; black \; and \; 2nd \; ball \; white, \; or \; 1st \; ball \; white \; and \; 2nd \; ball \; black})$.

Because the first ball is put back into the bag before the second ball is chosen, then these two events are independent. Thus, from the formula given above, we can replace the ‘and‘ with a $\times$ symbol.

Similarly, either the first ball is black and the second white, or the first ball is white and the second is black. These two events cannon occur at the same time; hence, from the formula above, we can replace the ‘or‘ with a $+$ symbol.

So: $P(\mathrm{1st \; ball \; black \; and \; 2nd \; ball \; white, \; or \; 1st \; ball \; white \; and \; 2nd \; ball \; black})$

becomes; $P(\mathrm{1st \; ball \; black \; and \; 2nd \; ball \; white}) + P(\mathrm{1st \; ball \; white \; and \; 2nd \; ball \; black})$.

Which we can again simplify to: $P(\mathrm{1st \; ball \; black}) \times P(\mathrm{2nd \; ball \; white}) + P(\mathrm{1st \; ball \; white}) \times P(\mathrm{2nd \; ball \; black})$.

Remembering our BODMAS rules, this gives: $\frac{12}{20} \times \frac{8}{20} + \frac{8}{20} \times \frac{12}{20} = \frac{96}{400} + \frac{96}{400} = 0.48$.

You’v now seen how using the AND OR Rules correctly can simplify difficult sounding questions down to a few simple sums.

Now take a look at our article on Probability Spaces and Probability Trees for some further examples of multiple events.