We now give a few examples of simple probability calculations.

Example

Consider a raffle of 100 tickets in which Anne Brown bought seven tickets. Tony Ng also bought four of these 100 tickets. What is the probability that either Anne or Tony win?

Since all 100 raffle tickets are equally likely to win and Anne bought 7 of them, her probability of winning is 0.07.

P(Anne wins)   =   0.07

Similarly, the probability of Tony winning is

P(Tony wins)   =   0.04

Since these two events are mutually exclusive,

P(Anne wins or Tony wins)   =   0.07 + 0.04   =   0.11

The next example is based on a real data set.

Example

The following contingency table describes all 121,690 couples who were married in Australia in 2011.

Previous marital status of couple Both partners
born in Australia
Both partners
born in same
overseas country
Born partners in
different countries
First marriage both partners 51,663 10,522 24,630
First marriage one partner 9,450 2,910 8,050
Remarriage both partners 6,769 1,889 5,807

In a randomly selected marriage from 2011, what is the probability that either it was the first marriage of both partners or that both partners were born in Australia?

Firstly note that all nine cells in this contingency table are mutually exclusive events and that they include all possibilities for a marriage — each marriage was in exactly one cell in the table.

Since the couple is randomly selected from the 121,690 marriages that year, the probabilities for each combination of Previous marital status and Origin are the proportions of marriages in each cell:

Previous marital status of couple Both partners
born in Australia
Both partners
born in same
overseas country
Born partners in
different countries
First marriage both partners \(\frac {51,663} {121,690}\) \(\frac {10,522} {121,690}\) \(\frac {24,630} {121,690}\)
First marriage one partner \(\frac {9,450} {121,690}\) \(\frac {2,910} {121,690}\) \(\frac {8,050} {121,690}\)
Remarriage both partners \(\frac {6,769} {121,690}\) \(\frac {1,889} {121,690}\) \(\frac {5,807} {121,690}\)

There are two ways to approach the problem. The event that either it was the first marriage of both partners or that both partners were born in Australia corresponds to the highlighted cells in the table,

Previous marital status of couple Both partners
born in Australia
Both partners
born in same
overseas country
Born partners in
different countries
First marriage both partners \(\frac {51,663} {121,690}\) \(\frac {10,522} {121,690}\) \(\frac {24,630} {121,690}\)
First marriage one partner \(\frac {9,450} {121,690}\) \(\frac {2,910} {121,690}\) \(\frac {8,050} {121,690}\)
Remarriage both partners \(\frac {6,769} {121,690}\) \(\frac {1,889} {121,690}\) \(\frac {5,807} {121,690}\)

The probability is therefore

\[ \frac {51,663} {121,690} + \frac {10,522} {121,690} + \frac {24,630} {121,690} + \frac {9,450} {121,690} + \frac {6,769} {121,690} = \frac {103,034} {121,690} = 0.8467 \]

Alternatively, we could find the probability that it was the first marriage for both partners from the marginal totals of the table

Previous marital status of couple Both partners
born in Australia
Both partners
born in same
overseas country
Born partners in
different countries
Total
First marriage both partners 51,663 10,522 24,630 86,815
First marriage one partner 9,450 2,910 8,050 20,410
Remarriage both partners 6,769 1,889 5,807 14,465
Total 67,882 15,321 38,487 121,690

Therefore

\[ \text {P(first marriage for both)} = \frac {86,815} {121,690} = 0.7134 \]

and similarly,

\[ \text {P(both born in Australia)} = \frac {67,882} {121,690} = 0.5578 \]

We cannot however simply add these two probabilities since they are not mutually exclusive. Their sum would also be greater than one which is clearly wrong. We must subtract from the sum the joint probability of both events, \(\frac {51,663} {121,690} = 0.4245\)

P(Both first marriage or Both Australian)   =   0.7134 + 0.5578 − 0.4245  =   0.8467