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