Section PC.5 – “Or” Probabilities
The previous examples looked at the probability of both events occurring. Now we will look at the probability of either event occurring.
| Example 1 |
| Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin or a 6 on the die. |
| Here, there are still 12 possible outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
By simply counting, we can see that 7 of the outcomes have a head on the coin or a 6 on the die or both – we use or inclusively here (these 7 outcomes are H1, H2, H3, H4, H5, H6, T6), so the probability is [latex]\frac{7}{12}[/latex]. How could we have found this from the individual probabilities? As we would expect, [latex]\frac{1}{2}[/latex] of these outcomes have a head, and [latex]\frac{1}{6}[/latex] of these outcomes have a 6 on the die. If we add these, [latex]\frac{1}{2}[/latex] +[latex]\frac{1}{6}[/latex] = [latex]\frac{6}{12}[/latex] + [latex]\frac{2}{12}[/latex] = [latex]\frac{8}{12}[/latex], which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head and rolling a 6 is [latex]\frac{1}{12}[/latex]. If we subtract out this double count, we have the correct probability: [latex]\frac{8}{12}[/latex] – [latex]\frac{1}{12}[/latex] = [latex]\frac{7}{12}[/latex]. |
| P(A or B) |
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The probability of either A or B occurring (or both) is P(A or B) = P(A) + P(B) – P(A and B) |
| Example 2 |
| Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King? |
| There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:
P (King or Queen) = [latex]\frac{8}{52}[/latex] Note that in this case, there are no cards that are both a Queen and a King, so P (King or Queen) = [latex]\frac{8}{52}[/latex] = 0. Using our probability rule, we could have said: P (King or Queen) = P (King) + P (Queen) – P (King or Queen) = [latex]\frac{4}{52}[/latex] + [latex]\frac{4}{52}[/latex] – 0 = [latex]\frac{8}{52}[/latex] |
| Example 3 |
| Suppose we draw one card from a standard deck. What is the probability that we get a red card or a King? |
| Half the cards are red, so P (red) = [latex]\frac{26}{52}[/latex]
There are four kings, so P (king) = [latex]\frac{4}{52}[/latex] There are two red kings, so P (Red and King) = [latex]\frac{2}{52}[/latex] We can calculate: P (Red and King) = P (Red) + P (King) – P (Red and King) = [latex]\frac{26}{52}[/latex] + [latex]\frac{4}{52}[/latex] – [latex]\frac{2}{52}[/latex] = [latex]\frac{28}{52}[/latex] |
| Example 4 | ||||||||||||||||||||||||||||||||||||
| Consider again that we roll two die. What is the probability that the sum of the two die is odd or 4? | ||||||||||||||||||||||||||||||||||||
Below is a list of all possible outcomes in the experiment of rolling 2 die:
Note: The table gives the values of each die, not the sum of the two die. Since it is not possible for the sum to be both odd and 4, these events are mutually exclusive. P(sum is odd) = [latex]\frac{18}{36}[/latex] = [latex]\frac{1}{2}[/latex] P(sum is 4) = [latex]\frac{3}{36}[/latex] = [latex]\frac{1}{12}[/latex] P(sum is odd and 4) = [latex]\frac{0}{36}[/latex] P(sum is odd or 4) = P(sum is odd) + P(sum is 4) – P(sum is odd and 4) = [latex]\frac{18}{36}[/latex] + [latex]\frac{3}{36}[/latex] – [latex]\frac{0}{36}[/latex] = [latex]\frac{21}{36}[/latex] = [latex]\frac{7}{12}[/latex] |
| You Try PC.5.A |
| Consider again that we roll two die. What is the probability that the sum of the two die is odd or one of the die is a 1? |
| Example 5 | ||||||||||||||||
| The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:
a) Has a red car and got a speeding ticket
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| a. We can see that 15 people of the 665 surveyed had both a red car and got a speeding ticket, so the probability is
P(red car and speeding ticket) = [latex]\frac{15}{665}[/latex] ≈ 0.0226 Notice that having a red car and getting a speeding ticket are not independent events, so the probability of both of them occurring is not simply the product of probabilities of each one occurring. b. We could answer this question by simply adding up the numbers: 15 people with red cars and speeding tickets + 135 with red cars but no ticket + 45 with a ticket but no red car = 195 people. So the probability is P(red car or speeding ticket) = [latex]\frac{195}{665}[/latex] We also could have found this probability by: P(red car) + P(speeding ticket) – P(red car and speeding ticket) = [latex]\frac{150}{665}[/latex] + [latex]\frac{60}{665}[/latex] – [latex]\frac{15}{665}[/latex] = [latex]\frac{195}{665}[/latex] ≈ 0.2932 |
Section PC.5 Answers to You Try Problems
PC.5.A
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