Section PC.2 – Complementary Events; Calculating Odds
Complementary Events
Now let us examine the probability that an event does not happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is P(six) =1/6. Now consider the probability that we do not roll a six: there are 5 outcomes that are not a six, so the answer is P(not a six) = [latex]\frac{5}{6}[/latex]. Notice that P(six) + P(not six) = [latex]\frac{1}{6}[/latex] + [latex]\frac{5}{6}[/latex] = [latex]\frac{1}{1}[/latex] = 1. This is not a coincidence.
| Complement of an Event |
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The complement of an event is the event “E does NOT happen”. The notation E is used for the complement of event E. You should recognize this notation from the last chapter. For any event E, P(E) + P(E) = 1 We can compute the probability of the complement using P(E) = 1 – P(E) Notice also that P(E) = 1 – P(E) |
| Example 1 |
| If you pull a random card from a deck of playing cards, what is the probability it is not a heart? |
| There are 13 hearts in the deck, so P(heart) = [latex]\frac{13}{52}[/latex] = [latex]\frac{1}{4}[/latex].
The probability of not drawing a heart is the complement: P(not heart) = 1 – P(heart) = 1 – [latex]\frac{1}{4}[/latex] = [latex]\frac{3}{4}[/latex] |
| Example 2 |
| A jar contains 28 marbles, 12 of which are red. If you pick 1 marble out of the jar, what is the probability that it is not red? |
| There are 12 red marbles out of 28 total marbles, so P(red) = [latex]\frac{12}{28}[/latex] = [latex]\frac{3}{7}[/latex].
The probability of not drawing a red marble is the complement: P(not red) = 1 – [latex]\frac{3}{7}[/latex] = [latex]\frac{4}{7}[/latex] |
| You Try P.2.A |
| Your favorite basketball player is an 84% free throw shooter. Find the probability that he does NOT make his next free throw. |
Calculating Odds
Another way besides probability to talk about the chances of an event occurring is with odds. You may have heard of the phrases “fifty-fifty” or “even odds” to describe an unpredictable situation like the chances of getting heads when you toss a coin. The phrase means that the coin is as likely to come up tails as it is heads (each event occurring 50% of the time), but odds are not generally expressed as a fraction or a percentage.
The odds of an event are given by the ratio of the number of times the event occurs to the number of times the event does not occur.
| Odds For (in favor of) an Event |
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Odds = number of ways event can occur : number of ways event cannot occur We can also calculate odds using probabilities: Odds = [latex]\frac{P(E)}{P(not~E)}[/latex] |
To avoid confusion with probability, odds are usually left as a ratio such as 1:5, which would be read as “one to five”. When probability is read as a ratio, it’s usually written as a fraction like , which would usually be read as “one in five.”
| Example 3 |
| Find the odds for the event of tossing a coin and getting heads. |
| Solution 1:
The key to finding odds is looking at how many outcomes result in the event and how many do not.
The sample space consists of 2 outcomes: heads or tails. The event we are interested is the event that the coin lands on heads. There is 1 way the event can occur (the coin lands on heads) and there is 1 way the event can NOT occur (the coin lands on tails).
The odds for getting heads are 1 : 1 We call those odds one to one, or even. We are just as likely to get a head as to not get a head.
Solution 2 (Using the probability method): We are going to answer the question this time using the probability definition for finding odds.
The odds for getting heads = [latex]\frac{P(heads)}{P(not~heads)}[/latex] = [latex]\frac{\frac{1}{2}}{\frac{1}{2}}[/latex] = [latex]\frac{1}{1}[/latex] = 1 We would write the odds for getting heads as 1 to 1, or 1 : 1 (one to one). |
| Example 4 |
| Find the odds for rolling a die and getting a 3. |
| Odds for getting a 3 = number of ways to get a 3 : number of ways to not get a 3
Odds for getting a 3 = 1 : 5 Using the probability method: Odds for getting a 3 = [latex]\frac{P(3)}{P(not~3)}[/latex] = [latex]\frac{\frac{1}{6}}{\frac{5}{6}}[/latex] = [latex]\frac{1}{5}[/latex] = 1 : 5 (one to five) |
Look carefully at the example above. This illustrates the need to avoid confusion between odds and probability. We know that the probability of getting a 3 is P(3) = [latex]\frac{1}{6}[/latex] or “one in six”, but the odds describes the same event with the ratio 1 : 5 or “one to five”.
| You Try PC.2.B |
| A board game has a spinner that is divided into 8 different colored sections. The sections are red, orange, yellow, green, blue, purple, black, and white. Find the odds of the spinner landing on a primary color (red, yellow, or blue). |
| Example 5 |
| You are told that the probability of a tornado hitting your hometown during the month of May is 0.15. Find the odds for a tornado hitting your hometown during May. |
| Odds for a tornado = [latex]\frac{P(tornado)}{P(no~tornado)}[/latex] = [latex]\frac{0.15}{(1~-~0.150}[/latex] = [latex]\frac{0.15}{0.85}[/latex] = 0.18 (18 hundreths)
Odds for a tornado = 18 : 100 which can be reduced to 9 : 50 (nine to fifty). |
| Example 6 | ||||||||||||||||
| Find the odds for tossing 4 coins and getting exactly 3 tails. | ||||||||||||||||
In this example, we first need to determine what all of the possible outcomes are when you toss 4 coins. Here is a list of all possible outcomes when tossing 4 coins:
There are 4 outcomes where exactly 3 tails came up and 12 outcomes when they did not. So the odds for getting 3 tails are 4 : 12 = 1 : 3 (one to three). |
| Finding Odds Against an Event |
Most gambling applications involve calculating and reporting the odds against an event. The odds against an event are given by a ratio, just like the odds for an event.
Since we now want the odds against an event occurring, we create a ratio of the number of times the event does not occur to the number of times the event does occur.
| Odds Against an Event |
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Odds = number of ways event cannot occur : number of ways event can occur We can also calculate odds using probabilities: Odds = [latex]\frac{P(not~E)}{P(E)}[/latex] |
| Example 7 |
| Find the odds against tossing 4 coins and getting exactly 3 tails. |
| In the previous example we calculated the odds for this event. Now we want to calculate the odds against it. Recall, the sample space contains 4 outcomes where exactly 3 tails came up and 12 outcomes when they did not.
Since we are looking for the odds against tossing 4 coins and getting exactly 3 tails, we need to take the ratio of the number of ways we do not get exactly 3 tails to the number of ways we do get exactly 3 tails.
Odds against getting exactly 3 tails = 12 : 4 = 3 : 1 (three to one).
Notice this is the reverse of the odds for getting exactly 3 tails (1 : 3) that was calculated in Example 6. |
| Example 8 |
| Experts calculate the probability of a particular horse winning the Kentucky Derby to be 0.2. Calculate the odds against the horse winning the race. |
| Odds against the horse winning:
[latex]\frac{P(not~winning)}{P(winning)}[/latex] = [latex]\frac{1~-~0.2}{0.2}[/latex] = [latex]\frac{0.8}{0.2}[/latex] = [latex]\frac{4}{1}[/latex] = 4 : 1 (four to one) |
| You Try PC.2.C |
| a. Find the odds for drawing a club from a standard 52-card deck.
b. Find the odds against drawing a club from a standard 52-card deck. |
Section PC.2 Answers to You Try Problems
PC.2.A
1 – 0.84 = 0.16
PC.2.B
3 : 5
PC.2.C
a. 1 : 3
b. 3 : 1
