3.2: Combining Probabilities with “And” and “Or”
Many probabilities in real life involve more than one outcome. If we draw a single card from a deck we might want to know the probability that it is either red or a jack. If we look at a group of students, we might want to know the probability that a single student has brown hair and blue eyes. When we combine two outcomes to make a single event we connect the outcomes with the word “and” or the word “or.” It is very important in probability to pay attention to the words “and” and “or” if they appear in a problem. The word “and” restricts the field of possible outcomes to only those outcomes that simultaneously satisfy more than one event. The word “or” broadens the field of possible outcomes to those that satisfy one or more events.
Suppose a teacher wants to know the probability that a single student in her class of 30 students is taking either Art or English. She asks the class to raise their hands if they are taking Art and counts 13 hands. Then she asks the class to raise their hands if they are taking English and counts 21 hands. The teacher then calculates
\[P(\text{Art or English}) = \dfrac{13+21}{30} = \dfrac{33}{30} \nonumber \]
The teacher knows that this is wrong because probabilities must be between zero and one, inclusive. After thinking about it she remembers that nine students are taking both Art and English. These students raised their hands each time she counted, so the teacher counted them twice. When we calculate probabilities we have to be careful to count each outcome only once.
Mutually Exclusive Events
An experiment consists of drawing one card from a well shuffled deck of 52 cards. Consider the events E : the card is red, F : the card is a five, and G : the card is a spade. It is possible for a card to be both red and a five at the same time but it is not possible for a card to be both red and a spade at the same time. It would be easy to accidentally count a red five twice by mistake. It is not possible to count a red spade twice.
Two events are mutually exclusive if they have no outcomes in common.
Two fair dice are tossed and different events are recorded. Let the events E , F and G be as follows:
- E = {the sum is five} = {(1, 4), (2, 3), (3, 2), (4, 1)}
- F = {both numbers are even} = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
- G = {both numbers are less than five} = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4,1), (4, 2), (4, 3), (4,4)}
- Are events E and F mutually exclusive?
Yes. E and F are mutually exclusive because they have no outcomes in common. It is not possible to add two even numbers to get a sum of five.
- Are events E and G mutually exclusive?
No. E and G are not mutually exclusive because they have some outcomes in common. The pairs (1, 4), (2, 3), (3, 2) and (4, 1) all have sums of 5 and both numbers are less than five.
- Are events F and G mutually exclusive?
No. F and G are not mutually exclusive because they have some outcomes in common. The pairs (2, 2), (2, 4), (4, 2) and (4, 4) all have two even numbers that are less than five.
Addition Rule for “Or” Probabilities
The addition rule for probabilities is used when the events are connected by the word “or”. Remember our teacher in Example \(\PageIndex{1}\) at the beginning of the section? She wanted to know the probability that her students were taking either art or English. Her problem was that she counted some students twice. She needed to add the number of students taking art to the number of students taking English and then subtract the number of students she counted twice. After dividing the result by the total number of students she will find the desired probability. The calculation is as follows:
\[ \begin{align*} P(\text{art or English}) &= \dfrac{\# \text{ taking art + } \# \text{ taking English - } \# \text{ taking both}}{\text{total number of students}} \\[4pt] &= \dfrac{13+21-9}{30} \\[4pt] &= \dfrac{25}{30} \approx {0.833} \end{align*} \nonumber \]
The probability that a student is taking art or English is 0.833 or 83.3%.
When we calculate the probability for compound events connected by the word “or” we need to be careful not to count the same thing twice. If we want the probability of drawing a red card or a five we cannot count the red fives twice. If we want the probability a person is blonde-haired or blue-eyed we cannot count the blue-eyed blondes twice. The addition rule for probabilities adds the number of blonde-haired people to the number of blue-eyed people then subtracts the number of people we counted twice.
If A and B are any events then
\[P(A\, \text{or}\, B) = P(A) + P(B) – P(A \,\text{and}\, B). \nonumber \]
If A and B are mutually exclusive events then \(P(A \,\text{and}\, B) = 0\), so then
\[P(A \, \text{or}\, B) = P(A) + P(B). \nonumber \]
A single card is drawn from a well shuffled deck of 52 cards. Find the probability that the card is a club or a face card.
Solution
There are 13 cards that are clubs, 12 face cards (J, Q, K in each suit) and 3 face cards that are clubs.
\[ \begin{align*} P(\text{club or face card}) &= P(\text{club}) + P(\text{face card}) - P(\text{club and face card}) \\[4pt] &= \dfrac{13}{52} + \dfrac{12}{52} - \dfrac{3}{52} \\[4pt] &= \dfrac{22}{52} = \dfrac{11}{26} \approx {0.423} \end{align*} \nonumber \]
The probability that the card is a club or a face card is approximately 0.423 or 42.3%.
An experiment consists of tossing a coin then rolling a die. Find the probability that the coin lands heads up or the number is five.
Solution
Let H represent heads up and T represent tails up. The sample space for this experiment is S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.
- There are six ways the coin can land heads up, {H1, H2, H3, H4, H5, H6}.
- There are two ways the die can land on five, {H5, T5}.
- There is one way for the coin to land heads up and the die to land on five, {H5}.
\[ \begin{align*} P(\text{heads or five}) &= P(\text{heads}) + P(\text{five}) - P(\text{both heads and five}) \\[4pt] &= \dfrac{6}{12} + \dfrac{2}{12} - \dfrac{1}{12} \\[4pt] &= \dfrac{7}{12} = \approx {0.583} \end{align*} \nonumber \]
The probability that the coin lands heads up or the number is five is approximately 0.583 or 58.3%.
Two hundred fifty people who recently purchased a car were questioned and the results are summarized in the following table.
| Satisfied | Not Statisfied | Total | |
|---|---|---|---|
| New Car | 92 | 28 | 120 |
| Used Car | 83 | 47 | 130 |
| Total | 175 | 75 | 250 |
Find the probability that a person bought a new car or was not satisfied.
Solution
\[\begin{align*} P(\text{new car or not satisfied}) &= P(\text{new car}) + P(\text{not satisfied}) - P(\text{new car and not satisfied}) \\[4pt] &= \dfrac{120}{250} + \dfrac{75}{250} - \dfrac{28}{250} = \dfrac{167}{250} \approx 0.668 \end{align*} \nonumber \]
The probability that a person bought a new car or was not satisfied is approximately 0.668 or 66.8%.
Independent Events
Sometimes we need to calculate probabilities for compound events that are connected by the word “and.” We have two methods to choose from, independent events or conditional probabilities (Section 3.3). Tossing a coin multiple times or rolling dice are independent events. Each time you toss a fair coin the probability of getting heads is ½. It does not matter what happened the last time you tossed the coin. It’s similar for dice. If you rolled double sixes last time that does not change the probability that you will roll double sixes this time. Drawing two cards without replacement is not an independent event. When you draw the first card and set it aside, the probability for the second card is now out of 51 cards not 52 cards.
Two events are independent events if the occurrence of one event has no effect on the probability of the occurrence of the other event.
If events A and B are independent events, then \( P(\text{A and B}) = P(A) \cdot P(B)\).
Suppose a fair coin is tossed four times. What is the probability that all four tosses land heads up?
Solution
The tosses of the coins are independent events. Knowing a head was tossed on the first trial does not change the probability of tossing a head on the second trial.
\(P(\text{four heads in a row}) = P(\text{1st heads and 2nd heads and 3rd heads and 4th heads})\)
\( = P(\text{1st heads}) \cdot P(\text{2nd heads}) \cdot P(\text{3rd heads}) \cdot P(\text{4th heads})\)
\( = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2}\)
\( = \dfrac{1}{16}\)
The probability that all four tosses land heads up is \(\dfrac{1}{16}\).
A bag contains five red and four white marbles. A marble is drawn from the bag, its color recorded and the marble is returned to the bag. A second marble is then drawn. What is the probability that the first marble is red and the second marble is white?
Since the first marble is put back in the bag before the second marble is drawn these are independent events.
\[\begin{align*} P(\text{1st red and 2nd white}) &= P(\text{1st red}) \cdot P(\text{2nd white}) \\[4pt] &= \dfrac{5}{9} \cdot \dfrac{4}{9} = \dfrac{20}{81}\end{align*} \nonumber \]
The probability that the first marble is red and the second marble is white is \(\dfrac{20}{81}\).
Abby has an important meeting in the morning. She sets three battery-powered alarm clocks just to be safe. If each alarm clock has a 0.03 probability of malfunctioning, what is the probability that all three alarm clocks fail at the same time?
Solution
Since the clocks are battery powered we can assume that one failing will have no effect on the operation of the other two clocks. The functioning of the clocks is independent.
\[\begin{align*} P(\text{all three fail}) &= P(\text{first fails}) \cdot P(\text{second fails})\cdot P(\text{third fails}) \\[4pt] &= (0.03)(0.03)(0.03) \\[4pt] &= 2.7 \times 10^{-5} \end{align*} \nonumber \]
The probability that all three clocks will fail is approximately 0.000027 or 0.0027%. It is very unlikely that all three alarm clocks will fail.
At Least Once Rule for Independent Events
Many times we need to calculate the probability that an event will happen at least once in many trials. The calculation can get quite complicated if there are more than a couple of trials. Using the complement to calculate the probability can simplify the problem considerably. The following example will help you understand the formula.
The probability that a child forgets her homework on a given day is 0.15. What is the probability that she will forget her homework at least once in the next five days?
Solution
Assume that whether she forgets or not one day has no effect on whether she forgets or not the second day.
If P (forgets) = 0.15, then P (not forgets) = 0.85.
\[\begin{align*} P(\text{forgets at least once in 5 tries}) &= P(\text{forgets 1, 2, 3, 4 or 5 times in 5 tries}) \\[4pt] & = 1 - P(\text{forgets 0 times in 5 tries}) \\[4pt] &= 1 - P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \\[4pt] &= 1 - (0.85)(0.85)(0.85)(0.85)(0.85) \\[4pt] & = 1 - (0.85)^{5} = 0.556 \end{align*} \nonumber \]
The probability that the child will forget her homework at least one day in the next five days is 0.556 or 55.6%
The idea in Example \(\PageIndex{9}\) can be generalized to get the At Least Once Rule.
If an experiment is repeated n times, the n trials are independent and the probability of event A occurring one time is P(A) then the probability that A occurs at least one time is: \(P(\text{A occurs at least once in n trials}) = 1 - P(\overline{A})^{n}\)
The probability of seeing a falcon near the lake during a day of bird watching is 0.21. What is the probability that a birdwatcher will see a falcon at least once in eight trips to the lake?
Solution
Let A be the event that he sees a falcon so P(A) = 0.21. Then, \(P(\overline{A}) = 1 - 0.21 = 0.79\).
\(P(\text{at least once in eight tries}) = 1 - P(\overline{A})^{8}\)
\( = 1 - (0.79)^{8}\)
\( = 1 - (0.152) = 0.848\)
The probability of seeing a falcon at least once in eight trips to the lake is approximately 0.848 or 84.8%.
A multiple choice test consists of six questions. Each question has four choices for answers, only one of which is correct. A student guesses on all six questions. What is the probability that he gets at least one answer correct?
Solution
Let A be the event that the answer to a question is correct. Since each question has four choices and only one correct choice, \(P(\text{correct}) = \dfrac{1}{4}\).
That means \(P(\text{not correct}) =1 - \dfrac{1}{4} = \dfrac{3}{4}\).
\[ \begin{align*} P(\text{at least one correct in six trials}) &= 1 - P(\text{not correct})^{6} \\[4pt] &= 1 - \left(\dfrac{3}{4}\right)^{6} \\[4pt] &= 1 - (0.178) = 0.822 \end{align*} \nonumber \]
The probability that he gets at least one answer correct is 0.822 or 82.2%.
“And” Probabilities from Two-Way Tables
“And” probabilities are usually done by one of two methods. If you know the events are independent you can use the rule \(P(A \text{and } B) = P(A) \cdot P(B)\). If the events are not independent you can use the conditional probabilities in Section 3.3. There is an exception when we have data given in a two-way table. We can calculate “and” probabilities without knowing if the events are independent or not.
Continuation of Example \(\PageIndex{5}\):
Two hundred fifty people who recently purchased a car were questioned and the results are summarized in the following table.
| Satisfied | Not Statisfied | Total | |
|---|---|---|---|
| New Car | 92 | 28 | 120 |
| Used Car | 83 | 47 | 130 |
| Total | 175 | 75 | 250 |
A person is chosen at random. Find the probability that the person:
- bought a new car and was satisfied.
\[\begin{align*} P(\text{new car and satisfied}) &= \dfrac{\text{number of new car and satisfied}}{\text{number of people}} \\[4pt] &= \dfrac{92}{250} = 0.368 = 36.8 \% \end{align*} \nonumber \]
- bought a used car and was not satisfied.
\[\begin{align*} P(\text{used car and not satisfied}) &= \dfrac{\text{number of used and not satisfied}}{\text{number of people}} \\[4pt] &= \dfrac{47}{250} = 0.188 = 18.8 \% \end{align*} \nonumber \]