3.4: Working with Events
The Complement Rule
In the sets and logic chapters, we saw the complement of set \(A\) and the negation of statement \(A\). We defined the complement of set \(A\) as another set containing the elements in the universal set that are not in set \(A\), and we wrote the complement of \(A\) as \(A^{\prime}\). Similarly, we defined the negation of statement \(A\) as exactly the opposite of statement \(A\) and wrote the negation of \(A\) as \(\sim A\).
In probability, we can think of the complement in a similar way. The complement of event \(A\) contains all outcomes in the sample space that are not in event \(A\). We write the probability of the complement as \(P(A^{\prime})\) or \(P(\text{not } A)\).
Consider the experiment of rolling a regular six-sided die. The sample space is \(S=\{1,2,3,4,5,6\}\). If event \(A\) is " rolling a number less than 3, " then \(A=\{1,2\}\). The complement of event \(A^{\prime}\), " not rolling a number less than 3 ," contains all the outcomes in the sample space that are not in \(A\). Thus, \(A^{\prime}=\{3,4,5,6\}\).
A Venn diagram that illustrates the relationship between \(S\), \(A\), and \(A^{\prime}\) is shown below:
Here, \(P(A)=\frac{2}{6}=\frac{1}{3}\) and \(P(A^{\prime})=\frac{4}{6}=\frac{2}{3}\).
When we combining all outcomes in event \(A\) with all outcomes in its complement \(A^{\prime}\), the result is all the outcomes in the sample space of the experiment. Therefore, the probabilities of an event and its complement must add to 1. This formula and its variations are known as the Complement Rule of Probability.
\(P(A)+P(A^{\prime})=1\) or \(P(A)+P(\text{not }A)=1\)
Alternatively, \(P(A)= 1 - P(A^{\prime})\) or \(P(A)= 1 - P(\text{not }A)\)
\(P(A^{\prime})=1-P(A)\) or \(P(\text{not }A) = 1-P(A)\)
As shown in the box, you may find that it is sometimes easier to calculate the probability of an event's complement than it is to calculate the probability of the event itself. Once this is done, the probability of the event, \(P(A)\), is found using the relationship \(1-P(A^{\prime})\).
Suppose you know that the probability of getting the flu this winter is 0.43. What is the probability that you will not get the flu?
Solution
The probability of getting the flu is \(P(F) = 0.43\).
The probability of not getting the flu is \(P(F^{\prime})=1-P(F)=1 - 0.43 = 0.57\).
The probability that a person does not live in an industrialized country of the world is \(\frac{4}{5}\). Find the probability that a person lives in an industrialized country.
Solution
The probability that a person does not live in an industrialized country is \(P(I^{\prime})=\frac{4}{5}\).
The probability that a person lives in an industrialized country is \(P(I) = 1 - P(I^{\prime}) =1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5} \).
A card is drawn from a standard deck of cards. What is the probability that the card is not a King?
Solution
The probability of getting a King is \(P(K) = \frac{4}{52}=\frac{1}{13}\).
Therefore, the probability of not getting a King is \(P(K^{\prime}) = 1 - \frac{1}{13} = \frac{13}{13} - \frac{1}{13}= \frac{12}{13} \).
A card is drawn from a standard deck of cards. What is the probability that the card is not a club?
- Answer
-
\(\frac{3}{4}\)
Combining Events
Many probabilities in real life involve more than one event. 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 “or” or the word “and.” 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 satisfy more than one event at the same time or in succession. The word “or” broadens the field of possible outcomes to those that satisfy one or more events.
Combining Events with "Or"
The event "\(A \text{ or } B\)" refers to an event that includes outcomes of \(A\), or \(B\), or both. As we saw with sets and truth tables, the union \(A \cup B\) and disjunction \(A \vee B\) refer to where at least one of \(A\) or \(B\) occur. Let’s begin the development of a rule for \(P(A \text{ or } B)\) by exploring an example.
A box contains 12 slips of paper, each numbered from 1 to 12. The sample space is \(\{1,2,3,4,5,6,7,8,9,10,11,12\}.\) These twelve outcomes are all equally likely. What is the probability of selecting an even number or a number divisible by 3?
Solution
The event in question must meet one or more of the two criteria: selecting an even number or a number divisible by 3. This means we can select a 2, 4, 6, 8, 10, or 12 (an even number) or select a 3, 6, 9, or 12 (a number divisible by 3.) Here,
\(P(\text{even number}) = \frac{6}{12}\) and \(P(\text{divisible by 3}) = \frac{4}{12}\).
It is tempting to add \(P(\text{even number}) + P(\text{number divisible by 3}) = \frac{6}{12} + \frac{4}{12} = \frac{10}{12}\). However, there is an error in this thinking.
To see this error, look at the sample space of this experiment represented by a Venn diagram. We are interested in finding \(P(A \text{ or } B)\), which is the union of two sets in the diagram.
The outcomes 6 and 12 meet both of these criteria because they are even numbers and divisible by 3. Consequently, we have included these outcomes in both \(P(\text{even number})\) and \(P(\text{number divisible by 3})\). To compensate for over-counting the outcomes 6 and 12, we must subtract the probability of these two over-counted outcomes, \(P(\text{even number and divisible by 3}) = \frac{2}{12}\). So,
\(P(\text{even number or number divisible by 3}) = \frac{6}{12} + \frac{4}{12} -\frac{2}{12} = \frac{8}{12} =\frac{2}{3}.\)
Based on the previous example, it appears that \(P(A\ \text{ or } B) = P(A) + P(B) \; – P(A \text{ and } B)\). This thinking should not be new or a surprise. Joining two sets with "or" is a union of those two sets. And, in Chapter 1, we found that \(n(A \cup B) = n(A)+n(B)-n(A \cap B)\).
But, before we state this general rule for finding the probability of \(A \text{ or } B\), we should consider a special case of a Venn diagram with event \(A\) and event \(B\). The figure below is a Venn diagram showing the events are mutually exclusive . The circles do not overlap because there are no outcomes that are common to both events.
In this Venn diagram, event \(A\) could represent all the outcomes of drawing a red card from a deck of cards, and event \(B\) could represent all the outcomes of drawing a club from a deck of cards. These two events have no outcomes in common because there are no cards in a deck that are both red and a club. Therefore, the events have no outcomes in common.
Event \(A\) and event \(B\) are mutually exclusive if the events have no common outcomes.
That is, event \(A\) and event \(B\) are mutually exclusive if and only if \(A \cap B = \text{ { } }\) and \(P(A \text{ and } B) = 0\).
In the case of mutually exclusive events, case \(P(A \text{ and } B) = 0\). There is no over-counting outcomes because there are no outcomes that are in both \(A\) and \(B\). So, when two mutually exclusive events are joined by "or," the formula we use for \(P(A\text{ or }B)\) becomes much simpler:
\( \begin{align*} P(A\text{ or }B) &= P(\text{A}) + P(\text{B}) - P(A \text{ and B}) \\[4pt] &= P(A) + P(B) - 0 \\[4pt] &= P(A) + P(B) \end{align*}\)
The formula for \(P(A \text{ or }B)\) when \(A\) and \(B\) are mutually exclusive is just a special case of the general formula. Since the two events are mutually exclusive, there is no double counting and no need to subtract their overlap.
The box gives the general formula for \(P(A \text { or } B)\) and a more specialized formula when the events are mutually exclusive.
If \(A\) and \(B\) are any events then
\(P(A\ \text{ or } B) = P(A) + P(B) \; – P(A,\text{ and } B)\) or \(P(A \cup B) = P(A) + P(B) \; – P(A \cap B).\)
If \(A\) and \(B\) are mutually exclusive events then
\(P(A \text{ or } B) = P(A) + P(B)\) or \(P(A \cup B) = P(A) + P(B)\)
To apply the probability formulas for "or" scenarios successfully, we should be careful to distinguish whether the two events are mutually exclusive or not. We now look at several examples.
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 (King, Queen, Jack.)
Solution
The events "club" and "face card" are not mutually exclusive because a card can be a club and also a face card. There are 13 cards that are clubs, 12 face cards (J, Q, K in each suit), and 3 cards that are both face cards and clubs.
\( \begin{align*} P(\text{club or face card}) &= P(\text{club}) + P(\text{face card}) - P(\text{club and face card}) \\[4pt] &= \dfrac{13}{52} + \dfrac{12}{52} - \dfrac{3}{52} \\[4pt] &= \dfrac{22}{52} = \dfrac{11}{26} \approx {0.4231} \end{align*}\)
The probability that the card is a club or a face card is approximately 0.4231 or 42.31%.
A single card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is a King or an Ace.
Solution
The event "King" and the event "Ace" have no common outcomes so these are mutually exclusive events.
\( \begin{align*} P(\text{King or Ace}) &= P(\text{King}) + P(\text{Ace}) \\[4pt] &= \dfrac{4}{52} + \dfrac{4}{52} \\[4pt] &= \dfrac{8}{52} \\ &= \dfrac{2}{13} \end{align*}\)
The probability of selecting a King or Ace is \(\dfrac{2}{13} \approx 0.1538\), or about \(15.39\%\).
A regular six-sided die is rolled. What's the probability that the die lands on an odd number or a number less than 5?
- Answer
-
\(\frac{5}{6}\)
On a game spinner, the probability of spinning red is \(\frac{1}{3}\), the probability of spinning white is \(\frac{2}{5}\), and the probability of spinning blue is \(\frac{4}{15}\). What is the probability of spinning once and getting white or blue?
- Answer
-
\(\frac{2}{3}\)
An experiment consists of tossing a coin and 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, \(\{H5\}\).
\( \begin{align*} P(\text{heads or five}) &= P(\text{heads}) + P(\text{five}) - P(\text{both heads and five}) \\[4pt] &= \dfrac{6}{12} + \dfrac{2}{12} - \dfrac{1}{12} \\[4pt] &= \dfrac{7}{12} = \approx {0.5833} \end{align*}\)
The probability that the coin lands heads up or the number is five is approximately 0.5833 or 58.33%.
There is a 90% chance of snow, a 20% chance of strong winds, and a 15% chance of both snow and strong winds. What is the probability of snow or strong winds?
Solution
\( \begin{align*} P(\text{snow or strong winds}) &= P(\text{snow}) + P(\text{strong winds}) - P(\text{snow and strong winds}) \\[4pt] &= 0.90 + 0.20 - 0.15 \\[4pt] &= 0.95 \end{align*}\)
The probability of snow or strong winds is 0.95 or 95%.
In a subdivision, there is a 67% chance that a home has a garage and a 32% chance that a home has a patio. There is a 13% chance that a home has both a garage and a patio. What's the chance that a home has a garage or patio?
- Answer
-
86%
The table shows a group of cyclists and the routes they prefer. One cyclist is randomly chosen from this group.
| Gender | Lake Path, \(L\) | Hilly Path, \(H\) | Wooded Path, \(W\) | Total |
|---|---|---|---|---|
| Male, \(M\) | 45 | 38 | 27 | 110 |
| Female, \(F\) | 26 | 52 | 12 | 90 |
| Total | 71 | 90 | 39 | 200 |
- What is the probability of selecting a cyclist who is a female or prefers the wooded path?
- What is the probability of selecting a cyclist or prefers the lake path or the hilly path?
- What is the probability of selecting a cyclist who does not prefer the lake path?
Solution
- This question asks for \(P(F \text{ or } W)\). These events are not mutually exclusive because there are cyclists who are both female and prefer the hilly path:
\(\begin{aligned} P(F \text{ or } W) &= P(F) + P(W) - P(F \text{ and } W) \\ &= \frac{90}{200} + \frac{39}{200} - \frac{12}{200} \\ &= \frac{117}{200} = 0.585 \end{aligned}\)
- This question asks for \(P(L \text{ or } H)\). These events are mutually exclusive because there are no cyclists who prefer both the Lake path and the Hilly path:
\(\begin{aligned} P(L \text{ or } H) &= P(L) + P(H) \\ &= \frac{71}{200} + \frac{90}{200} \\ &= \frac{161}{200} = 0.805 \end{aligned}\)
- This question asks for \(P(L^{\prime})\):
\(\begin{aligned} P(L^{\prime}) &= 1 - P(L) \\ &= 1-\frac{71}{200} \\ &= \frac{129}{200} = 0.645 \end{aligned}\)
A group of people were surveyed about the type of movies they prefer. Suppose a person is chosen at random from this group.
| Gender | Romantic, \(R\) | Action, \(A\) | Horror, \(H\) | Total |
|---|---|---|---|---|
| Male, \(M\) | 8 | 25 | 6 | 39 |
| Female, \(F\) | 12 | 10 | 3 | 25 |
| Total | 20 | 35 | 9 | 64 |
- What is the probability of selecting a person who is male or prefers horror movies?
- What is the probability of selecting a person who prefers romantic or action movies?
- What is the probability of selecting a person who does not prefer romantic movies?
- Answer
-
- \(\frac{42}{64}= \frac{21}{32}\)
- \(\frac{55}{64}\)
- \(\frac{44}{64} = \frac{11}{16}\)
Combining Events with "And"
Sometimes we need to calculate probabilities for compound events that are connected by the word “and.” This requires that we look for outcomes that are common to both events. As we saw with sets and truth tables, the intersection \(A \cap B\) and conjunction \(A \wedge B\) refer to where both \(A\) and \(B\) occur. When working with probabilities, these events may occur at the same time or they could happen in a sequence, such as \(A\) followed by \(B\).
How we calculate the theoretical probability of the event "\(A \text{ and } B\)" depends on whether the two events are independent or dependent . So, we must discuss these concepts before developing a formula.
Two events \(A\) and \(B\) are independent if the probability that \(B\) occurs is the same whether or not \(A\) occurs. If the probability of \(B\) is affected by the occurrence of \(A\), then we say that the events are dependent .
Coin tosses and die rolls are common examples of independent events – tossing heads does not change the probability of tossing heads the next time. The probability of tossing a heads on the second toss is still 1/2 no matter if there was a heads or tails on the first toss. Likewise, rolling a six on a first die does not change the probability that the next roll will be a six.
Another example of independent events is randomly selecting items from a container when the items are selected with replacement . By replacing the item after the first selection, we reset the probability back to what it was before we made the selection. Since the probabilities in the second selection are the same as the first selection, the events are independent.
If we select without replacement , however, we change the total number of possible outcomes, thereby changing the probabilities in subsequent selections. For example, if we select two people from a group for a committee, the first member selected cannot be selected again so the sample space (and therefore the probabilities) for the second selection has changed. So, if we draw without replacement, the events will be dependent.
What about drawing two cards from a deck? This can be a bit tricky. When you draw the first card and put it back before drawing a second card, the sample space for drawing the second card remains 52 cards just as it was for the first card. These events are independent. However, if you draw the first card and put it aside before drawing a second card, the sample space for drawing the second card changes to 51 because one card has been removed. This is an example of events that are dependent.
Two events are independent events if the occurrence of one event has no effect on the probability of the occurrence of the other event. Examples of independent events include tossing coins, rolling dice, spinning spinners, and selecting items from a group when the items are replaced after each selection.
Two events are dependent events if the occurrence of one event has an effect on the probability of the occurrence of the other event. Examples of dependent events include selecting items from a group when the items are not replaced after each selection.
Determine whether each pair of events are independent or dependent.
- Flipping a coin twice and getting tails both times.
- Selecting a president and then a vice president at random from a pool of five equally qualified individuals.
- The event that it will be cold in Washington, DC tomorrow and the event that it will be cold in Baltimore tomorrow.
- Wearing your lucky socks and getting an A on your exam.
Solution
- The probability of getting tails on the first flip is \(\frac{1}{2}\). After flipping tails, the probability of getting tails on the second flip is still \(\frac{1}{2}\). Since the probability of flipping tails on the second flip did not change because we flipped tails on the first flip, the events are independent.
- Since two different people will be put in the role of president and vice president, we are drawing without replacement and the events are therefore dependent.
- If it is cold in Baltimore it is more likely that it will be cold in Washington, so the events are dependent.
- The socks you wear do not have a direct effect on how well you do on your exam, so these events are independent.
Determine whether each pair of events are independent or dependent.
- Rolling a die twice and getting '2' both times.
- Getting two hearts when selecting one card from a well-shuffled deck, setting it aside, and then selecting another card.
- Drawing two red marbles one-by-one from a bucket of marbles where the first marble is replaced before drawing the second marble.
- Answer
-
- independent
- dependent
- independent
To calculate "and" probabilities we always multiply, but we need to determine whether the events are independent or dependent. If they are independent, then we can multiply the individual probabilities of the events because one does not affect the other. If the events are dependent, we must remember that the first event has had an effect on the probability of the second event. A general formula for finding \(P(A \text{ and } B)\) is given below:
For events \(A\) and \(B\),
\( P(\text{A and B}) = P(A) \cdot P(B)^{*}\).
\(^{*}\) If events \(A\) and \(B\) are dependent, \(P(B)\) depends on the fact that \(A\) happened.
So why does this multiplication rule work? Consider a box containing three balls -- one red, one blue, and one white. One ball is selected, its color is observed, and then the ball is placed back in the box. The balls are scrambled, and a ball is selected again. The tree diagram shows the sample space for the experiment.
From the tree diagram, there are 9 equally likely outcomes in the sample space is \(S =\{RR, RB, RW, BR, BB, BW, WR, WB, WW\}\). By counting, we can see that the probability of selecting a red ball followed by a white ball \(P(RW)\) is \(\frac{1}{9}\).
We could have reached this same answer by thinking another way and without listing a sample space.
- We know there are 9 outcomes in the sample space because there are 3 options for the first selection of ball and 3 options for the second selection of ball. This means there are \(3 \times 3=9\) ways to make the selection of two balls together.
- There is only one ball that is red during the first selection and one ball that is white during the second selection. This means there is \(1 \times 1=1\) way to to make the selection of these two colors.
- Therefore, the probability of selecting red ball followed by a white ball is \(\frac{1\times1}{3\times3}\) or \(\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\). We have multiplied \(P(\text{red ball on 1st selection}) \times P(\text{white ball on 2nd selection})\) to get \(P(\text{red ball on 1st selection and white ball on 2nd selection})\).
Suppose a fair coin is tossed.
- If the coin is tossed twice, what is the probability that both tosses land heads up?
- If the coin is tossed four times, what is the probability that all four tosses and heads up?
Solution
The tosses of the coins are independent events. We will use the formula for \(P(A \text{ and } B)\) when \(A\) and \(B\) are independent.
- \(P(\text{two heads in a row}) = P(\text{1st is heads and 2nd is heads})\)
\( = P(\text{1st is heads}) \cdot P(\text{2nd is heads}) \)
\( = \dfrac{1}{2} \cdot \dfrac{1}{2} \)
\( = \dfrac{1}{4}\)
- \(P(\text{four heads in a row}) = P(\text{1st is heads and 2nd is heads and 3rd is heads and 4th is heads})\)
\( = P(\text{1st is heads}) \cdot P(\text{2nd is heads}) \cdot P(\text{3rd is heads}) \cdot P(\text{4th is heads})\)
\( = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2}\)
\( = \left(\dfrac{1}{2}\right)^4\)
\( = \dfrac{1}{16}\)
The probability that all four tosses land heads up is \(\dfrac{1}{16}\).
The game spinner has 6 equal-sized sections. Jeremy will spin the spinner 3 times in a row.
- Find the probability that Jeremy gets blue on all three spins.
- Find the probability that Jeremy gets red on the first spin, blue on the second spin, and red on the third spin.
- Answer
-
- \(\frac{1}{27}\)
- \(\frac{1}{12}\)
A bag contains 5 red and 4 white marbles. Two marbles are selected one by one and colors are recorded. What is the probability that the first marble is red and the second marble is white if
- the first marble is returned to the bag before the second marble is selected.
- the first marble is not returned to the bag before the second marble is selected.
Solution
- Since the first marble is returned to the bag before the second marble is selected, these are independent events.
\[\begin{align*} P(\text{1st red and 2nd white}) &= P(\text{1st red}) \cdot P(\text{2nd white}) \\[4pt] &= \dfrac{5}{9} \cdot \dfrac{4}{9} = \dfrac{20}{81}\end{align*}\]
The probability that the first marble is red and the second marble is white is \(\dfrac{20}{81}\).
- Since the first marble not returned to the bag before the second marble is selected, these are dependent events.
\[\begin{align*} P(\text{1st red and 2nd white}) &= P(\text{1st red}) \cdot P(\text{2nd white}) \\[4pt] &= \dfrac{5}{9} \cdot \dfrac{4}{8} = \dfrac{20}{72} = \dfrac{5}{18}\end{align*}\]
The probability that the first marble is red and the second marble is white is \(\dfrac{5}{18}\).
A box contains 6 red blocks, 4 green blocks, and 3 black blocks. What is the probability of selecting
- two red blocks in a row if the first red block is put back in the box?
- two red blocks in a row if the first red block is not put back in the box?
- a red block and then a green block if the red block is not put back in the box?
- Answer
-
- \(\frac{36}{169}\)
- \(\frac{5}{26}\)
- \(\frac{2}{13}\)
Abby has an important meeting in the morning. She sets 3 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.
\[\begin{align*} P(\text{all three fail}) &= P(\text{first fails}) \cdot P(\text{second fails})\cdot P(\text{third fails}) \\[4pt] &= (0.03)(0.03)(0.03) \\[4pt] &= (0.03)^3 \\[4pt] &= 2.7 \times 10^{-5} \end{align*}\]
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.
Approximately 80% of U.S. children are covered by some type of health insurance. Assume that whether one child has health insurance or not has no effect on whether the next child has health insurance. If 4 children are selected at random, find the probability that
- all four of the children have insurance.
- none of the four children have insurance
Solution
- We will assume independence in this scenario.
\(\begin{aligned} P(\text{all four have insurance}) &= P(\text{1st has insurance}) \cdot P(\text{2nd has insurance}) \cdot P(\text{3rd has insurance}) \cdot P(\text{4th has insurance}) \\ &= (0.80)(0.80)(0.80)(0.80) \\ &= (0.80)^4 \\&= 0.4096 \end{aligned}\)
The probability that all four children are covered by insurance is 0.4096 or 40.96%.
- To find the probability that none of the four children have insurance, we will need to use the Complement Rule. If 80% of children are covered by insurance, then 100% - 80% = 20% are not covered by insurance.
\(\begin{aligned} P(\text{none of the four have insurance}) &= P(\text{1st no insurance}) \cdot P(\text{2nd no insurance})\cdot P(\text{3rd no insurance}) \cdot P(\text{4th no insurance}) \\&= (0.20)(0.20)(0.20)(0.20) \\&= (0.20)^5)\\ &= 0.0016 \end{aligned}\)
The probability that none of the four children are covered by insurance is 0.0016 or 0.16%.
According to some studies, it is believed that 15% of people are left-handed. Suppose 3 people are selected at random.
- Find the probability that all three are left-handed.
- Find the probability that none of the three are left-handed.
- Find the probability that only the first one selected is left-handed (and the others are right-handed.)
- Answer
-
- 0.003375 = 0.3375%
- 0.614125 = 61.4125%
- 0.108375 = 10.8375%
'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. That is, we will find the probability of what we don't want to happen first, and then subtract that probability from 1. 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. The events are independent.
If \(P(\text{forgets}) = 0.15\), then \(P(\text{not forget}) = 0.85\).
\(\begin{aligned} P(\text{forgets at least once in 5 tries}) &= P(\text{forgets 1, 2, 3, 4 or 5 times in 5 tries}) \\[4pt] & = 1 - P(\text{forgets 0 times in 5 tries}) \\[4pt] &= 1 - P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \\[4pt] &= 1 - (0.85)(0.85)(0.85)(0.85)(0.85) \\[4pt] & = 1 - (0.85)^{5} \\[4pt] &\approx0.5563 \end{aligned}\)
The probability that the child will forget her homework at least one day in the next five days is 0.5563 or 55.63%
The idea in the previous example can be generalized as the At Least Once Rule .
If an experiment is repeated \(n\) times where the \(n\) trials are independent and the probability of event \(A\) occurring one time is \(P(A)\),
\(P(A \text{ occurs at least once}) = 1 - P(\text{not } A)^{n}\).
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
If \(P(\text{sees falcon}) = 0.21\), then \(P(\text{doesn't see a falcon}) = 1- 0.21 =0.79\).
\(\begin{aligned} P(\text{sees falcon at least once in 8 trips}) &= 1 - P(\text{doesn't see a falcon})^8 \\[4pt] &= 1 - (0.79)^{8} \\[4pt] &\approx 0.8483 \end{aligned}\)
The probability of seeing a falcon at least once in eight trips to the lake is approximately 0.8483 or 84.83%.
A multiple choice test consists of six questions. Each question has four choices for answers (A, B, C, or D), 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 \(P(\text{not correct}) =1 - \dfrac{1}{4} = \dfrac{3}{4}\).
\(\begin{aligned} P(\text{at least one correct in six trials}) &= 1 - P(\text{not correct})^{6} \\[4pt] &= 1 - \left(\dfrac{3}{4}\right)^{6} \\[4pt] &\approx 0.8220 \end{aligned}\)
The probability that he gets at least one answer correct is 0.8220 or 82.20%.
Mary is in the archery club. The probability that she hits the target on an attempt is 65%. She will make 4 attempts to hit the target. What's the probability that Mary hits the target at least once on her attempts?
- Answer
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0.9850 or 98.50%