3.7: Probability with Counting Methods
Recall our definition of theoretical probability: If outcomes of an experiment are equally likely, then
\(P(E)=\dfrac{\text { number of outcomes in event } E }{\text {number of outcomes in the sample space }S}\).
Earlier in this chapter, we determined the number of outcomes in event \(E\) and in the sample space \(S\) by listing them to count them. However, some experiments are so complex that it would take too long to list all the outcomes and then count them. When that is the case, we depend on counting methods.
Probabilities Involving Counting Methods
As we begin to find probabilities with counting methods, we must continue to focus on whether order is important in distinguishing one outcome from another in the event.
There is a box that contains 13 marbles: 8 blue and 5 red. We will select two marbles from the box. What's the probability that both marbles are blue?
Solution
The order in which we select the two blue marbles is not important to the answer. We just want both to be blue, so this problem can be considered a combination probability problem. We can use combinations to find the number of outcomes in the event "select 2 blue marbles." Then, we use combinations to find the number of outcomes in the sample space of "selecting 2 marbles of any color."
- The number of ways to select 2 blue marbles from the 8 blue marbles in the box is
\( _{8}C_2 =\frac{8 !}{(8-2)! \; 2!} = \frac{8 !}{6! \; \times \; 2!} = \frac{8 \times 7 \times 6! }{6! \; \times \; 2!} = \frac{8 \times 7 \times \cancel{6!} }{\cancel{6!} \; \times \; 2!} = \frac{8 \times 7 }{ 2 \times 1} = \frac{56}{2}= 28 \).
- The number of ways to select 2 marbles of any color from the 13 marbles in the box is
\( _{13}C_2 =\frac{13 !}{(13-2)! \; 2!} = \frac{13 !}{11! \; \times \; 2!} = \frac{13 \times 12 \times 11! }{11! \; \times \; 2!} = \frac{13 \times 12 \times \cancel{11!} }{\cancel{11!} \; \times \; 2!} = \frac{13 \times 12 }{ 2 \times 1} = \frac{156}{2}= 78 \).
- So, \(P(\text{selecting 2 blue marbles}) = \frac{_8C_2}{_{13}C_2}=\frac{28}{78}=\frac{14}{39} \approx 0.3590\).
The probability of selecting 2 blue marbles from the box is 0.3590 or 35.90%.
There are 8 finalists in the 100-meter sprint at the Olympic games. Suppose 4 of the runners are from the United States and that all the runners have an equal chance of winning. What's the probability that runners from the United States finish in 1st, 2nd, and 3rd place?
Solution
Order is important in counting the number of ways the runners could finish 1st, 2nd, and 3rd place, so this problem can be considered a permutation probability problem.
We can use permutations to find the number of orders in which 3 of the 4 U.S. runners could finish the race. Then, we use permutations to find the number of orders in which 3 of the 8 runners could finish the race.
- The number of orders in which any 3 of the 4 U.S. runners can finish the race is
\( _{4}P_3 = \frac{4 !}{(4-3)!} = \frac{4 !}{1!} = \frac{4\times 3 \times 2 \times 1 }{1} = \frac{24}{1}= 24 \).
- The number of orders in which any 3 of the 8 runners can finish the race is
\( _{8}P_3 = \frac{8 !}{(8-3)!} = \frac{8 !}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = \frac{8 \times 7 \times 6 \times \cancel{5!}}{\cancel{5!}}= 8 \times 7 \times 6= 336 \).
- So, \(P(\text{3 runners from U.S. win 1st, 2nd, and 3rd place }) = \frac{_4P_3}{_{8}P_3}=\frac{24}{336}=\frac{1}{14} \approx 0.0714\).
The probability that U.S. runners finish in 1st, 2nd, and 3rd place is 0.0714 or 7.14%.
From this point on, we will use technology to count permutations and combinations rather than writing out all steps.
There are 20 students in a class, of which 12 are females. The names of the students are put into a hat and five names are drawn. What is the probability that all of the students chosen are females?
Solution
We need to find \(P(\text { select all females })=\frac{\text {number of ways to select } 5 \text { females }}{\text{number of ways to select } 5 \text { students }}\).
This is a combination problem because the order of selecting the students is not important in counting the outcomes as different. We just want them all to be females.
The number of ways to select 5 females is \(_{12}C_5=792\). The number of ways to select 5 students from the class is \(_{20}C_5=15,504\).
\(P(\text { select all females })=\dfrac{\text {number of ways to select } 5 \text { females }}{\text {number of ways to select } 5 \text { students }}=\dfrac{_{12}C_5}{_{20}C_5}=\dfrac{792}{15,504} \approx0.0511\)
The probability that all the students chosen are females is 0.0511 or 5.11%.
A research laboratory requires a four-digit passcode to enter. A passcode may contain any four digits 0, 1, 2, 3, ... 8, 9 but no digit can be repeated. What is the probability that a scientist is assigned a passcode with the digits 6, 7, 8, and 9 in any order?
Solution
The order of the digits in a passcode is meaningful. That is, a passcode of 6-7-8-9 is different than 7-6-9-8 even though the same four digits are used. Therefore, we need to count permutations in this problem.
The number of ordered passcodes that can be formed with the digits 6, 7, 8, and 9 is \(_{4}P_4=24\), or \(4!=24\). The number of ordered passcodes that can be formed using any 4 digits is \(_{10}P_4=5,040\).
\(P(\text {passcode contains the digits 6, 7, 8, and 9 })=\dfrac{\text {number of passcodes containing the digits 6, 7, 8, and 9 } }{\text {number of passcodes containing any 4 digits} }=\dfrac{_{4}P_4}{_{10}P_4}=\dfrac{24}{5,040} =\dfrac{1}{210} \approx0.0048\)
The probability that a scientist is assigned a passcode with the digits 6, 7, 8, and 9 is 0.0048 or 0.48%.
There are 10 canned drinks in a cooler: 6 are colas and 4 are fruit drinks. Danielle will reach into the cooler and pull out 5 drinks without looking to see what she gets. Find the probability that all 5 drinks are colas.
- Answer
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\(\frac{1}{42}\)
A motorcycle license plate consists of 5 digits that are randomly selected. No digit is repeated. What is the probability of getting a license plate with all odd numbers?
- Answer
-
\(\frac{1}{252}\)
There are 12 males and 8 females in a jury pool. A group of 8 of them will be selected to sit on a jury. What's the probability of selecting 5 males and 3 females for the jury?
Solution
\(P(\text { select 5 males and 3 females })=\frac{\text {ways to select } 5 \text { male jurors and }3 \text { female jurors}}{\text {ways to select } 8 \text { jurors }}\)
This problem involves combinations because the order in which jurors are selected is not important to counting a jury as different.
First, we count the number of outcomes in the event "select 5 male jurors and 3 female jurors." The number of ways to select 5 male jurors is \(_{12}C_5 =792\). The number of ways to select 3 females jurors is \(_8C_3=56\). Using the Fundamental Counting Principle as we did at the end of Section 3.6, the number of ways of selecting 5 males and 3 females is \(_{12}C_5 \times \; _8C_3 = 792 \times 56 = 44,352\).
Next, we count the number of outcomes in the sample space of the experiment "select any 8 jurors." The number of ways to select 8 jurors from the group of \(12 + 8 = 20\) people is \(_{20}C_8 = 125,970\) ways.
Finally,
\(P(\text { 5 males and 3 females })=\dfrac{\text {ways to select } 5 \text { male jurors and }3 \text { female jurors}}{\text {ways to select } 8 \text { jurors }} = \dfrac{_{12}C_5 \times \; _8C_3 }{_{20}C_8}= \dfrac{44,352}{125,970} \approx 0.3521.\)
The probability the jury will contain 5 males and 3 females is 0.3521 or 35.21%.
There are 10 canned drinks in a cooler: 6 are colas and 4 are fruit drinks. Danielle will reach into the cooler and pull out 5 drinks without looking to see what she gets. Find the probability that 3 drinks are colas and 2 drinks are fruit drinks.
- Answer
-
\(\frac{10}{21}\)
A technician is launching fireworks near the end of a show. Of the remaining 15 fireworks, 8 are blue, 4 are red, and 3 are white. If she launches 5 of them in a random order, what is the probability that she launches no blue fireworks?
Solution
This problem involves combinations because the order in which fireworks are selected is not important to counting the outcomes as different.
If no blue fireworks are launched, then only red and white fireworks are to be selected. There are \(4+3=7\) fireworks that are red or white. The number of ways 5 fireworks can be selected from this group of 7 red or white fireworks is \(_7C_5 = 21\).
The total number of possible outcomes for the experiment of selecting 5 fireworks is \(_{15}C_5 = 3,003\).
\(P(\text {select no blue fireworks})=\dfrac{\text {ways to select } 5 \text { red or white fireworks }}{\text {ways to select } 5 \text { fireworks }} = \dfrac{_7C_5 }{_{15}C_5}= \dfrac{21}{3,003}= \dfrac{1}{143} \approx 0.0070.\)
The probability that no blue fireworks are launched is 0.0070 or 0.70%.
Compute the probability that a 5-card poker hand is dealt to you that contains exactly two hearts.
Solution
This problem involves combinations because the order in which cards are dealt is not important to counting poker hands as different. In solving this problem, it is important to understand that the event must be made up of 5 cards: 2 of the cards must be hearts and the remaining 3 cards can be anything else except hearts.
First, count the number of outcomes in the event "2 hearts and 3 non-hearts." The number of ways to select 2 hearts is \(_{13}C_2 =78\). The number of ways to select 3 non-hearts is \(_{39}C_3=9,139\). Using the Fundamental Counting Principle , the number of ways to get 2 hearts and 3 non-hearts is \(_{13}C_2 \times \; _{39}C_3 = 78 \times 9,139 = 712,842\).
Next, we count the number of outcomes in the sample space of a 5-card poker hand. The number of different poker hands is \(_{52}C_5 = 2,598,960\).
Finally,
\(P(\text { exactly 2 hearts})=\dfrac{\text {ways to select } 2 \text { hearts and }3 \text { non-hearts}}{\text {ways to select } 5 \text { cards }} = \dfrac{_{13}C_2 \times \; _{39}C_3 }{_{52}C_5}= \dfrac{78 \times 9,139}{2,598,960}= \dfrac{712,842}{2,598,960} \approx 0.2743.\)
The probability of being dealt exactly 2 hearts in a five-card poker hand is 0.2743 or 27.43%.
A bag contains 6 real diamonds and 5 fake diamonds. If 4 diamonds are picked from the bag at random, what is the probability that
- none of them are fake.
- exactly 2 of them are real.
- Answer
-
- \(\frac{1}{22}\)
- \(\frac{5}{11}\)
A basket contains 6 good apples and 4 bad apples. A distracted shopper reaches into the basket and picks 3 apples without looking. What is the probability he gets at least one bad apple?
Solution
This is a combination since the order in which the apples are picked is not important. Getting "at least one bad apple" when selecting 3 apples means there could be 1 bad apple, 2 bad apples, or 3 bad apples. This is an "at least one" problem so we can use the Complement Rule:
\(P(\text{select at least 1 bad apple})=1-P(\text{select no bad apples})\).
Selecting no bad apples means selecting 3 good apples. There are 6 good apples from which to choose. The number of ways of selecting 3 good apples is \(_6C_3 = 20\). The number of outcomes in the experiment of selecting any of the 3 apples from the basket is \(_{10}C_3 = 120\).
Therefore,
\(P(\text { select 3 good apples })=\dfrac{\text {ways to select } 3 \text { good apples }}{\text {ways to select } 3 \text { apples }} = \dfrac{_6C_3 }{_{10}C_3}= \dfrac{20}{120}= \dfrac{1}{6}\)
Finally,
\(\begin{align*}P(\text{at least 1 bad apple}) &=1-P(\text{select no bad apples}) \\ &=1-P(\text{select 3 good apples}) \\ &= 1 - \frac{1}{6} \\&= \frac{5}{6} \approx 0.8333 \end{align*}\)
The probability that the customer gets at least one bad apple is 0.8333 or 83.33%.
A jar contains 5 black buttons and 4 brown buttons. If 3 buttons are picked at random, what is the probability that at least one of them is brown?
- Answer
-
\(\frac{37}{42}\)