Careful counting of combinations is needed to solve probability problems. In the case where objects are grouped, it is often easiest to divide the number of ways to form a group using the given restrictions by the total number of groups that can be formed without restrictions. Let us consider a word game example as shown below.
Example 1: A word game requires players to select 4 tiles from a bag containing 26 tiles, each with one of the letters \(A\) through \(Z\) written on it. If each letter appears once and only once, what is the probability that a player will be able to spell CATS with his tiles?
Solution:
Since a player needs a \(C\), an \(A\), a \(T\), and an \(S\) in any order, we are looking at a combination calculation. We first need to determine how many combinations there are, choosing 4 letters from a total of 26 :
Choosing 4 from 26 :
\(
\begin{aligned}
{ }^{26} C_4 & =\frac{26 !}{(26-4) ! 4 !}=\frac{26 !}{22 ! 4 !} \\
& =\frac{26 \times 25 \times 24 \times 23}{4 \times 3 \times 2 \times 1} \\
& =59,800 \text { combinations. }
\end{aligned}
\)
Since only one combination allows a player to spell CATS, the probability of getting that combination is \(\frac{1}{59,800}\).
Example 2: A funfair game consists of pulling numbered chips from a bag. The game starts with nine chips numbered 1 through 9, and players are allowed to pull out three chips. A player wins by drawing the number 7 chip. What is the probability that a player will win?
Solution:
To find the probability of winning this game we need to know two pieces of information:
(a) the total number of combinations for the game and
(b) the number of combinations that contain a 7.
To find the total number of combinations for the game, use the formula \({ }^n C_r\) with \(n=9\) and \(r=3\) :
\(
{ }^9 C_3=\frac{9 !}{(9-3) ! 3 !}=\frac{9 !}{6 ! 3 !}=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}=84 \text { combinations }
\)
Now we need to determine how many combinations contain a 7. We can figure this out by using reasoning: given that there MUST be a 7 in the list, the number of combinations that contain a seven is the same as the number of combinations of choosing any two numbers out of the eight chips that don’t include a 7. In other words, if we imagine that we got to pick the 7 chip on purpose, how many ways would there be of picking the other two chips?
To find that number, we use the formula \({ }^n C_r\) with \(n=8\) and \(r=2\) :
\(
{ }^8 C_2=\frac{8 !}{(8-2) ! 2 !}=\frac{8 !}{6 ! 2 !}=\frac{8 \times 7}{2 \times 1}=28 \text { combinations }
\)
So the probability is given by:
\(
P(\text { getting a } 7)=\frac{84}{28}=\frac{1}{3}=\text { or one in three. }
\)
Example 3: Calculate the probability of being dealt four aces in a five-card poker hand.
Solution:
First, we will calculate, using the formula, the number of combinations for choosing 5 cards from a 52-card deck.
Choosing 5 from 52 :
\({ }^{52} C_5=\frac{52 !}{(52-5) ! 5 !}=\frac{52 !}{47 ! 5 !}=\frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1}=2,598,960\) unique hands.
Next, we need to calculate how many hands there are that contain four aces.
Since the four aces are accounted for, there are 48 (that’s 52 – 4) cards left in the deck.
Then the number of hands which has 4 aces is 48 (because the 5th card can be any of 48 other cards).
Since a unique hand is independent of the order in which the cards are dealt, there must be 48 unique hands that contain four aces (one unique hand for every non-ace card in the deck).
There are 48 possible hands that contain four aces. So the probability of being dealt four aces in poker is:
\(
P(\text { four aces })=\frac{48}{2,598,690}=\frac{1}{54,145}
\)
Example 4: Five boys and seven girls, are randomly assigned to three groups of four students each. What is the probability that one of the three groups will be all girls?
Solution:
Total number of students = 5 + 7 = 12.
Selecting 4 students from a group of 12 students, the number of combinations \({ }^{12} C_4=495\). Therefore there are 495 ways to select 4 students from a group of 12.
Similarly selecting 4 girls from 7, the number of ways we can do is \({ }^7 C_4=35\). The probability of a randomly formed group consisting of all girls is therefore 35/495 = 7/99.
Because there are 3 groups formed, the probability of any one of those groups consisting of all girls is:
\(
3 \cdot \frac{7}{99}=\frac{21}{99}=\frac{7}{33}
\)
Alternate way:
This problem could have also been solved using compound probability by finding the probability of one group having four girls and then multiplying by three:
\(
\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} \cdot \frac{4}{9} \cdot 3=\frac{7}{99} \cdot 3=\frac{7}{33} .
\)
Example 5: The 26 letters of the alphabet are placed on tiles and randomly separated into two equal piles. What is the probability that the word MATH can be found in one of the two piles?
Solution:
First, we consider the number of 13 letter piles selected from 26 letters. The number of ways we can do is \({ }^{26} C_{13}=10,400,600\). Then we consider the number of those piles that contain the letters MATH. This leaves room in the group for 9 more letters(13-4=9) selected from a set of 22(26-4=22), or \({ }^{22} C_9=497,420\). Dividing and simplifying we get 497,420/10,400,600 =11/230. There are two piles of letters. So the probability of MATH being in one of the two piles is twice that 2(11/230)=11/115.
Alternate way:
we can use compound probability:
\(
{ }^{13} C_4 \cdot \frac{4}{26} \cdot \frac{3}{25} \cdot \frac{2}{24} \cdot \frac{1}{23} \cdot 2=\frac{11}{115}
\)
Example 6: You have ten marbles: four blue and six red. You choose three marbles without looking. What is the probability that all three marbles are blue?
Solution:
There are \({ }^4 C_3\) ways to choose the blue marbles.
There are \({ }^{10} C_3\) total number of combinations.
\(
P(\text { all } 3 \text { marbles are blue })=\frac{{ }^4 C_3}{{ }^{10} C_3}
=\frac{4}{120}=\frac{1}{30}
\)
There is approximately a \(3.33 \%\) chance that all three marbles drawn are blue.
Example 7: There are 45 dogs at the shelter and you want to choose 2 of them to adopt. How many different pairs could you adopt? If the dogs were chosen at random, would it be possible for you to find the probability of adopting a particular pair?
Solution:
Since we do not care what order the dogs are chosen in, this is a combination problem. We want to know how many groups of 2 we can choose from 45.
\(
\begin{aligned}
{ }^{45} C_2 & =\frac{45 !}{2 !(45-2) !} \\
& =\frac{45 !}{2 ! 43 !} \\
& =990
\end{aligned}
\)
There are 990 ways to adopt 2 of the 45 dogs.
To find the probability that a particular pair of dogs is adopted, note that there is only 1 way to successfully pick that pair. There are 990 ways to pick any pair. Therefore, the probability is \(\frac{1}{990}\).
Example 8: The U.S. Senate is made up of 100 people, two per state. How many different four-person committees are possible? what is the probability that the committee will only have members from two states?
Solution:
This question does not care how the committee members are chosen; we will use the formula for combinations.
\(
{ }^{100} C_4=\frac{100 !}{4 !(100-4) !}=3,921,225 \text { ways }
\)
If there are only members from two states, that means two are from one state and two are from another. This problem is simply about how many ways you can choose 2 states out of 50 states.
\(
{ }^{50} C_2=\frac{50 !}{2 !(50-2) !}=1,225 \text { ways }
\)
\(
P(\text { only two states represented })=\frac{1225}{3,921,225}=0.00031
\)
The probability that only two states will be represented on the committee is \(0.031 \%\), which is a very small chance!
Example 9: A 4-digit PIN number is selected. What is the probability that there are no repeated digits?
Solution:
There are 10 possible values for each digit of the PIN (namely: \(0,1,2,3,4,5,6,7,8,9\) ), so there are \(10 \cdot 10 \cdot 10 \cdot 10=10^4=10000\) total possible PIN numbers.
To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute \(10 \cdot 9 \cdot 8 \cdot 7\), or notice that this is the same as the permutation \({ }^{10} P_4=5040\).
The probability of no repeated digits is the number of 4 digit PIN numbers with no repeated digits divided by the total number of 4-digit PIN numbers. This probability is \(\frac{{ }^{10} P_4}{10^4}=\frac{5040}{10000}=0.504\)
Example 10: In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.
Solution:
n order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player’s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is \({ }^{48} C_6=12,271,512\). Of these possible outcomes, only one would match all six numbers on the player’s ticket, so the probability of winning the grand prize is:
\(
\frac{{ }^6 C_6}{{ }^{48} C_6}=\frac{1}{12271512} \approx 0.0000000815
\)
Example 11: In the state lottery, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.
Solution:
As above, the number of possible outcomes of the lottery drawing is \({ }^{48} C_6=12,271,512\). In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by \({ }^{6} C_5=6\) and the number of ways to choose 1 out of the 42 losing numbers is given by \({ }^{42} C_1=42\). Thus the number of favorable outcomes is then given by the Basic Counting Rule: \({ }^{6} C_5 \times { }^{42} C_1=6 \times 42=252\). So the probability of winning the second prize is.
\(
\frac{\left({ }^6 C_5\right)\left({ }^{42} C_1\right)}{{ }^{48} C_6}=\frac{252}{12271512} \approx 0.0000205
\)
Example 12: Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.
Solution:
The total number of possible outcomes(drawing 5 cards from a deck of 52 cards): \({ }^{52} C_5\)
we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Since there are four Aces and we want exactly one of them, there will be \({ }^{4} C_1\) ways to select one Ace. since there are 48 non-Aces and we want 4 of them, there will be \({ }^{48} C_4\) ways to select the four non-Aces.
Now we use the Basic Counting Rule to calculate that there will be \({ }^{4} C_1 \times { }^{48} C_4\) ways to choose one ace and four non-Aces.
Putting this all together, we have
\(
P(\text { one Ace })=\frac{{ }^{4} C_1 \times { }^{48} C_4}{{ }^{52} C_5}=\frac{778320}{2598960} \approx 0.299
\)
Example 13: Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.
Solution:
Now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:
\(
P \text { (two Aces })==\frac{{ }^{4} C_2 \times { }^{48} C_3}{{ }^{52} C_5}=\frac{103776}{2598960} \approx 0.0399
\)
Example 14: Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people?
Solution:
There are a lot of ways there could be at least one shared birthday. Fortunately, there is an easier way. We ask ourselves “What is the alternative to having at least one shared birthday?” In this case, the alternative is that there are no shared birthdays. In other words, the alternative to “at least one” is having none. In other words, since this is a complimentary event,
\(
\mathrm{P}(\text { at least one })=1-\mathrm{P}(\text { none })
\)
We will start, then, by computing the probability that there is no shared birthday. Let’s imagine that you are one of these three people. Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday? There are 365 days in the year (let’s ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is \(364 / 365\). Now we move to the third person. What is the probability that this third person does not have the same birthday as either you or the second person? There are 363 days that will not duplicate your birthday or the second person’s, so the probability that the third person does not share a birthday with the first two is \(363 / 365\).
We want the second person not to share a birthday with you and the third person not to share a birthday with the first two people, so we use the multiplication rule:
\(
P(\text { no shared birthday })=\frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \approx 0.9918
\)
and then subtract from 1 to get
\(
P(\text { shared birthday })=1-P(\text { no shared birthday })=1-0.9918=0.0082 .
\)
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