Probability Examples
1 Examples of probability and counting problems
A fruit packaging company is preparing “Mixed Fruit Boxes” for customers. Each box contains \(15\) pieces of fruit, which can be selected from \(5\) different types of fruit: apples, bananas, oranges, mangoes, and kiwis. The packers can include any combination of these fruits in each box.
Answers
Fruits can be repeated and the order does not matter, so we are dealing with a combination with replacement problem:
Let the number of fruit types be \(n = 5\) and the number of fruits to be packed is \(m = 15\).
The number of possible combinations is given as:
\[ \binom{n + m - 1}{m} = \binom{5 + 15 - 1}{15} = \binom{19}{15} = 3,876 \]
To ensure that each type of fruit is included in the box, we need to assign one fruit from each type to a slot in the box.
This means that we use \(5\) slots leaving \(10\) slots to be filled with any fruit combination.
This is also a combination with replacement problem but now \(n = 5\) and \(m = 10\):
This is is given as:
\[ \binom{5+10-1}{10} = \dbinom{14}{10} = 1,001 \]
So, the probability that a random box contains at least one fruit from each type is:
\[ \frac{1001}{3876} = 0.258 \]
The Olympic Games final of the \(100\) m sprint consists of \(8\) runners. How many combinations for the top \(3\) finishers exist when:
Answers
We want to find the number of arrangements for the top \(3\) finishers from \(8\) runners.
This is a permutation problem where order matters but without replacement because a runner cannot finish in more than one position.
The number of permutations is given as:
\[ _8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 336 \]
When order does not matter, we are dealing with a combination problem and again without replacement.
The number of combinations is given as:
\[ _8C_3 = \frac{8!}{3! \cdot\ (8-3)!} = \frac{8!}{3!\cdot\ 5!} = 56 \]
Given three pairwise disjoint and exhaustive events \(A\), \(B\), and \(C\). \[ P(A) = 0.1, \quad P(B) = 0.3 \] Calculate:
Answers:
\(P(A \cup B \cup C) = P(A) + P(B) + P(C) = 1\)
\(P(C) = 1 - P(A) - P(B) = 1 - 0.1 - 0.3 = 0.6\)
- \(P(A \cup B) = P(A) + P(B) = 0.1 + 0.3 = 0.4\)
- \(P(A \cap B) = 0\)
A restaurant surveyed its customers for complaints about food quality and service quality. The restaurant found that: \(30\%\) of customers complain about food quality, \(19\%\) of customers complain about service quality, and \(12\%\) of customers complain about both food and service.
If a customer is randomly selected, what is the probability that this customer:
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You have a set of \(4\) red, \(5\) orange, and \(2\) blue balls.
Answers:
To find the distinct arrangements, we are dealing with a permutation of multiple distinguishable groups with indistinguishable objects.
The number of arrangements is given as:
\[ \frac{11!}{4! \cdot\ 5! \cdot\ 2!} = 6,930 \]
A school has \(7\) teachers, each is preassigned \(2\) classrooms by the principal. The teachers are asked to select \(2\) classrooms for the upcoming school year. What is the probability that a specific teacher:
Answer
This is a combination without replacement problem, where the order does not matter.
The total number of ways to select \(2\) classrooms from \(14\) is:
\[ _{14}C_2 = \binom{14}{2} = \frac{14!}{2! \cdot\ 12!} = 91 \]
There is only \(1\) way to select both preassigned classrooms, so its probability is:
\[ \frac{1}{91} = 0.011 \]
The number of ways to select \(2\) classrooms from \(12\) is:
\[ _{12}C_2 = \binom{12}{2} = \frac{12!}{2! \cdot\ 10!} = 66 \]
So, the probability that a teacher selects none of the preassigned classrooms is:
\[ \frac{66}{91} = 0.725 \]
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