Probability Power Sheet Reference Sheet Probability Cheat Sheet
Unit Summary: Topics A & B
Essential Formulas
Probability Rules
Addition Rule (General)
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Multiplication Rules
General \( P(A \cap B) = P(A) \cdot P(B|A) \)
Indep. \( P(A \cap B) = P(A) \cdot P(B) \)
Conditional Probability: The likelihood of B given A occurred. If independent, P(B|A) = P(B) .
Counting Rules
Principle
\( n_{1} \cdot n_{2} \cdot n_{3} \dots \)
Factorials
\( n! \text{ (where } 0! = 1) \)
Permutations (Order Matters)
\( _{n}P_{r} = \frac{n!}{(n - r)!} \)
Positions, Wettbewerbe, Passwords
Combinations (No Order)
\( _{n}C_{r} = \frac{n!}{r!(n - r)!} \)
Committees, Groups, Toppings
Random Variables
Discrete Isolated Values (Counts)
Ex: Number of siblings, heads on coins.
Continuous Interval Values (Measures)
Ex: Height, weight, time, distance.
Tables & Graphs
Probability Table
\( x_{i} \) \( P(x_{i}) \) Value Prob
\( P(x) = \frac{\text{Freq}}{\text{Total } n} \)
• Dot Plot: Every dot marks one trial outcome.
• Histogram: Area of bar = Probability.
Expected Value
The Mean
\( E(x) = \sum x \cdot P(x) \)
Calculation
Multiply every **x** by its probability **p**, then sum.
Concept
Theoretical average balance point.
\( \sum \) Summation
\( \cup \) "Or"
\( \cap \) "And"
Order: \( P \) | Groups: \( C \)
Probability Practice Review Worksheet Probability Practice Showdown
Mid-Module Review: Topics A & B
Name:
Date:
1
In a volleyball match, a player gets two chances to land a serve in bounds. If the first serve is out, they are allowed a second serve. Suppose a player, Jamie, has a \( 0.55 \) probability of the first serve landing in bounds and a \( 0.82 \) probability of the second serve landing in bounds. Once a serve lands in bounds, the rally begins. For Jamie, if the first serve is in, there is a \( 68\% \) chance of winning the point. If the first serve is out but the second serve is in, Jamie has a \( 28\% \) chance of winning the point.
a. Calculate the probability that Jamie lands the first serve in bounds and then goes on to win the point for a randomly selected point.
b. Calculate the probability that Jamie misses the first serve, lands the second serve, and then wins the point.
c. Calculate the probability that Jamie wins a randomly selected point.
2
In a local forest, there are approximately \( 150 \) species of spiders, but only about \( 15 \) are considered venomous to humans. Suppose a researcher randomly selects one species from each of four different habitats to study. What is the probability that exactly one of the four species selected is venomous?
3
Standard parking permits in a university city are currently of the form 2 letters followed by 4 numbers. Assume the letters are A–Z and the numbers are 0–9.
a. In theory, how many different possible parking permits are there, assuming we can repeat letters and numbers?
b. How many different possible parking permits are there if we are not allowed to repeat any letters or numbers?
c. For part (b), did you use permutations or combinations to carry out the calculation? Explain how you know.
4
A grocery store manages its wait times by opening different numbers of checkout registers. Let the random variable \( X \) be the number of checkout registers open during the peak evening hours. The probability distribution of \( X \) is given in the table below:
<table class="w-full border-collapse border-2 border-slate-800 mb-8"><tbody><tr><th class="border border-slate-300 p-2 bg-slate-100 font-mono italic">x</th><td class="border border-slate-300 p-2 text-center">0</td><td class="border border-slate-300 p-2 text-center">1</td><td class="border border-slate-300 p-2 text-center">2</td><td class="border border-slate-300 p-2 text-center">3</td><td class="border border-slate-300 p-2 text-center">4</td><td class="border border-slate-300 p-2 text-center">5</td></tr><tr><th class="border border-slate-300 p-2 bg-slate-100 font-mono italic">p(x)</th><td class="border border-slate-300 p-2 text-center">0.10</td><td class="border border-slate-300 p-2 text-center">0.25</td><td class="border border-slate-300 p-2 text-center">0.30</td><td class="border border-slate-300 p-2 text-center">0.20</td><td class="border border-slate-300 p-2 text-center">0.10</td><td class="border border-slate-300 p-2 text-center">0.05</td></tr></tbody></table>
a. Produce a graph of the probability distribution for this random variable, including all relevant labels.
0.40
0.30
0.20
0.10
0.00
0
1
2
3
4
5
Number of Registers (x)
b. Calculate the expected value of the random variable.
c. Explain how to interpret this expected value in context.
5
The following table lists the number of households in a city survey with 0 smartphones, 1 smartphone, 2 smartphones, or 3 or more smartphones.
No Smartphone
One Smartphone
Two Smartphones
Three or More
3,250
14,800
21,450
10,500
a. Use these data to create a table of relative frequencies that could be used as estimates of the probability distribution of the number of smartphones for a randomly selected household.
No Smartphone
One Smartphone
Two Smartphones
Three or More
b. Suppose you want to examine the distribution of the number of smartphones in all households. Define a random variable that corresponds to the probability distribution in part (a).
c. Assume for the moment that the last column corresponds to exactly three smartphones. Calculate the expected number of smartphones per household.
d. Now, reconsider the last category. Suppose we were to find the information for the actual number of smartphones for these households. Would the expected number of smartphones per household with this new information be larger or smaller than the expected value you found in (c)? Explain your reasoning.
e. Suppose a town has about 6,200 households. What is a good estimate for the total number of smartphones in town? Explain how you determined your answer.
Probability Practice Answer Key Answer Key
Probability Practice Showdown
Problem 1: Volleyball Serves
a. \( P(\text{1st In and Win}) \)
\( 0.55 \cdot 0.68 = 0.374 \)
b. \( P(\text{1st Out, 2nd In, and Win}) \)
\( (1 - 0.55) \cdot 0.82 \cdot 0.28 = 0.45 \cdot 0.82 \cdot 0.28 = 0.10332 \)
c. \( P(\text{Total Win}) \)
\( 0.374 + 0.10332 = 0.47732 \)
Problem 2: Spider Species
Calculated using combination of selections:
\( P(\text{exactly 1}) = \frac{\binom{15}{1} \cdot \binom{135}{3}}{\binom{150}{4}} \approx \frac{15 \cdot 401,310}{20,482,975} \approx 0.2939 \)
Problem 3: Parking Permits
a. With Repetition
\( 26^2 \cdot 10^4 = 676 \cdot 10,000 = 6,760,000 \)
b. Without Repetition
\( (26 \cdot 25) \cdot (10 \cdot 9 \cdot 8 \cdot 7) = 650 \cdot 5,040 = 3,276,000 \)
c. Explanation
Permutations. Order matters because license plate "AB 1234" is distinct from "BA 1234".
Problem 4: Checkout Registers
b. Expected Value Calculation
\( E(X) = (0 \cdot 0.10) + (1 \cdot 0.25) + (2 \cdot 0.30) + (3 \cdot 0.20) + (4 \cdot 0.10) + (5 \cdot 0.05) \)
c. Interpretation
In the long run, the store expects to have an average of 2.1 registers open during peak evening hours.
Problem 5: Smartphone Survey
a. Relative Frequencies (Total \( n = 50,000 \))
0: 0.065 | 1: 0.296 | 2: 0.429 | 3+: 0.210
c. Expected Value
\( E(X) = (0 \cdot 0.065) + (1 \cdot 0.296) + (2 \cdot 0.429) + (3 \cdot 0.210) = 1.784 \)
d. Reasoning
Larger. If the last category includes households with 4 or 5 phones, these higher values would pull the weighted average (expected value) upward.
e. Town Estimate
\( 1.784 \cdot 6,200 = 11,060.8 \rightarrow \text{Approx 11,061 phones} \)
