Math · Problem Solving and Data Analysis

Probability and conditional probability

Explanation

Probability measures the likelihood of an event occurring, expressed as a number between 00 (impossible) and 11 (certain), or as a percentage from 0%0\% to 100%100\%.

Basic Probability Principles

The probability of an event AA, written as P(A)P(A), is calculated by comparing desired outcomes to the total possible outcomes.

  • Basic Probability Formula:

    P(A)=Number of Favorable OutcomesTotal Number of Possible OutcomesP(A) = \dfrac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}

    Example: Rolling a 44 on a standard six-sided die   ⟹  P(rolling 4)=16\implies P(\text{rolling } 4) = \frac{1}{6}.

  • Complement Rule: The probability that an event does not happen is 11 minus the probability that it does happen.

    P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)

    Example: If the probability of rain is 0.30.3, the probability of no rain is 1−0.3=0.71 - 0.3 = 0.7.

  • Mutually Exclusive Events: Two events that cannot happen at the same time.

    P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)

    Example: Rolling a 22 or a 55 on a single die roll   ⟹  16+16=26=13\implies \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}.

Conditional Probability and Two-Way Tables

Conditional probability measures the likelihood of an event occurring given that another condition has already happened.

  • Conditional Probability Formula: The notation P(A∣B)P(A \mid B) reads "the probability of AA given BB."

    P(A∣B)=P(A and B)P(B)=Number of outcomes in both A and BTotal number of outcomes in condition BP(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{\text{Number of outcomes in both } A \text{ and } B}{\text{Total number of outcomes in condition } B}
  • Two-Way Tables: SAT probability questions frequently present data in two-way tables. To find conditional probability, restrict your sample space to only the row or column specified by the condition.

    Example: Consider the table below:

    PassedFailedTotal
    Studied4040554545
    Did Not Study101015152525
    Total505020207070
    • Probability of selecting someone who passed given that they studied:

      • Condition: Look only at the "Studied" row (total =45= 45).

      • Calculation: 4045=89\frac{40}{45} = \frac{8}{9}.

Example 1

Type of bicyclePriced under $500Priced $500 or moreTotal
Road61016
Mountain9514
Total151530
The table shows information about 3030 bicycles listed for sale at a shop.
If one of these bicycles is selected at random, what is the probability that it will be a mountain bike priced under $500?
A
310\frac{3}{10}
B
16\frac{1}{6}
C
914\frac{9}{14}
D
13\frac{1}{3}
Choice A is correct. There are 99 mountain bikes priced under 500500 and 3030 bicycles total, so the probability of selecting a mountain bike priced under 500500 is 930=310\frac{9}{30} = \frac{3}{10}. Choice B is incorrect. This is the probability of selecting a mountain bike priced 500500 or more. Choice C is incorrect. This is the probability, when choosing randomly from only the mountain bikes, of selecting one priced under 500500. Choice D is incorrect. This is the probability of selecting a road bike priced $500 or more.

Example 2

Has parkingNo parkingTotal
Has elevator15520
No elevator61420
Total211940
The table shows information about 4040 office buildings in a business district.
If one of these office buildings is selected at random, what is the probability that it does not have parking, given that it has an elevator?
A
14\frac{1}{4}
B
18\frac{1}{8}
C
540\frac{5}{40}
D
34\frac{3}{4}
Choice A is correct. Since the probability is conditioned on the building having an elevator, the denominator is restricted to only the 2020 buildings with an elevator (15+515+5), not all 4040 buildings. Of those 2020 buildings, 55 do not have parking, so the probability is 520=14\frac{5}{20} = \frac{1}{4}. Choice B is incorrect and may result from an arithmetic error in reducing the fraction. Choice C is incorrect; it uses the full 4040 buildings as the denominator instead of restricting to only the 2020 buildings with an elevator. Choice D is incorrect; this is the probability that a building with an elevator does have parking, not the probability that it does not.

Tips for solving

  1. Use row/column totals for conditional probability versus overall totals for general probability.

    Example: Based on the study group data table below:

    Passed ExamFailed ExamTotal
    Attended Review Session3232443636
    Did Not Attend181816163434
    Total505020207070
    • General probability (selected at random from all students):

      • Fraction who attended review and passed: 3270=1635\frac{32}{70} = \frac{16}{35}

    • Conditional probability (given they attended):

      • Fraction who passed out of attendees: 3236=89\frac{32}{36} = \frac{8}{9}

  2. Solve "at least one" problems using the complement rule to save computation steps.

    Example: Based on the vehicle inventory table below:

    SedanSUVTotal
    New151525254040
    Used303010104040
    Total454535358080
    • Question: What fraction of the SUVs are new?

      • "Given" group: All SUVs (Column total =35= 35)

      • Target: New SUVs (2525)

      • Probability: 2535=57\frac{25}{35} = \frac{5}{7}

  3. Determine if sequential events require multiplying probabilities.

    Example: Based on the voter survey table below:

    In FavorOpposedTotal
    Group A404010105050
    Group B202030305050
    Total60604040100100
    • Question: Among the voters who opposed, what fraction belonged to Group B?

      • Condition phrase: "Among the voters who opposed" (Opposed column total =40= 40)

      • Target: Group B (3030)

      • Probability: 3040=34\frac{30}{40} = \frac{3}{4}

Practice questions

00:00
Question 1 · easy
Each vertex of an 1818-sided polygon is labeled with a different letter from AA through RR. If one vertex is selected at random, what is the probability that the letter FF will be at the selected vertex? (Express your answer as a decimal or fraction, not as a percent.)
Question 2 · easy
A bag contains 2020 marbles: 66 red, 55 blue, and 99 green. If one marble is selected at random, what is the probability that it is red?
Question 3 · easy
A jar contains 5050 candies, and 40%40\% of them are red. If one candy is selected at random, what is the probability that it is red?
Question 4 · easy
Type of laptopPrice \$800 or lessPrice more than \$800Total
Windows laptops5914
Mac laptops6410
Total111324
The table shows information about 2424 laptops for sale at an electronics store. If one of these laptops is selected at random, what is the probability that it will be a Mac laptop priced $800\$800 or less?
Question 5 · easy
Type of laptopPrice \$800 or lessPrice more than \$800Total
Windows laptops5914
Mac laptops6410
Total111324
Using the same table of 2424 laptops, if one of the Mac laptops is selected at random, what is the probability that it is priced $800\$800 or less?
Question 6 · easy
A spinner is divided into 88 equal sections labeled 11 through 88. If the spinner is spun once, what is the probability that it lands on an even number?
Question 7 · easy
In a raffle, the probability of a ticket being a winning ticket is 14\frac{1}{4}. If there are 6060 tickets total, how many of them are winning tickets?
Question 8 · easy
Has seatingNo seatingTotal
Has Wi-Fi12416
No Wi-Fi314
Total15520
The table shows information about 2020 coffee shops in a city. If one of these coffee shops is selected at random, what is the probability that it has seating but does not have Wi-Fi?
Question 9 · easy
In a class of 2424 students, 99 play soccer. If one student from the class is selected at random, what is the probability that the student plays soccer?
Question 10 · easy
At a conference, there are 4040 attendees. Each attendee is assigned to group A, group B, or group C. If one attendee is selected at random, the probability of selecting an attendee in group A is 0.30.3, and the probability of selecting an attendee in group B is 0.450.45. How many attendees are in group C?