Skip to content
NumeriveMATH
Make learning comfortable
Changing style, motion, or reading speed keeps your current question, answers, homework, and saved progress. The school band still chooses your task layout. Saved to your learner profile across devices. All lessons and help stay available.
← Introduction to Counting and Probability levels

Introduction to Counting and Probability / LEVEL 3 · DIFFICULTY 3/5

Probability Models

Count outcomes and handle dependent draws.

3 stages · 24 practice problems · two 6-question assessment forms

STAGE CP 3.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Inclusion and Exclusion

Useful preparation: Count the Complement

Goal: Understand and apply inclusion and exclusion.

Before you begin: Count the Complement

Understand the idea

Adding two set sizes counts their intersection twice. Subtracting the intersection once leaves every member of the union counted exactly once.

|A∪B|=∣A∣+∣B∣\displaystyle \left|A\right| + \left|B\right|−|A∩B|

Choose and carry out a method

Identify the two sets, their overlap and whether the problem asks for the union or an exclusive category. Apply the matching count.

Check the reasoning

The union must be at least as large as either set and at most their sum. Draw a labeled Venn diagram to check.

WORKED EXAMPLE 1

Sets A and B have 13 and 10 elements, with 4 in both. Find |A∪B|.

  1. Adding the two set sizes counts the overlap twice.
  2. |A∪B|=13+10−4\displaystyle 13 + 10 - 4.
  3. There are 19 distinct members in the union.

19

WORKED EXAMPLE 2

Sets A and B have 14 and 11 elements, with 5 in both. Find |A∪B|.

  1. Adding the two set sizes counts the overlap twice.
  2. |A∪B|=14+11−5\displaystyle 14 + 11 - 5.
  3. There are 20 distinct members in the union.

20

Common pitfalls

Possible mix-up: Add both set sizes without adjustment.

Members in both sets are counted twice.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

How would you count members belonging to exactly one set?

Preview the eight practice prompts
  1. Problem 1

    Sets A and B have 16 and 13 elements, with 7 in both. Find |A∪B|.

    Practise this exact problem →
  2. Problem 2

    Sets A and B have 17 and 14 elements, with 8 in both. Find |A∪B|.

    Practise this exact problem →
  3. Problem 3

    Sets A and B have 18 and 15 elements, with 9 in both. Find |A∪B|.

    Practise this exact problem →
  4. Problem 4

    Sets A and B have 19 and 16 elements, with 10 in both. Find |A∪B|.

    Practise this exact problem →
  5. Problem 5

    Sets A and B have 20 and 17 elements, with 11 in both. Find |A∪B|.

    Practise this exact problem →
  6. Problem 6

    Sets A and B have 21 and 18 elements, with 12 in both. Find |A∪B|.

    Practise this exact problem →
  7. Problem 7New context

    22 students joined chess and 19 joined robotics; 13 joined both clubs. How many joined at least one of the two clubs?

    Practise this exact problem →
  8. Problem 8New context

    23 students joined chess and 20 joined robotics; 14 joined both clubs. How many joined at least one of the two clubs?

    Practise this exact problem →
Open stage CP 3.1 in the student workspace →

STAGE CP 3.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Equally Likely Outcomes

Useful preparation: Inclusion and Exclusion

Goal: Understand and apply equally likely outcomes.

Before you begin: Inclusion and Exclusion

Understand the idea

Probability is a ratio of favorable to possible outcomes only when the counted elementary outcomes are equally likely. Choosing a physical token uniformly makes individual tokens the outcomes.

P(A)=favorable outcomes/total outcomes

Choose and carry out a method

Define one elementary outcome, count all possible tokens, then count tokens with the requested property. Reduce the resulting fraction if possible.

Check the reasoning

A probability lies between zero and one. Equal color categories do not imply equal probabilities when their token counts differ.

WORKED EXAMPLE 1

A bag has 3 red and 6 blue equally likely tokens. What is the probability one draw is red?

  1. For equally likely outcomes, divide favorable outcomes by all outcomes.
  2. Probability=33+6\displaystyle \frac{3}{3 + 6}.
  3. The probability is 13\displaystyle \frac{1}{3}, a value between 0 and 1.

13\displaystyle \frac{1}{3}

WORKED EXAMPLE 2

A bag has 4 red and 7 blue equally likely tokens. What is the probability one draw is red?

  1. For equally likely outcomes, divide favorable outcomes by all outcomes.
  2. Probability=44+7\displaystyle \frac{4}{4 + 7}.
  3. The probability is 411\displaystyle \frac{4}{11}, a value between 0 and 1.

411\displaystyle \frac{4}{11}

Common pitfalls

Possible mix-up: Two colors always mean probability one-half.

Different numbers of tokens create different color probabilities.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

What assumption about the draw makes token counting valid?

Preview the eight practice prompts
  1. Problem 1

    A bag has 6 red and 9 blue equally likely tokens. What is the probability one draw is red?

    Practise this exact problem →
  2. Problem 2

    A bag has 7 red and 10 blue equally likely tokens. What is the probability one draw is red?

    Practise this exact problem →
  3. Problem 3

    A bag has 8 red and 11 blue equally likely tokens. What is the probability one draw is red?

    Practise this exact problem →
  4. Problem 4

    A bag has 9 red and 12 blue equally likely tokens. What is the probability one draw is red?

    Practise this exact problem →
  5. Problem 5

    A bag has 10 red and 13 blue equally likely tokens. What is the probability one draw is red?

    Practise this exact problem →
  6. Problem 6

    A bag has 11 red and 14 blue equally likely tokens. What is the probability one draw is red?

    Practise this exact problem →
  7. Problem 7New context

    A random selector chooses one of 27 equally likely tickets. Exactly 12 tickets win a small prize. What is the probability of a prize?

    Practise this exact problem →
  8. Problem 8New context

    A random selector chooses one of 29 equally likely tickets. Exactly 13 tickets win a small prize. What is the probability of a prize?

    Practise this exact problem →
Open stage CP 3.2 in the student workspace →

STAGE CP 3.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Dependent Draws

Useful preparation: Equally Likely Outcomes

Goal: Understand and apply dependent draws.

Before you begin: Equally Likely Outcomes

Understand the idea

Removing a token changes both the population and, after success, the favorable count. The second probability must describe the updated situation conditioned on the first result.

P(A then B)=P(A)P(B|A)

Choose and carry out a method

Multiply the first success probability by the probability of a second success after the first token is removed. Track numerator and denominator separately.

Check the reasoning

Compare with choosing two tokens as an unordered subset. Both correct methods must yield the same probability.

WORKED EXAMPLE 1

A bag holds 4 red and 5 blue tokens. Two are drawn without replacement. Find the probability both are red.

  1. Multiply the probability of the first success by the conditional probability of the second.
  2. P=49⋅38\displaystyle P = \frac{\frac{4}{9} \cdot 3}{8}.
  3. The exact probability is 16\displaystyle \frac{1}{6}; both numerator and denominator change after the first draw.

16\displaystyle \frac{1}{6}

WORKED EXAMPLE 2

A bag holds 5 red and 6 blue tokens. Two are drawn without replacement. Find the probability both are red.

  1. Multiply the probability of the first success by the conditional probability of the second.
  2. P=511⋅410\displaystyle P = \frac{\frac{5}{11} \cdot 4}{10}.
  3. The exact probability is 211\displaystyle \frac{2}{11}; both numerator and denominator change after the first draw.

211\displaystyle \frac{2}{11}

Common pitfalls

Possible mix-up: Square the original success probability.

Without replacement, the second draw has a changed sample space.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

How would the calculation change if the first token were replaced?

Preview the eight practice prompts
  1. Problem 1

    A bag holds 7 red and 8 blue tokens. Two are drawn without replacement. Find the probability both are red.

    Practise this exact problem →
  2. Problem 2

    A bag holds 8 red and 9 blue tokens. Two are drawn without replacement. Find the probability both are red.

    Practise this exact problem →
  3. Problem 3

    A bag holds 9 red and 10 blue tokens. Two are drawn without replacement. Find the probability both are red.

    Practise this exact problem →
  4. Problem 4

    A bag holds 10 red and 11 blue tokens. Two are drawn without replacement. Find the probability both are red.

    Practise this exact problem →
  5. Problem 5

    A bag holds 11 red and 12 blue tokens. Two are drawn without replacement. Find the probability both are red.

    Practise this exact problem →
  6. Problem 6

    A bag holds 12 red and 13 blue tokens. Two are drawn without replacement. Find the probability both are red.

    Practise this exact problem →
  7. Problem 7New context

    A 27-card deck contains 13 marked cards. Two cards are dealt without returning the first. What is the probability both dealt cards are marked?

    Practise this exact problem →
  8. Problem 8New context

    A 29-card deck contains 14 marked cards. Two cards are dealt without returning the first. What is the probability both dealt cards are marked?

    Practise this exact problem →
Open stage CP 3.3 in the student workspace →
Browse another level →
Numerive