Inclusion–Exclusion

  • 12th Grade
Reviewed by Cierra Henderson
Cierra Henderson, MBA |
K-12 Expert
Review Board Member
Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.
 , MBA
By Thames
T
Thames
Community Contributor
Quizzes Created: 8156 | Total Attempts: 9,588,805
| Questions: 10 | Updated: Dec 10, 2025
Please wait...
Question 1 / 11
🏆 Rank #-- ▾
Score 0/100

1) Given P(A)=0.58, P(B)=0.46, P(A∩B)=0.31, find P(A∪B).

Explanation

P(A∪B)=0.58+0.46−0.31=0.73.

Submit
Please wait...
About This Quiz
Inclusionexclusion - Quiz

Overlapping sets often confuse probability and counting problems. The inclusion–exclusion principle clears up the confusion! In this quiz, you’ll practice applying the principle to handle multiple intersections. Try this quiz to unlock one of the most powerful tools in combinatorics.

2)

What first name or nickname would you like us to use?

You may optionally provide this to label your report, leaderboard, or certificate.

2) |A|=240, |B|=180, |A∩B|=95, total n=500. Find P(A∪B).

Explanation

(240+180−95)/500=325/500=0.65.

Submit

3) P(A)=0.40, P(B)=0.35, P(C)=0.30, P(A∩B)=0.18, P(A∩C)=0.12, P(B∩C)=0.11, P(A∩B∩C)=0.05. Find P(A∪B∪C).

Explanation

0.40+0.35+0.30−0.18−0.12−0.11+0.05=0.69.

Submit

4) |A|=100, |B|=90, |C|=60, |A∩B|=40, |A∩C|=30, |B∩C]=20, |A∩B∩C|=0, total n=200. Find P(A∪B∪C).

Explanation

(100+90+60−40−30−20+0)/200=160/200=0.80.

Submit

5) P(A)=0.50, P(B)=0.40, P(A∪B)=0.72. Find P(A∩B).

Explanation

P(A∩B)=P(A)+P(B)−P(A∪B)=0.50+0.40−0.72=0.18.

Submit

6) |A|=120, |B|=100, |C|=80, |A∩B|=50, |A∩C|=40, |B∩C|=35, total n=200, and |A∪B∪C|=180. Find |A∩B∩C|.

Explanation

120+100+80−(50+40+35)+x=180 ⇒ 300−125+x=180 ⇒ x=5.

Submit

7) P(A)=0.50, P(B)=0.40, P(A∩B)=0.20. Find P(exactly one of A or B).

Explanation

P(A only)+P(B only)=P(A)+P(B)−2P(A∩B)=0.50+0.40−0.40=0.50.

Submit

8) P(A)=0.40, P(B)=0.35, P(C)=0.30, P(A∩B)=0.18, P(A∩C)=0.12, P(B∩C)=0.11, P(A∩B∩C)=0.02. Find P(none of A, B, or C).

Explanation

First P(union)=0.40+0.35+0.30−0.18−0.12−0.11+0.02=0.66, so P(none)=1−0.66=0.34.

Submit

9) P(A)=0.50, P(B)=0.35, P(C)=0.25, P(A∩B)=0.20, P(A∩C)=0.15, P(A∩B∩C)=0.08. Find P(A only).

Explanation

P(A only)=P(A)−P(A∩B)−P(A∩C)+P(A∩B∩C)=0.50−0.20−0.15+0.08=0.23.

Submit

10) P(A∩B)=0.18, P(A∩C)=0.14, P(B∩C)=0.12, P(A∩B∩C)=0.05. Find P(exactly two of A, B, C).

Explanation

Exactly two events = (0.18 + 0.14 + 0.12) – 3(0.05) = 0.29.

Submit
×
Saved
Thank you for your feedback!
View My Results
Cierra Henderson |MBA |
K-12 Expert
Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.
Cancel
  • All
    All (10)
  • Unanswered
    Unanswered ()
  • Answered
    Answered ()
Given P(A)=0.58, P(B)=0.46, P(A∩B)=0.31, find P(A∪B).
|A|=240, |B|=180, |A∩B|=95, total n=500. Find P(A∪B).
P(A)=0.40, P(B)=0.35, P(C)=0.30, P(A∩B)=0.18, P(A∩C)=0.12,...
|A|=100, |B|=90, |C|=60, |A∩B|=40, |A∩C|=30, |B∩C]=20,...
P(A)=0.50, P(B)=0.40, P(A∪B)=0.72. Find P(A∩B).
|A|=120, |B|=100, |C|=80, |A∩B|=50, |A∩C|=40, |B∩C|=35,...
P(A)=0.50, P(B)=0.40, P(A∩B)=0.20. Find P(exactly one of A or B).
P(A)=0.40, P(B)=0.35, P(C)=0.30, P(A∩B)=0.18, P(A∩C)=0.12,...
P(A)=0.50, P(B)=0.35, P(C)=0.25, P(A∩B)=0.20, P(A∩C)=0.15,...
P(A∩B)=0.18, P(A∩C)=0.14, P(B∩C)=0.12,...
play-Mute sad happy unanswered_answer up-hover down-hover success oval cancel Check box square blue
Alert!