En anglais

Sample spaces, events and the three axioms

Mathematics · Probability · ages 18-19

Name ______________________ Date ____________

P(A) = 0.6, P(B) = 0.35 and P(A and B) = 0.2. Find P(A or B).
Answer: ______________
A menu has 3 starters and 5 mains, and one of each is chosen. How many outcomes are in the sample space?
- 8
- 12
- 15
- 30
A coin is tossed and a six-sided die is rolled. How many outcomes are in the sample space?
- 6
- 12
- 36
The chance of rain tomorrow is 0.35. What is the chance it does not rain?
Answer: ______________
Two events A and B overlap. Which statement gives P(A or B)?
- P(A) + P(B), because the overlap is counted once by adding
- P(A) + P(B) - P(A and B), because the overlap was counted twice
- P(A) + P(B) + P(A and B), because the overlap needs extra weight
- P(A) × P(B), because both events are involved
A die is rolled. Event E is an even roll, {2, 4, 6}. Which set is the complement of E?
- {2, 4, 6}
- {1, 2, 3}
- {1, 3, 5}
Sam assigns P(A) = 0.6, P(B) = 0.7 and P(A or B) = 1.5. Sam says the axioms allow any number between 0 and 2. Is Sam right?
Circle one: True False
Exactly one of outcomes a, b and c must happen. Someone gives them chances 0.1, 0.5 and 0.2 and says no axiom is broken, since each chance is between 0 and 1. Is that right?
Circle one: True False
P(A) = 0.5, P(B) = 0.4 and P(both) = 0.2. Jo skips the overlap and reports 0.9. What is the correct P(A or B)?
Answer: ______________
Dee says the sample space for rolling two dice has 12 outcomes. Is Dee right?
Circle one: True False

Answer key

For grown-ups. Fold this page away before handing over the rest.

0.75 · 0.6 + 0.35 - 0.2 = 0.75, because the overlap was counted twice.
15 · Each of the 3 starters pairs with each of the 5 mains, so 3 times 5 = 15.
12 · Two coin faces times six die faces gives 2 times 6, which is 12 possible pairs. Counting possible pairs does not require them to be equally likely.
0.65 · The complement rule gives 1 minus 0.35, which is 0.65.
P(A) + P(B) - P(A and B), because the overlap was counted twice · Adding P(A) and P(B) counts the overlap twice, so P(A and B) is subtracted once.
{1, 3, 5} · The complement holds every outcome outside E, which is {1, 3, 5}.
False · The axioms force every probability to be at most 1. A union probability of 1.5 would give its complement probability 1 - 1.5 = -0.5, contradicting nonnegativity.
False · No. Each chance is allowed, but the three outcomes cover the whole space, so their chances must total 1. They total 0.8.
0.7 · Subtract the overlap once: 0.5 plus 0.4 minus 0.2 is 0.7, so Jo double counted.
False · Dee is wrong: 6 faces times 6 faces gives 36 pairs, not 12.