Probability of Compound Events — Challenge
Two or more things happening: multiply along a path, add across paths, and watch whether the first event changes the second.
What these problems cover
These questions use 2 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Sample space — The set of all possible outcomes.
- Complement — Everything except the event. P(not A) = 1 - P(A).
Problem 1
Two fair coins are flipped. What is the probability that both show heads?
- 18
- 14
- 12
- 34
- 1
Worked answer
14
The four equally likely outcomes are HH, HT, TH and TT.
Only one of the four is two heads, so the probability is 14.
Multiplying the separate probabilities gives the same thing: 12·12 = 14.
Problem 2
Two fair six-sided dice are rolled. What is the probability that the two numbers add to 7?
- 112
- 19
- 16
- 536
- 736
Worked answer
16
There are 6· 6 = 36 equally likely ordered outcomes.
The pairs adding to 7 are (1,6),(2,5),(3,4),(4,3),(5,2),(6,1), which is six of them.
So the probability is 636 = 16, and 7 is the most likely total.
Problem 3
A bag holds 4 red and 6 blue counters. Two are drawn without replacement. What is the probability that both are red?
- 115
- 215
- 425
- 15
- 310
Worked answer
215
The first draw is red with probability 410.
After a red is removed, 3 reds remain among 9 counters, so the second is red with probability 39.
Multiply: 410·39 = 1290 = 215. With replacement it would be 425 instead.
Problem 4
A fair die is rolled and a fair coin is flipped. What is the probability of a 6 together with heads?
- 112
- 18
- 16
- 14
- 712
Worked answer
112
The two actions do not affect each other, so multiply their probabilities.
16·12 = 112.
Problem 5
Three fair coins are flipped. What is the probability that exactly two show heads?
- 18
- 14
- 38
- 12
- 58
Worked answer
38
There are 23 = 8 equally likely outcomes.
Exactly two heads happens as HHT, HTH and THH, which is three outcomes.
So the probability is 38. Note that three heads is a separate outcome and is not counted.
Problem 6
A spinner has four equal sectors numbered 1 to 4 and is spun twice. What is the probability of getting a 1 both times?
- 116
- 18
- 14
- 38
- 12
Worked answer
116
Each spin lands on 1 with probability 14, and the spins are independent.
Multiply: 14·14 = 116.
Problem 7
A fair coin is flipped twice. What is the probability of getting at least one head?
- 14
- 12
- 34
- 78
- 1
Worked answer
34
It is easier to find the opposite case: no heads at all means both tails, with probability 12·12 = 14.
At least one head is everything else: 1-14 = 34.
Adding 12+12 would give 1, which double-counts the outcome with two heads.
Problem 8
Two fair dice are rolled. What is the probability that both numbers are even?
- 19
- 16
- 14
- 13
- 12
Worked answer
14
Each die is even with probability 36 = 12.
The dice are independent, so multiply: 12·12 = 14.
Problem 9
Five cards are numbered 1 to 5. Two are drawn without replacement. What is the probability that both are odd?
- 110
- 310
- 925
- 25
- 12
Worked answer
310
The odd cards are 1, 3 and 5, so three of the five are odd.
The first card is odd with probability 35; then two odds remain among four cards, so the second is odd with probability 24.
Multiply: 35·24 = 620 = 310.
Problem 10
Two independent events have probabilities 0.5 and 0.2. What is the probability that both happen?
- 0.1
- 0.2
- 0.3
- 0.4
- 0.7
Worked answer
0.1
For independent events the probability that both happen is the product of the two probabilities.
0.5· 0.2 = 0.1. Adding them would give 0.7, which is the wrong operation here.