Intro to Probability - Extra Practice
These problems provide additional practice with the material from the Intro to Probability notes. Unless stated otherwise, assume that all outcomes in the stated outcome space are equally likely.
Try each problem before selecting Show answer to check your reasoning. Assume repeated coin tosses and die rolls are independent.
General questions
Exercise 1
True or False: If something has probability 1,000%, it is sure to happen!
Show answer
False. A probability must be between 0 and 1 (0% and 100%). A probability of 1,000% is not valid and therefore has no well-defined meaning.
Exercise 2
A fair coin is tossed six times. Which of these sequences is most likely? (H = Heads, T = Tails)
H H H H H HH T H T T TH T H T H T
Show answer
They are all equally likely. Assuming independent tosses, every particular sequence of six results has probability \((1/2)^6=1/64\). There are more sequences with three heads and three tails than with six heads, but every specific sequence is equally likely here.
Outcome spaces, events, and equally likely outcomes
Exercise 3
A fair six-sided die is rolled once. Write the outcome space \(\Omega\) and define the event \(A\) that the result is greater than 4 by listing its outcomes. Then find \(P(A)\).
Show answer
\(\Omega=\{1,2,3,4,5,6\}\) and \(A=\{5,6\}\). Thus \(P(A)=2/6=1/3\).
Exercise 4
A fair coin is tossed twice. Write the complete outcome space, being careful to distinguish the two possible orders in which one head and one tail can occur. Let \(A\) be the event that at least one toss is heads. Find \(P(A)\).
Show answer
\(\Omega=\{HH,HT,TH,TT\}\), where the first letter records the first toss. With a fair coin and independent tosses, these four outcomes are equally likely. Since \(A=\{HH,HT,TH\}\), \(P(A)=3/4\).
Exercise 5
An eight-sided die has sides labeled \(1,2,3,\ldots,8\). Let \(A\) be the event that the result is a multiple of 3, and let \(B\) be the event that the result is odd. List \(A\), \(B\), \(A\cap B\), and \(A\cup B\).
Show answer
\(A=\{3,6\}\) and \(B=\{1,3,5,7\}\).
The intersection contains outcomes in both events: \(A\cap B=\{3\}\).
The union contains outcomes in either event: \(A\cup B=\{1,3,5,6,7\}\).
Exercise 6
A box contains five tickets labeled \(\boxed{1}, \boxed{2}, \boxed{2}, \boxed{3}, \boxed{4}\). One ticket is drawn at random. Let \(A\) be the event that the number drawn is even. Find \(P(A)\). Why is it incorrect to treat the four distinct labels \(1,2,3,4\) as equally likely outcomes?
Show answer
Three of the five tickets have even labels (2, 2, and 4), so \(P(A)=3/5\).
The individual tickets are equally likely, but the distinct labels are not: label 2 has probability \(2/5\), while each other label has probability \(1/5\).
Exercise 7
A fair six-sided die is rolled twice. What is the chance of both rolls being six? What about the chance of not rolling two sixes? What about _neither _roll being six?
Show answer
Assuming independent rolls:
- Both six: \((1/6)^2=1/36\).
- Not both six: \(1-1/36=35/36\). This includes exactly one six or no sixes.
- Neither six: \((5/6)^2=25/36\).
Axioms, unions, intersections, and complements
Exercise 8
Let \(A\) be the event that a fair die roll is even and \(B\) be the event that a fair die roll is greater than 3. List \(A\cap B\) and \(A\cup B\), and find the probability of each.
Show answer
\(A=\{2,4,6\}\) and \(B=\{4,5,6\}\).
Thus \(A\cap B=\{4,6\}\), with probability \(2/6=1/3\), and \(A\cup B=\{2,4,5,6\}\), with probability \(4/6=2/3\).
Exercise 9
A pair of fair six-sided dice is rolled, and the spots are summed. Let \(A\) be the event that the sum is at least 5. Use the complement rule to find \(P(A)\) by first finding \(P(A^C)\).
Show answer
The complement is a sum less than 5. Of the 36 equally likely ordered pairs, six give sums of 2, 3, or 4:
\[(1,1),\ (1,2),\ (2,1),\ (1,3),\ (2,2),\ (3,1).\]
So \(P(A^C)=6/36=1/6\), and \(P(A)=1-1/6=5/6\).
Exercise 10
A fair coin is tossed three times. Let \(A\) be the event that all three tosses are tails. Use the complement rule to find the probability of getting at least one head. You may list the outcome space to justify your answer.
Show answer
The complement of all tails is at least one head. For independent tosses, \(P(A)=(1/2)^3=1/8\), so
\[P(\text{at least one head})=1-P(A)=7/8.\]
Exercise 11
Suppose \(P(A)=0.55\), \(P(B)=0.40\), and \(P(A\cap B)=0.15\). Find \(P(A\cup B)\). Then find the probability that neither \(A\) nor \(B\) occurs.
Show answer
By the addition rule,
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.55+0.40-0.15=0.80.\]
The probability of neither event is \(1-P(A\cup B)=0.20\).
Exercise 12
An outcome space has events \(A\) and \(B\) with \(P(A)=0.6\), \(P(B)=0.5\), and \(P(A\cup B)=0.8\). What do you think is the value of \(P(A\cap B)\)?
Show answer
Rearranging the addition rule gives
\[P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.6+0.5-0.8=0.3.\]
Box models and repeated draws
Exercise 13
Four draws are going to be made at random, with replacement, from a box containing six tickets: \(\boxed{1}, \boxed{2}, \boxed{4}, \boxed{4}, \boxed{5}, \boxed{5}\). Find the probability that the ticket marked 4 is drawn at least once.
Show answer
Two of the six tickets are marked 4, so a single draw has probability \(2/6=1/3\) of being a 4, and therefore (by the complement rule) a \(2/3\) chance of not being a 4. The complement of “at least one 4” is “no 4s in all four draws.”
Replacement makes the draws independent, so
\[P(\text{at least one 4})=1 - P(\text{no 4s at all})=1-(2/3)^4=1-\frac{16}{81}=\frac{65}{81}\approx 0.8025.\]
Exercise 14
A box contains \(\boxed{1}\,\boxed{2}\,\boxed{2}\,\boxed{3}\). One ticket is drawn. Let \(A\) be the event that the draw is at least 2. Find \(P(A)\) and \(P(A^C)\), and verify that they add to 1.
Show answer
Three of the four tickets are at least 2, so \(P(A)=3/4\). The complement consists of drawing the ticket labeled 1, so \(P(A^C)=1/4\). Their probabilities add to \(3/4+1/4=1\).
Exercise 15
An American roulette wheel has 18 red pockets, 18 black pockets, and 2 green pockets. Find the probability of a red result, the probability of a non-red result, and the probability of winning a one-dollar net gain when betting on red.
Show answer
There are 38 equally likely pockets.
- Red: \(18/38=9/19\).
- Non-red: \((18+2)/38=20/38=10/19\).
- A one-dollar net gain on a one-dollar bet on red: \(18/38=9/19\), since the bet wins exactly when the ball lands in a red pocket.
Probability and simulation
Exercise 16
A student writes the following R code to simulate a fair die roll:
sample(1:6, size = 1)Explain what the code does and why each of the six values should be equally likely.
Show answer
1:6 creates the values 1 through 6, and sample(..., size = 1) randomly selects one of them. By default, sample() gives each entry equal probability, so each value has probability \(1/6\).
Exercise 17
Write R code using sample() to simulate drawing one ticket from the box \(\boxed{1}\,\boxed{2}\,\boxed{2}\,\boxed{3}\). Then change your sample function call to instead simulate 1,000 draws. Use the results estimate the probability of drawing a 2.
Show answer
box = c(1, 2, 2, 3)
draws = sample(box, size = 1000, replace = TRUE)
mean(draws == 2)Each call samples from the full box, so these are draws with replacement. draws == 2 records whether each draw is a 2; its mean is the proportion of draws that are 2. The estimate should be near \(2/4=1/2\), with random variation.
Exercise 18
The theoretical probability of rolling a 6 on a fair die is \(1/6\). A simulation of 20 rolls produces 7 sixes. Is the simulation result impossible? Explain the difference between a theoretical probability and an empirical proportion.
Show answer
No, seven sixes in 20 rolls is possible. The theoretical probability, \(1/6\), describes the chance of a six on each roll. You’d typically get about \(20*(1/6) \approx 3.33\) sixes, but anything can happen! All 20 rolls could be sixes, it’s just unlikely (not impossible).
The empirical proportion is the fraction actually observed, here \(7/20=0.35\), versus our theoretical probability of \(1/6\). The empirical proportion varies from simulation to simulation and need not equal the theoretical probability.
Exercise 19
A student wants to simulate the sum of two fair dice and writes sample(2:12, size = 1). Explain why this does not give every possible sum the correct probability. Describe, in words or R code, a better simulation using two separate die rolls.
Show answer
sample(2:12, size = 1) gives all 11 sums equal probability. Actual dice sums are not equally likely: a sum of 2 comes only from \((1,1)\), while a sum of 7 comes from six ordered pairs: \((1,6)\;(2,5)\;(3,4)\;(4,3)\;(5,2)\;(6,1)\).
Simulate two independent die rolls and add them:
sum(sample(1:6, size = 2, replace = TRUE))replace = TRUE lets both dice show the same value and makes the two sampled values independent.
Exercise 20
A fair six-sided die is rolled once. Let \(A\) be the event that the result is odd and \(B\) the event that the result is prime. List \(A\), \(B\), \(A\cap B\), and \(A\cup B\), and calculate the probability of each event.
Show answer
\(A=\{1,3,5\}\) and \(B=\{2,3,5\}\); 1 is not prime.
- \(P(A)=3/6=1/2\).
- \(P(B)=3/6=1/2\).
- \(A\cap B=\{3,5\}\), so \(P(A\cap B)=2/6=1/3\).
- \(A\cup B=\{1,2,3,5\}\), so \(P(A\cup B)=4/6=2/3\).