The Lumenary · Grades 6–8

Probability

Name: ______________________

Date: ______________________

You will learn to

List the sample space of an experiment, using a table or tree diagram for two stages
Compute probabilities of events with equally likely outcomes as fractions
Prove the complement rule, the addition rule and the product rule for two independent stages
Compare theoretical probabilities with relative frequencies from repeated trials and simulations

1.
A coin is tossed and a die is rolled. How many outcomes are in the sample space? (For example, (H, 3) is one outcome.)
2.
Two fair dice are rolled. What is the probability that the sum is $8$ ?
- A. $\tfrac{5}{36}$
- B. $\tfrac{1}{11}$
- C. $\tfrac{3}{36}$
- D. $\tfrac{8}{36}$
3.
A die is rolled three times. What is the probability of getting at least one six?
- A. $1 - \left(\tfrac56\right)^3 = \tfrac{91}{216}$
- B. $\tfrac16 + \tfrac16 + \tfrac16 = \tfrac12$
- C. $\left(\tfrac16\right)^3 = \tfrac1{216}$
- D. $\tfrac{3}{18}$
4.
A number is chosen at random from $1, 2, \ldots, 20$ (all equally likely). What is the probability that it is a multiple of $2$ or a multiple of $5$ ?
- A. $\tfrac{12}{20}$
- B. $\tfrac{14}{20}$
- C. $\tfrac{2}{20}$
- D. $\tfrac{10}{20}$
5.
A fair coin has landed tails $6$ times in a row. What is the probability that the next toss is heads?
- A. $\tfrac12$
- B.More than $\tfrac12$ , because heads is due
- C.Less than $\tfrac12$ , because tails is on a streak
- D. $\left(\tfrac12\right)^7$
6.
In 1754 d'Alembert argued: “Toss a coin up to twice, stopping when heads appears. The possible results are H, TH and TT; two of the three give a head, so the probability of getting a head is $\tfrac23$ .” What is wrong?
- A.The three results are not equally likely: H has probability $\tfrac12$ , TH and TT $\tfrac14$ each
- B.Nothing — the answer $\tfrac23$ is right
- C.There should be only two results, H and T
- D.He should have multiplied instead of counting
7.
Neema rolls a fair die $60$ times and gets $14$ sixes. Which statements are true? (Choose all that apply.)
- A.The relative frequency of a six was $\tfrac{14}{60}$
- B.The die must be unfair, because she should have got exactly 10 sixes
- C.The probability of a six is still $\tfrac16$ for a fair die
- D.With 6,000 rolls, the relative frequency would very probably be closer to 1/6
8.
Three dice are rolled. In how many of the $216$ ordered outcomes is the sum equal to $10$ ?
9.
In the Lean file, prob S A is the probability of the event A in the list of outcomes S. Which statement is the complement rule, correctly stated?
- A.S ≠ [] → prob S (fun x => !A x) = 1 - prob S A
- B.prob S (fun x => !A x) = 1 - prob S A
- C.S ≠ [] → prob S (fun x => !A x) = prob S A - 1
- D.S ≠ [] → prob S A = 1
10.
Write prob_sum(dice, total) that returns the exact probability (as a Fraction) that dice fair dice add up to total. List every ordered outcome with itertools.product(range(1, 7), repeat=dice) and count. Then check Galileo: is a total of $10$ more likely than $9$ with three dice?
Computational — Practice

Answer key — Probability

1.
The outcomes are the pairs (coin, die): $2 \times 6 = 12$ of them, from (H, 1) to (T, 6).
2.
The ordered pairs with sum $8$ are $(2,6), (3,5), (4,4), (5,3), (6,2)$ , so $P = \tfrac{5}{36}$ .
3.
$P(\text{at least one six}) = 1 - P(\text{no six}) = 1 - \left(\tfrac56\right)^3 = 1 - \tfrac{125}{216} = \tfrac{91}{216} \approx 0.42$ .
4.
Multiples of $2$ : $10$ . Multiples of $5$ : $4$ . Multiples of both ( $10$ and $20$ ): $2$ . By the addition rule, $P = \tfrac{10 + 4 - 2}{20} = \tfrac{12}{20} = \tfrac35$ .
5.
$\tfrac12$ . Successive tosses are independent, so the six tails already seen do not affect the next toss.
6.
$P(H) = \tfrac12$ , $P(TH) = \tfrac14$ , $P(TT) = \tfrac14$ . So $P(\text{a head}) = \tfrac12 + \tfrac14 = \tfrac34$ . Counting results gives the right answer only when they are equally likely.
7.
The relative frequency is $\tfrac{14}{60} \approx 0.23$ , while the probability is $\tfrac16 \approx 0.17$ . Such differences are normal in $60$ rolls; with many more rolls, the relative frequency is very likely to be close to $\tfrac16$ . A fair die does not guarantee exactly $10$ sixes.
8.
$1+3+6$ , $1+4+5$ and $2+3+5$ have $6$ orders each ( $18$ ); $2+2+6$ , $2+4+4$ and $3+3+4$ have $3$ orders each ( $9$ ). In total $27$ ordered outcomes, so $P(10) = \tfrac{27}{216} = \tfrac18$ .
9.
S ≠ [] → prob S (fun x => !A x) = 1 - prob S A (the theorem prob_compl). Without S ≠ [] the statement is false, because for an empty list all probabilities are $0$ .

10.

from fractions import Fraction
from itertools import product

def prob_sum(dice, total):
    count = 0
    for roll in product(range(1, 7), repeat=dice):
        if sum(roll) == total:
            count += 1
    return Fraction(count, 6 ** dice)

prob_sum(3, 10) is $\tfrac{27}{216}$ and prob_sum(3, 9) is $\tfrac{25}{216}$ , so $10$ is slightly more likely — Galileo's answer. This is exact for the model of fair, independent dice.