3.3 Two Basic Rules of Probability

Read the unconditional rate straight off the setup. "The probability that a customer will order a pizza is 87%" is a plain probability about every customer, with no condition attached.

$$P(Z) = 0.87$$

Answer: \(P(Z) = 0.87\).

Same reading, second rate. "The probability that a customer will order a salad is 32%" is also unconditional — it covers all customers, not just the ones who ordered pizza.

$$P(S) = 0.32$$

Answer: \(P(S) = 0.32\).

Spot the phrase that signals a condition. "Of the customers who order pizzas, 55% of them also order a salad" restricts attention to pizza orderers first, then asks about salad. That is a conditional probability with \(Z\) on the right of the bar:

$$P(S \mid Z) = 0.55$$

Notice it is not 0.32. The filter changed the answer, which is the first hint that \(Z\) and \(S\) are not independent.

Answer: \(P(S \mid Z) = 0.55\).

Translate the bar back into a sentence. \(S \mid Z\) means the event \(S\) evaluated inside the shrunken world where \(Z\) already happened.

In words: \(S \mid Z\) is the event that a customer orders a salad, given that the customer ordered a pizza. Its probability, \(P(S \mid Z)\), is the proportion of pizza orderers who also order a salad.

Answer: the event that a randomly chosen customer orders a salad given that the customer has ordered a pizza.

Use the multiplication rule in the order the conditional was given. You were handed \(P(S \mid Z)\), so pair it with \(P(Z)\):

$$P(Z \text{ AND } S) = P(Z) \, P(S \mid Z) = (0.87)(0.55) = 0.4785$$

Answer: \(P(Z \text{ AND } S) = 0.4785\).

AND means both, on the same trial. \(Z \text{ AND } S\) is the event that one randomly selected customer orders a pizza and orders a salad — the same visit, both items.

Compare this with problem 3.3.4: \(S \mid Z\) is restricted to pizza orderers and asks what fraction of them add a salad, while \(Z \text{ AND } S\) is about all customers and asks what fraction do both.

Answer: the event that a randomly chosen customer orders both a pizza and a salad.

Run the independence test both ways; either one settles it.

Test 1 — compare the conditional to the plain probability:

$$P(S \mid Z) = 0.55 \qquad P(S) = 0.32$$

These are not equal, so the events are not independent.

Test 2 — compare the AND to the product:

$$P(Z) \, P(S) = (0.87)(0.32) = 0.2784$$

But \(P(Z \text{ AND } S) = 0.4785\), and \(0.2784 \neq 0.4785\). Same conclusion.

Knowing a customer ordered a pizza raises the chance of a salad from 32% to 55%, so the events carry information about each other.

Answer: no — \(Z\) and \(S\) are not independent, because \(P(S \mid Z) = 0.55 \neq 0.32 = P(S)\) (equivalently \(P(Z)P(S) = 0.2784 \neq 0.4785\)).

Addition rule, subtracting the overlap you already computed in 3.3.5.

$$P(Z \text{ OR } S) = P(Z) + P(S) - P(Z \text{ AND } S) = 0.87 + 0.32 - 0.4785 = 0.7115$$

Adding 0.87 and 0.32 alone would give 1.19, which is impossible for a probability — the double-counted pizza-and-salad customers are exactly the excess the subtraction removes.

Answer: \(P(Z \text{ OR } S) = 0.7115\).

OR is inclusive in probability. \(Z \text{ OR } S\) is the event that a randomly chosen customer orders a pizza, or orders a salad, or orders both. It fails only for a customer who orders neither.

Answer: the event that a randomly chosen customer orders a pizza or a salad or both.

Mutually exclusive means the AND is impossible. Check whether \(P(Z \text{ AND } S)\) is zero:

$$P(Z \text{ AND } S) = 0.4785 \neq 0$$

Nearly half of all customers order both, so the two events plainly co-occur.

Note that this is a different question from problem 3.3.7. Independence asks whether one event changes the odds of the other; mutual exclusivity asks whether they can happen together at all. Here the answer is "no" to both, for different reasons.

Answer: no — \(Z\) and \(S\) are not mutually exclusive, because \(P(Z \text{ AND } S) = 0.4785 \neq 0\).

Set up your symbol list before computing anything. The survey gives four numbers, and two of them are conditionals:

• 61% of all CA registered voters approve → \(P(C) = 0.61\)
• "Six in ten" say the ruling is important → \(P(B) = 0.60\)
• among 18-to-39 year olds, approval is 78% → \(P(C \mid A) = 0.78\)
• of those who support same-sex marriage, 75% say the ruling is important → \(P(B \mid C) = 0.75\)

a. \(P(C) = 0.61\)

b. \(P(B) = 0.60\)

c. \(P(C \mid A) = 0.78\) — the 78% figure is already restricted to the 18-to-39 group, so it is a conditional, not a plain probability.

d. \(P(B \mid C) = 0.75\) — "out of those CA registered voters who support same-sex marriage" is the phrase that puts \(C\) on the right of the bar.

e. \(C \mid A\) in words: the event that a California registered voter supports same-sex marriage, given that the voter is 18 to 39 years old.

f. \(B \mid C\) in words: the event that a voter says the Supreme Court's ruling is very or somewhat important, given that the voter supports same-sex marriage.

g — multiplication rule with the conditional from part d.

$$P(C \text{ AND } B) = P(C) \, P(B \mid C) = (0.61)(0.75) = 0.4575$$

h. \(C \text{ AND } B\) in words: the event that a voter both supports same-sex marriage and says the ruling is very or somewhat important to them.

i — addition rule, subtracting the overlap from part g.

$$P(C \text{ OR } B) = P(C) + P(B) - P(C \text{ AND } B) = 0.61 + 0.60 - 0.4575 = 0.7525$$

j — check whether the AND is zero. \(P(C \text{ AND } B) = 0.4575\), which is not 0, so \(C\) and \(B\) are not mutually exclusive. A voter can easily do both — support same-sex marriage and consider the ruling important.

Answer: a. 0.61; b. 0.60; c. 0.78; d. 0.75; e. supports same-sex marriage given the voter is 18–39; f. says the ruling is important given the voter supports same-sex marriage; g. 0.4575; h. supports same-sex marriage and says the ruling is important; i. 0.7525; j. no, since \(P(C \text{ AND } B) = 0.4575 \neq 0\).

a — the sample size is the number of people polled.

$$n = 1{,}046$$

b — disapproval is the complement of approval. In Year 3 the approval rate was 42%, so

$$P(\text{disapprove}) = 1 - 0.42 = 0.58$$

c — turn a proportion back into a headcount. Multiply the Year 3 approval rate by the sample size:

$$(0.42)(1{,}046) = 439.32 \approx 439 \text{ people}$$

You cannot poll a fraction of a person, so round to the nearest whole respondent.

d — Year 2 approval, read as a probability. A randomly chosen respondent from the Year 2 poll approved with probability

$$P(\text{approve in Year 2}) = 0.57$$

e — Year 1 approval.

$$P(\text{approve in Year 1}) = 0.60$$

Answer: a. 1,046; b. 0.58; c. about 439 people; d. 0.57; e. 0.60.

a — list the sample space. All 38 pockets:

$$S = \{0,\; 00,\; 1,\; 2,\; 3,\; \ldots,\; 36\}$$

That is the two green pockets plus the numbers 1 through 36, so \(n(S) = 38\).

b — count the red pockets. Eighteen of the 38 pockets are red:

$$P(\text{red}) = \frac{18}{38} = \frac{9}{19} \approx 0.4737$$

c — count the 1st Dozen. The 1st 12 range holds the numbers 1 through 12:

$$P(\text{1st 12}) = \frac{12}{38} = \frac{6}{19} \approx 0.3158$$

d — count the even numbers. Among 1 through 36 the even numbers are 2, 4, 6, …, 36, which is 18 pockets:

$$P(\text{even number}) = \frac{18}{38} = \frac{9}{19} \approx 0.4737$$

e — check whether odd and even fill the whole sample space. No. There are 18 odd pockets and 18 even pockets, but 0 and 00 are neither, so

$$P(\text{odd}) + P(\text{even}) = \frac{18}{38} + \frac{18}{38} = \frac{36}{38} \approx 0.9474 \neq 1$$

Complements must add to 1. Getting an odd number is therefore not the complement of getting an even number — the complement of "even" is "odd or green."

f — any two events that cannot co-occur. Red and Green work: a pocket has exactly one color, so \(P(\text{red AND green}) = 0\). So do 1st 12 and 2nd 12 — the ball lands in one pocket, which belongs to at most one dozen.

g — run the independence test. The even numbers inside the 1st Dozen are 2, 4, 6, 8, 10, 12, so six pockets are both:

$$P(\text{even AND 1st 12}) = \frac{6}{38} \approx 0.1579$$

If the events were independent this would equal the product of the separate probabilities:

$$P(\text{even}) \, P(\text{1st 12}) = \left(\frac{18}{38}\right)\left(\frac{12}{38}\right) = \frac{216}{1444} \approx 0.1496$$

Since \(0.1579 \neq 0.1496\), Even and 1st Dozen are not independent. The two green pockets are the culprit: they sit outside both events and tilt the totals just enough to break the product rule.

Answer: a. \(\{0, 00, 1, 2, \ldots, 36\}\); b. \(\frac{18}{38} \approx 0.4737\); c. \(\frac{12}{38} \approx 0.3158\); d. \(\frac{18}{38} \approx 0.4737\); e. no, because 0 and 00 are neither odd nor even, so the two probabilities sum to \(\frac{36}{38}\), not 1; f. e.g. Red and Green, or 1st 12 and 2nd 12; g. no, since \(\frac{6}{38} \approx 0.1579 \neq 0.1496 \approx \left(\frac{18}{38}\right)\left(\frac{12}{38}\right)\).

Every bet is the same computation: count the pockets it covers, divide by 38. The pockets are equally likely, so a bet covering \(k\) of them wins with probability \(\frac{k}{38}\).

a — two touching lines, 1-2-3-4-5-6. Six numbers:

$$P = \frac{6}{38} = \frac{3}{19} \approx 0.1579$$

b — three numbers in a line, 1-2-3. Three numbers:

$$P = \frac{3}{38} \approx 0.0789$$

c — one number.

$$P = \frac{1}{38} \approx 0.0263$$

d — four numbers forming a square, 10-11-13-14. Four numbers:

$$P = \frac{4}{38} = \frac{2}{19} \approx 0.1053$$

e — two touching numbers, 10-11 or 10-13. Two numbers:

$$P = \frac{2}{38} = \frac{1}{19} \approx 0.0526$$

f — 0-00-1-2-3. Five numbers, including both green pockets:

$$P = \frac{5}{38} \approx 0.1316$$

g — 0-1-2, or 0-00-2, or 00-2-3. Each of these is a three-number bet:

$$P = \frac{3}{38} \approx 0.0789$$

Notice the pattern across all seven: the probability rises in exact proportion to how many pockets you cover, which is why the casino adjusts the payout rather than the odds to keep its edge.

Answer: a. \(\frac{6}{38} \approx 0.1579\); b. \(\frac{3}{38} \approx 0.0789\); c. \(\frac{1}{38} \approx 0.0263\); d. \(\frac{4}{38} \approx 0.1053\); e. \(\frac{2}{38} \approx 0.0526\); f. \(\frac{5}{38} \approx 0.1316\); g. \(\frac{3}{38} \approx 0.0789\).