Introduction to Probability: Practice

A hand is a set of five cards, so the sample space is counted with combinations.

$\lvert S \rvert = \binom{52}{5} = 2{,}598{,}960$

To build a hand with exactly two aces, choose which two of the four aces appear, then fill the remaining three slots from the $48$ cards that are not aces. Those two choices are made in sequence, so multiply them.

$\lvert A \rvert = \binom{4}{2}\binom{48}{3} = 6 \times 17{,}296 = 103{,}776$

Dividing gives $P(A) = \frac{103{,}776}{2{,}598{,}960} \approx 0.0399$. Drawing the last three cards from all $50$ remaining cards would allow a third ace to sneak in, so the hand would no longer have exactly two.

An outcome assigns one day to each of the four people, so the multiplication principle sizes the sample space.

$\lvert S \rvert = 365^{4}$

To make all four birthdays different, hand out the days one person at a time, each new person avoiding every day already taken.

$\lvert D \rvert = 365 \times 364 \times 363 \times 362$

So $P(D) = \dfrac{365 \times 364 \times 363 \times 362}{365^{4}} \approx 0.9836$. Subtracting that from $1$ would answer the opposite question, the probability that at least two of the four birthdays match.

Turn the percentages into counts. Imagine $1000$ people, of whom $2\%$, that is $20$, have the disease and $980$ do not.

Among the $20$ sick people the test finds $90\%$, giving $0.9 \times 20 = 18$ positives. Among the $980$ healthy people it fires anyway $10\%$ of the time, giving $0.1 \times 980 = 98$ positives.

$\text{total positives} = 18 + 98 = 116$

A positive result shrinks the sample space to those $116$ people, and the sick ones among them number $18$.

$P(\text{disease} \mid \text{positive}) = \frac{18}{116} = \frac{9}{58} \approx 0.155$

The figure $0.90$ is $P(\text{positive} \mid \text{disease})$, the reversed conditional, and reading it as the answer is the classic base rate error.

Split $B$ into the part inside $A$ and the part outside it, since every outcome of $B$ is in exactly one of the two.

$P(A^{c} \cap B) = P(B) - P(A \cap B) = 0.6 - (0.3)(0.6) = 0.6 - 0.18 = 0.42$

Factoring the same computation shows why the answer is so tidy.

$P(B)\bigl(1 - P(A)\bigr) = 0.6 \times 0.7 = 0.42 = P(A^{c})\,P(B)$

So $A^{c}$ and $B$ are independent too, which is exactly what you would hope: if $B$ says nothing about $A$, it says nothing about "not $A$" either.

Write down what each condition says and set them equal. Mutual exclusivity forces the intersection to be empty, and independence names its probability.

$P(A \cap B) = 0 \qquad \text{and} \qquad P(A \cap B) = P(A)\,P(B)$

Therefore $P(A)P(B) = 0$, and a product of real numbers is zero only when one of its factors is zero.

$P(A) = 0 \quad \text{or} \quad P(B) = 0$

This is the one degenerate case in which the two conditions can hold together: at least one of the events must be impossible. Nothing forces the two probabilities to be equal or to sum to $1$, so those statements are not consequences.

Chain the three draws with the multiplication rule, shrinking the deck each time.

$P = \frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} = \frac{1716}{132600} = \frac{11}{850} \approx 0.0129$

Combinations give the same value, which is a good check on the arithmetic.

$\frac{\binom{13}{3}}{\binom{52}{3}} = \frac{286}{22100} = \frac{11}{850}$

The value $\frac{1}{64} = \left(\frac{1}{4}\right)^3$ would be right only with replacement, since it assumes the three draws are independent.

Compute all three quantities and compare. There are $36$ equally likely rolls, $18$ of which have an even first die, and $6$ of which sum to $7$.

$P(A) = \frac{18}{36} = \frac{1}{2}, \qquad P(B) = \frac{6}{36} = \frac{1}{6}$

The rolls in both events are $(2,5)$, $(4,3)$, and $(6,1)$, so $P(A \cap B) = \frac{3}{36} = \frac{1}{12}$. Now test the product rule.

$P(A)\,P(B) = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12} = P(A \cap B)$

The equation holds, so the events are independent. Whatever the first die shows, exactly one value of the second die completes a sum of $7$, which is why the first die carries no information about the sum being $7$.

Conditioning shrinks the sample space to the rolls that sum to $8$.

$B = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}, \qquad \lvert B \rvert = 5$

Of those, the rolls with an even first die are $(2,6)$, $(4,4)$, and $(6,2)$.

$P(\text{first even} \mid \text{sum} = 8) = \frac{3}{5}$

Since $P(\text{first even}) = \frac{1}{2}$ and $\frac{3}{5} \ne \frac{1}{2}$, these two events are dependent. Compare that with a sum of $7$, where the same pair of events turns out to be independent. Independence is a numerical accident of the particular events, not a property you can see at a glance.

The sample space has $2^6 = 64$ equally likely sequences. The complement of "at least one head" is "all six are tails", which is a single sequence.

$P(\text{no heads}) = \frac{1}{64}$

Subtract from $1$.

$P(\text{at least one head}) = 1 - \frac{1}{64} = \frac{63}{64} \approx 0.984$

Conditioning on the first draw changes the box. One defective item has been removed, so $2$ defectives remain among $9$ items.

$P(\text{2nd defective} \mid \text{1st defective}) = \frac{2}{9}$

The unconditional probability that the second item is defective is $\frac{3}{10}$, so learning about the first draw lowered it. That is precisely what it means for the draws to be dependent.

Rearrange the addition rule to isolate the intersection.

$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 1.3 - P(A \cup B)$

The overlap is smallest when the union is as large as possible, and no probability can exceed $1$, so $P(A \cup B) \le 1$.

$P(A \cap B) \ge 1.3 - 1 = 0.3$

The events cannot be disjoint: together they claim more than the whole sample space, so at least $0.3$ of it must be shared. The value $0.42$ is what independence would give, which is possible but is not the minimum.

The draw produces a set of six numbers, and the order they come out in does not affect whether your ticket matches, so the sample space is counted with combinations.

$\lvert S \rvert = \binom{49}{6} = 13{,}983{,}816$

Exactly one of those sets is the one on your ticket, so the event has a single favourable outcome.

$P(\text{jackpot}) = \frac{1}{13{,}983{,}816} \approx 7.2 \times 10^{-8}$

The expression $\frac{1}{49^6}$ would be right for six draws made with replacement in a fixed order, which is not how the draw works.