Statistics

Probability: Fundamentals

Basic probability, expected value, and Bayes' theorem — reasoning under uncertainty.

For equally likely outcomes, $P (A) = \frac{favourable}{total}$ , and $P (not A) = 1 - P (A)$ . The expected value of a payoff is $E [X] = \sum x_{i} P (x_{i})$ . Bayes' theorem updates a prior with evidence: $P (A ∣ B) = \frac{P ( B ∣ A ) P ( A )}{P ( B )}$ .

1. Complementary events
easy
Two fair dice are rolled. What is the probability that the sum is not $7$ ?
Show solution
There are $36$ equally likely outcomes. A sum of $7$ occurs in $6$ ways: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ , so $P (7) = \frac{6}{36} = \frac{1}{6}$ .

By the complement rule, $P (not 7) = 1 - \frac{1}{6} = \frac{5}{6} \approx 0.833$ .
2. Expected value of a bet
medium
A game costs $\$ 5 $t o pl a y . Y o u w in$ $20 $w i t h p r o babi l i t y$ 0.2$ and win nothing otherwise. What is the expected net value, and should you play?
Show solution
Expected winnings: $E[\text{win}] = 0.2 \times \$ 20 + 0.8 \times $0 = $4$.

Net of the $\$ 5 $cos t :$ E[\text{net}] = $4 - $5 = -$1$.

The expected net value is $-\$ 1$ per play, so on average you lose. A risk-neutral player should not play.
3. Bayes' theorem
hard
A disease affects $1%$ of a population. A test is $99%$ sensitive (detects the disease if present) and $95%$ specific (correctly negative if absent). A random person tests positive. What is the probability they actually have the disease?
Show solution
Let $D$ = has disease. $P (D) = 0.01$ , $P (pos ∣ D) = 0.99$ , $P (pos ∣ no D) = 0.05$ .

$P (pos) = 0.99 \times 0.01 + 0.05 \times 0.99 = 0.0099 + 0.0495 = 0.0594$ .

$P (D ∣ pos) = \frac{0.0099}{0.0594} \approx 0.167$ .

Only about 16.7% — despite an accurate test, a positive result mostly reflects false positives because the disease is rare. This is the base-rate fallacy.

Probability: Fundamentals

1. Complementary events

2. Expected value of a bet

3. Bayes' theorem

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