Conditional probability, Bayes and expected value

Given means a smaller world

$P(A \mid B)$, read "the probability of A given B", is the chance of A once you know that B has happened. Knowing B rules out every outcome outside B, so B becomes the new whole:

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$$

With counts, it is the number of cases with both A and B divided by the number of cases with B. Take 200 surveyed graduates:

	Works in finance	Works elsewhere	Total
Studied economics	48	72	120
Studied another subject	16	64	80
Total	64	136	200

$P(\text{finance} \mid \text{economics}) = \frac{48}{120} = 0.4$, because the economics row is the new whole. $P(\text{economics} \mid \text{finance}) = \frac{48}{64} = 0.75$, because now the finance column is. Same numerator, different questions, different answers. Before dividing, name the group the question puts you in; its total goes underneath.

Rearranging the definition gives the multiplication rule, $P(A \cap B) = P(B) \cdot P(A \mid B)$, which is what you apply along each branch of a probability tree. The draws without replacement in counting and probability used exactly this rule.

Independence

A and B are independent when knowing one leaves the chance of the other unchanged: $P(A \mid B) = P(A)$. The easiest test to apply is

$$P(A \cap B) = P(A) \cdot P(B).$$

In the table, $P(\text{economics}) \cdot P(\text{finance}) = 0.6 \times 0.32 = 0.192$, but $P(\text{both}) = \frac{48}{200} = 0.24$, so subject and job are not independent. Equivalently, $0.4 \ne 0.32$: economics graduates are more likely than the group as a whole to work in finance.

Independent does not mean mutually exclusive. Two events that cannot happen together have $P(A \cap B) = 0$, so if each has a positive chance they fail the test: learning that one happened tells you the other did not.

Bayes: imagine 1,000 cases

Bayes questions give the chance of the evidence given the cause and ask for the reverse. By hand, the safest route is to turn the percentages into counts for an imagined group of 1,000. Suppose 10% of borrowers default, and a credit model flags 90% of the borrowers who default and 10% of those who do not:

	Flagged	Not flagged	Total
Default	90	10	100
No default	90	810	900
Total	180	820	1,000

A flagged borrower is one of 180, and 90 of them default: $P(\text{default} \mid \text{flagged}) = \frac{90}{180} = \frac{1}{2}$. The model catches 90% of defaulters yet is right only half the time when it flags someone, because its 10% false-positive rate applies to a group nine times as large. When the cause is rare, false positives can match or outnumber true ones.

Bayes' theorem says the same in symbols, with $D$ for default and $F$ for flagged:

$$P(D \mid F) = \frac{P(F \mid D)\,P(D)}{P(F \mid D)\,P(D) + P(F \mid \text{not } D)\,P(\text{not } D)} = \frac{0.09}{0.09 + 0.09} = \frac{1}{2}.$$

The denominator is the flagged column of the table. If a percentage of 1,000 is not a whole number, imagine 10,000 instead.

Expected value

A payoff $X$ that takes the values $x_1, x_2, \ldots$ with probabilities $p_1, p_2, \ldots$ has expected value

$$E[X] = x_1 p_1 + x_2 p_2 + \cdots,$$

a weighted average with the probabilities as weights. It is the average result per play over many repetitions, not necessarily a result that can occur: one roll of a fair die has expected value 3.5. Check first that the probabilities add to 1. A cost paid in every case comes off once at the end: with a €10 ticket and an expected prize of €7, the expected net result is a loss of €3 per ticket. Expected values also add and scale: $E[X + Y] = E[X] + E[Y]$ and $E[aX + b] = aE[X] + b$.

A 2024 sample test on Bocconi's website, not linked from the current test page, included items on expected values. Bocconi publishes no topic list for quantitative reasoning, so treat this as a useful tool rather than a promised topic. The lesson on scoring, pacing and guessing applies expected value to guessing under negative marking.

The binomial distribution

Repeat the same trial $n$ times, independently, with success probability $p$ each time. The number of successes $X$ then follows the binomial distribution:

$$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}.$$

Any one order with $k$ successes and $n - k$ failures has probability $p^k (1 - p)^{n - k}$, and $\binom{n}{k}$ counts the orders, that is, which $k$ of the $n$ trials succeed. Exactly 3 heads in 5 tosses of a fair coin: $\binom{5}{3} \left(\frac{1}{2}\right)^5 = \frac{10}{32} = \frac{5}{16}$.

The mean and variance are

$$E[X] = np, \qquad \operatorname{Var}(X) = np(1 - p),$$

and the standard deviation is $\sqrt{np(1 - p)}$. For 100 tosses of a fair coin, the mean is 50, the variance 25 and the standard deviation 5.

The formula needs a fixed number of trials, two outcomes per trial, the same $p$ every time, and independence. Drawing from a small bag without replacement is not binomial, because each draw's chances depend on what came out before.