Conditional probability and independence

“Given that B happened” means: every outcome outside B is now ruled out. The outcomes left are the new sample space, and we ask what share of them are also in A. In the grid below, pick A, then pick the condition B, and watch the ruled-out outcomes fade.

The starting set-up is JNU 2014: two dice, and the sum is known to be even; what is the chance it is 8? Eighteen outcomes have an even sum, and five of them add to 8: 5/18, not the unconditional 5/36. Dividing the counts is the same as dividing the probabilities, which gives the definition.

With percentages instead of counts, nothing changes. Hyderabad 2021: 25% of students failed Economics, 15% failed English and 10% failed both. A student who failed English failed Economics with probability 0.10 / 0.15 = 2/3. Note that P(Economics | English) = 2/3 while P(English | Economics) = 0.10/0.25 = 2/5. The two directions are different questions; Lesson 4 is about turning one into the other.

Turn the definition round and it tells you how to find the chance that two things both happen, one after the other.

Two spades from a pack without replacement: the first is a spade with probability 13/52, and given that, the second with 12/51. So P = 13/52 × 12/51 = 1/17. JNU 2012 asked whether this is more or less than 1/16 = (13/52)²: it is less, because removing a spade makes the next one slightly less likely.

When an outcome can come about in several ways, split by the ways. Draw a tree: the first stage lists the ways (a partition), the second the outcome. Multiply along each path (the multiplication rule), then add the paths that end where you want (the addition rule, since the paths are mutually exclusive).

Sometimes learning B tells you nothing about A: P(A | B) = P(A). Then the multiplication rule simplifies, and that simpler form is the standard definition, because it is symmetric and works even when P(B) = 0.

• Complements inherit it. If A and B are independent, so are A and Bᶜ, Aᶜ and B, and Aᶜ and Bᶜ. So P(neither) = P(Aᶜ)P(Bᶜ). Hyderabad and JNU both test this.
• Independence is a fact about the numbers, not about stories. In the dice grid, “first die even” and “sum is even” look connected, yet P = 1/2 either way: they are independent. Try it.
• Pairwise is not enough for three. Toss two coins. A: head on the first; C: head on the second; D: the coins match. Each pair is independent (P = 1/4 = 1/2 × 1/2). But all three together: P(A ∩ C ∩ D) = P(HH) = 1/4, not 1/8. Mutual independence needs the triple product too. DSE 2014 used these very events, and also G = two heads: A and G are not independent, since P(A ∩ G) = 1/4 but P(A)P(G) = 1/8.

Mutually exclusive: A and B cannot happen together, P(A ∩ B) = 0. Independent: P(A ∩ B) = P(A)P(B). If both events have positive probability, P(A)P(B) > 0, so the two definitions can never hold at once. Knowing A happened tells you for certain that an exclusive B did not: that is the opposite of telling you nothing.

Tosses of a coin, rolls of a die, or shots at a target are usually independent of each other. Then the probability of a sequence is the product of the probabilities, and three patterns cover most questions.

• At least one success in n tries, each with chance p: $1-(1-p{)}^n$. At least one 6 among six dice (DSE 2019): $1-(5/6{)}^6\approx 0.665$. Not 1, however many dice you add.
• Everything must work (parts in series, such as two UPS batteries): multiply the chances that each works.
• The first success on a given try: fail every earlier try, then succeed. A fair coin first shows heads on an odd toss with probability ½ + ⅛ + 1/32 + ⋯ = (½)/(1 − ¼) = 2/3 (DSE 2012 and 2021).