Find conditional probability as a fraction of the given event and interpret it

Learning goal: Understand the conditional probability of A given B as P(A and B) / P(B), find it in a uniform model as the fraction of B's outcomes that also belong to A, and interpret the answer in terms of the model. Recognize that P(A | B) and P(B | A) answer different questions.

What "given" does to the sample space. A conditional probability answers a question of the form "if we already know that B happened, how likely is A?" The information that B happened rules out every outcome not in B. The outcomes that remain possible are exactly the outcomes of B, so B becomes the new, smaller sample space. Within it, the outcomes that make A happen are those in both A and B. The conditional probability of A given B, written P(A | B), is therefore

P(A | B) = P(A and B) / P(B), defined whenever P(B) > 0.

In a uniform model, where every outcome of S is equally likely, both numerator and denominator are counts divided by the size of S, and the size of S cancels. The formula becomes: P(A | B) is the fraction of B's outcomes that also belong to A, that is, (number of outcomes in A ∩ B) divided by (number of outcomes in B). This counting version is the one to use when the sample space is written out.

The dice model. Roll a red die and a blue die; S has 36 equally likely ordered pairs. Let A be "the sum is 7" and B be "the red die is even". Suppose you are told that the red die came up even, but nothing else. The possible outcomes are now the 18 pairs in B. Among them, the pairs with sum 7 are (2, 5), (4, 3) and (6, 1), which are the outcomes of A ∩ B. So P(A | B) = 3/18 = 1/6. With the formula: P(A and B) = 3/36 and P(B) = 18/36, and (3/36) / (18/36) = 3/18 = 1/6, the same number.

Now reverse the roles. Suppose you are told the sum is 7. The possible outcomes are the 6 pairs of A. Among them, the red die is even for (2, 5), (4, 3), (6, 1), three of the six. So P(B | A) = 3/6 = 1/2. The numerator is the same three outcomes in both calculations; the denominators differ, because the given event is different. P(A | B) = 1/6 and P(B | A) = 1/2 are different numbers because they answer different questions, and neither can be found from the other without more information.

Interpreting in terms of the model. A conditional probability is a statement about the model, not about a single roll. P(A | B) = 1/6 means: among all rolls in which the red die is even, the model assigns the sum 7 to one sixth of them; in a long run of rolls with an even red die, about one in six would have sum 7. It does not predict what any single roll will show, and it says nothing about rolls with an odd red die. Always say what set the fraction is a fraction of.

A second model with unequal categories. Consider the class of 30 students from the previous lesson: 18 play an instrument, 12 are on a team, 8 do both. Choose one student at random; each student is an equally likely outcome. P(team | instrument) is the fraction of instrument players who are also on a team: 8/18 = 4/9. P(instrument | team) is the fraction of team members who play an instrument: 8/12 = 2/3. Interpretation: if you learn that the chosen student plays an instrument, the chance they are on a team is 4/9, lower than 2/3, which is the chance an athlete plays an instrument. Both conditionals are larger than the unconditional P(team) = 12/30 = 2/5 and P(instrument) = 18/30 = 3/5, respectively; knowing one activity raises the chance of the other in this class. That comparison is the start of the independence idea developed in the next lesson: A and B are independent exactly when P(A and B) = P(A) · P(B); when P(B) > 0 this is the same as P(A | B) = P(A), and when P(A) > 0 it is the same as P(B | A) = P(B).

A fully worked model with cards. A standard deck has 52 cards in four suits of 13; each suit has an ace, the numbers 2 through 10, a jack, a queen and a king. Draw one card at random. Let K be "king" (4 cards) and F be "face card", meaning jack, queen or king (12 cards). P(K | F) is the fraction of face cards that are kings: 4/12 = 1/3. Interpretation: if the drawn card is known to be a face card, one third of the possibilities are kings. P(F | K) is the fraction of kings that are face cards: 4/4 = 1. Every king is a face card, so given a king, being a face card is certain. Compare with the unconditional P(K) = 4/52 = 1/13: knowing the card is a face card raises the chance of a king from 1/13 to 1/3, because the information rules out the 40 non-face cards, none of which is a king.

The main misconception. The most common error is to divide by the size of the whole sample space instead of by the size of the given event: writing P(A | B) = 3/36 for the dice. That number is P(A and B), the probability that both happen when nothing is known. A conditional probability always has the given event as its denominator. A second error is to swap the two conditionals, reading "the fraction of B that is in A" when the question asks for "the fraction of A that is in B". Before computing, write the given event, count it, and count the part of it that lies in the other event; the answer is the second count over the first.

When P(B) = 0. The formula divides by P(B), so P(A | B) is undefined if B is impossible. In the dice model "the sum is 13" has no outcomes, and asking for a probability given that event has no meaning. In a uniform model this is the same as saying you cannot take a fraction of an empty set.

Worked conditional from a subset: Two dice; let D be "doubles" (6 outcomes) and G be "sum is at least 10". Given doubles, the possible outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6); of these, (5, 5) and (6, 6) have sum at least 10. So P(G | D) = 2/6 = 1/3. Interpretation: among doubles, one third have a sum of 10 or more. Unconditionally, G has 6 outcomes, (4, 6), (5, 5), (5, 6), (6, 4), (6, 5) and (6, 6), so P(G) = 6/36 = 1/6; knowing the roll is a double doubles the chance.

Worked reversal: With the same events, P(D | G) is the fraction of sum-at-least-10 outcomes that are doubles: (5, 5) and (6, 6) out of the six listed, 2/6 = 1/3. Here P(D | G) happens to equal P(G | D) because D and G have the same size; that is a coincidence of this pair of events, not a rule. Check with the formula: P(D and G) = 2/36, P(G) = 6/36, ratio 2/6.

Worked class model: 30 students, 18 instrument, 12 team, 8 both, so 22 do at least one and 8 do neither. Given that a chosen student is on a team, P(does not play an instrument | team) is the fraction of the 12 team members who are not among the 8 who do both: 4/12 = 1/3. This is 1 − P(instrument | team) = 1 − 2/3, as it must be: within the given event, an outcome either belongs to A or it does not.

Represent and record: On the 6-by-6 dice grid, outline the 18 cells of "red die is even" and shade the sum-7 cells inside that outline. Write P(sum 7 | red even) = 3/18 and P(red even | sum 7) = 3/6 with a one-sentence interpretation of each that names the set being divided. Repeat for "red die is even" and "sum is 8": find both conditionals (3/18 and 3/5) and explain in words why they differ.

Independent performance: Build a uniform model with at least 20 equally likely outcomes and define two events A and B that overlap but are not nested. Find P(A | B) and P(B | A) by counting, and again by the formula P(A and B) / P(B) with all probabilities over the full sample space; show the two methods agree. Then choose a third event C that is a subset of A and find P(A | C). Success criteria: each conditional names the given event and uses its size as the denominator; each answer is interpreted as "among the outcomes where … , the fraction with … is … "; the nested case gives exactly 1 and you explain why; the two conditionals for A and B are correctly identified as different questions.