Conditional Probability GCSE: P(A|B), Without Replacement and Tree Diagrams

Conditional probability asks: given that one event has already happened, what is the probability of another? At GCSE this appears in "without replacement" problems and Venn diagram questions. You write it as P(A | B) — the probability of A, given that B has occurred — and it often appears on tree diagrams.

What does P(A | B) mean?

The notation P(A | B) is read as "the probability of A given B." The vertical bar | means "given that." It tells you to restrict your sample space to only those outcomes where B has already occurred, then ask what fraction of those outcomes also include A.

The formal formula (not required to be quoted at GCSE, but useful to understand) is:

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

This says: of all the times B happens, in what proportion of those does A also happen?

Key point: P(A | B) is not the same as P(B | A). The order matters. "The probability it rained given that the ground is wet" is very different from "the probability the ground is wet given that it rained."

What is the difference between with and without replacement?

With replacement: after each selection, the item is put back. The total number of outcomes stays the same for every branch of the tree. The events are independent — the first draw does not affect the second.

Without replacement: the item is not returned. The total decreases by 1 for the second draw, and the number of the chosen category also decreases by 1. The events are dependent.

This is where conditional probability becomes essential: the probability on the second branch changes depending on what happened on the first branch.

How do you draw a tree diagram for "without replacement"?

Worked example: A bag contains 5 red balls and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red.

Setting up the tree:

• First draw: P(Red) = 5/8, P(Blue) = 3/8.
• Second draw if first was red: only 4 red and 3 blue remain (7 total). P(Red | 1st Red) = 4/7.
• Second draw if first was blue: 5 red and 2 blue remain (7 total). P(Red | 1st Blue) = 5/7.

P(both red) = 5/14.

Check: all four probabilities sum to 20/56 + 15/56 + 15/56 + 6/56 = 56/56 = 1 ✓

How do you find P(A | B) from a Venn diagram or frequency table?

Worked example: 30 students chose subjects. 18 study Maths (M), 14 study Science (S), 8 study both. One student is chosen at random. Given that the student studies Science, find the probability they also study Maths.

We want P(M | S): out of the 14 Science students, how many also study Maths?

8 students study both, and 14 study Science.

$$P(M \mid S) = \frac{8}{14} = \frac{4}{7}$$

Notice we restrict our denominator to 14 (the Science students), not 30 (everyone). This is the essence of conditional probability — we are working within a reduced sample space.

How do you calculate the probability of two dependent events using the multiplication rule?

The multiplication rule for dependent events uses conditional probability:

$$P(A \cap B) = P(A) \times P(B \mid A)$$

This is exactly the calculation you perform when you multiply along a branch of a tree diagram. Each "along the branch" step is a conditional probability.

Worked example: Three chocolates are in a box: 2 dark, 1 milk. Two are chosen without replacement. Find the probability of choosing one dark and one milk (in any order).

P(Dark then Milk) = 2/3 × 1/2 = 2/6 = 1/3 P(Milk then Dark) = 1/3 × 2/2 = 1/3 × 1 = 1/3

P(one of each) = 1/3 + 1/3 = 2/3

How do you find probabilities using the conditional probability formula?

Worked example: P(A ∩ B) = 0.12 and P(B) = 0.4. Find P(A | B).

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.4} = \mathbf{0.3}$$

This can also be used in reverse: if P(A | B) = 0.6 and P(B) = 0.25, then P(A ∩ B) = 0.6 × 0.25 = 0.15.

Frequently asked questions

How do I know when to add and when to multiply probabilities?

Multiply along a tree branch (both events happen — AND). Add separate branches (either route leads to the outcome — OR). A useful phrase: AND → multiply, OR → add. In the two-red-balls example, we multiply along the "Red then Red" branch; if we wanted "at least one red" we would add the probabilities of several branches.

Does the order of drawing matter in "without replacement" problems?

The order affects the structure of the tree, but the final probabilities for unordered outcomes should combine correctly. When a question asks for the probability of "one red and one blue" (not specifying order), add both routes: P(RB) + P(BR). Always draw the full tree to avoid missing routes.

What if there are more than two draws?

Extend the tree for three or more branches. The total remains the same (e.g. 8 balls for a first draw, 7 for second, 6 for third). The calculations become more complex but the method is identical — multiply along branches, add branches that lead to the required outcome.

Is conditional probability on Foundation tier?

Simple "without replacement" tree diagrams appear on both Foundation and Higher tiers. The formal notation P(A | B) and the formula P(A ∩ B)/P(B) are Higher tier. Venn diagram conditional probability questions are Higher. At Foundation, the focus is on correctly adjusting the numbers on the second branch after replacing or not replacing.

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