Unit 13: Probability and Geometric Modeling · ~15–30 min
Objective
Calculate conditional probabilities from two-way frequency tables and determine whether two events are independent.
Two-way frequency tables
A two-way table organizes counts across two categories at once — one along the rows, one along the columns — so you can see how they overlap. Row and column totals (and a grand total) make it easy to pull out probabilities directly from counts.
Likes sports
Doesn't like sports
Total
Grade 9
54
26
80
Grade 10
42
28
70
Total
96
54
150
Conditional probability
Conditional probability asks: given that one event has already happened, what's the probability of another? "P(B given A)" is written P(B | A), and it means you restrict your attention to only the outcomes where A happened, then ask how many of those also satisfy B.
P(B | A) = P(A and B) ÷ P(A)
With a two-way table, this is even more direct — just work with counts instead of probabilities:
P(B | A) = (count of A and B) ÷ (row or column total for A)
P(B | A) restricts the sample space to circle A, then asks what fraction of it overlaps with B.
Independence
Two events are independent if knowing one happened doesn't change the probability of the other — that is, P(B | A) = P(B). If the conditional probability differs from the plain probability, the events are dependent.
Worked Example 1 · Conditional probability from a table
Grade 10 row total is 70, with 42 who like sports. P = 42/70 = 3/5 = 0.6.
2.Using the two-way table above, find P(Grade 9 | doesn't like sports).Show answer
The "doesn't like sports" column total is 54, with 26 in Grade 9. P = 26/54 = 13/27 ≈ 0.481.
3.P(A) = 0.5 and P(A and B) = 0.2. Find P(B | A).Show answer
P(B | A) = 0.2 ÷ 0.5 = 0.4.
4.In a deck of 52 cards, let A = "card is a heart" and B = "card is a face card (J, Q, K)." There are 3 heart face cards. Find P(B | A) and compare it to P(B) to decide if A and B are independent.Show answer
P(B | A) = 3/13 ≈ 0.231 (3 heart face cards out of 13 hearts). P(B) = 12/52 = 3/13 ≈ 0.231 (12 face cards out of 52). Since these are equal, A and B are independent.
5.Explain, in your own words, why P(B | A) can be different from P(B).Show answer
P(B) considers the entire sample space, while P(B | A) narrows the sample space down to only the outcomes where A already happened. If B is more or less common within that narrower group than it is overall, the two probabilities will differ — that difference is exactly what makes two events dependent rather than independent.