L4 · Probability Clinic
“I can choose and combine the tools of this unit - unions, intersections (including intersections of two independent events, counted from a combined sample space), complements, and the addition rule - and verify each answer with a second method.”
A coin is flipped and a die is rolled. Let A = the coin lands heads, and B = the die shows a multiple of 3. Find P(A ∩ B). Step 1: Build the combined sample space: 2 x 6 = 12 equally likely pairs: (H,1)...(H,6), (T,1)...(T,6). The two parts are independent - the coin cannot influence the die. Step 2: The intersection "A and B" is the pairs satisfying both conditions: (H,3) and (H,6) - 2 pairs. Step 3: P(A ∩ B) = 2/12 = 1/6. Step 4: Confirm with the formula the curriculum allows: P(A) x P(B) = 1/2 x 1/3 = 1/6. Counting is the ground truth; the product agrees because the events are independent. Say it plainly: this is the Grade 7 table method wearing its proper name - the favourable pairs ARE the intersection of the two events.
Why can the pairs be counted directly here, with no extra machinery?
Same coin and die. Find P(tails and a number greater than 4). -> Favourable pairs: (T,5), (T,6) - 2 of 12, so P = 1/6. (Formula check: 1/2 x 2/6 = 1/6)
"A coin is flipped and a spinner with 4 equal sections (numbered 1-4) is spun. The combined sample space has 2 x 4 = ___ pairs. P(heads and section 3) = 1 / ___ = ___." -> 8 pairs; 1/8
10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.