Egypt Secondary 3 Statistics, Week 15: Review of Unit 3: events and the axioms, conditional probability and independence in mixed problems

Review of Unit 3: events and the axioms, conditional probability and independence in mixed problems

Statistics Sec 3 Term 1, Units 1–3, Review

GoalReview Unit 3 in mixed problems: sample spaces and events, the axioms, conditional probability, trees and independence.

New words

which tool? · list outcomes for small spaces, a table for two features, a tree for draws in order, a Venn picture for overlapsindependence test · P(A ∩ B) = P(A) × P(B)exclusive is not independent · exclusive events with positive probabilities are dependent

Explanation

Unit 3 gave a toolbox, and a problem is mostly the choice of the right tool. For a small experiment, list the equally likely outcomes in a table or a tree, then P(A) = n(A) ÷ n(S). With two events, find the four regions of a Venn picture: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and P(A′) = 1 − P(A) come from the axioms. When the words “given that” appear, shrink the sample space: P(A | B) = P(A ∩ B) ÷ P(B). For draws in order, a tree multiplies along a path, and without replacement the second branches change. To test independence compare P(A ∩ B) with P(A) × P(B); remember that exclusive events with positive probabilities are never independent. The three examples that follow mix these tools.

Situation	Tool
one experiment with few outcomes	list or table of S; P(A) = n(A) ÷ n(S)
two events with an overlap	Venn picture with four regions; P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
“given that …”	reduced sample space; P(A \| B) = P(A ∩ B) ÷ P(B)
draws in order	tree: multiply along a path; without replacement the second branches change
“are they independent?”	compare P(A ∩ B) with P(A) × P(B)

Worked examples

	1	2	3
H	H1	H2	H3
T	T1	T2	T3

A coin and a three-sector spinner (1, 2, 3, equally likely): S has 6 outcomes H1, H2, H3, T1, T2, T3. Let A = “head” and B = “the number is at least 2”. P(A) = 3/6, P(B) = 4/6 and A ∩ B = {H2, H3}, so P(A ∩ B) = 2/6 = 1/3 = 1/2 × 2/3: A and B are independent. P(A ∪ B) = 1/2 + 2/3 − 1/3 = 5/6, and counting H1, H2, H3, T2, T3 gives 5 outcomes of 6.

	Robotics	Art	Chess	Total
Girls	18	30	12	60
Boys	12	10	18	40
Total	30	40	30	100

A two-way table. 100 pupils each join one club. P(girl | Art) = 30/40 = 3/4, but P(Art | girl) = 30/60 = 1/2. Test independence: P(girl ∩ Robotics) = 18/100 = 0.18 = 0.6 × 0.3, so “girl” and “Robotics” are independent. But P(girl ∩ Art) = 0.30 while 0.6 × 0.4 = 0.24, so “girl” and “Art” are dependent. “Robotics” and “Art” are exclusive (one club each) and have positive probabilities, so they are dependent.

A tree. A bag holds 4 blue and 2 orange counters; two are drawn without replacement. P(BB) = 4/6 × 3/5 = 2/5; P(OO) = 2/6 × 1/5 = 1/15; one of each colour = 2 × 4/6 × 2/5 = 8/15. A conditional question: P(first blue | second orange) = P(B then O) ÷ P(second O) = (4/15) ÷ (4/15 + 1/15) = 4/5.

Class activity

Make up a small probability problem of your own that needs two tools, for example a table and a conditional probability, or a tree and an independence test. Write the question on one side of a card and the full solution on the other, then swap cards with a classmate and solve each other’s.

Practice

P(A) = 0.55, P(B) = 0.4 and P(A ∪ B) = 0.8. Find P(A ∩ B), P(A only), P(neither) and P(A | B). Are A and B independent?

Two dice are thrown. A = “the sum is 7” and B = “the first die is odd”. Are A and B independent?

In a class of 36 pupils, 20 play chess, 15 do art and 6 do both. One pupil is chosen at random. Find P(chess | art), P(art | chess) and P(neither).

Five cards numbered 1 to 5; two are drawn without replacement. Find P(both even), P(both odd) and P(the sum is even).

A and B are independent with P(A) = 0.3 and P(B) = 0.5. Find P(A ∪ B) and P(exactly one). What would P(A ∪ B) be if A and B were exclusive?

Review of Term 1

Review of Unit 3: events and the axioms, conditional probability and independence in mixed problems