Porisma

Worked solution

Bound the part of C lying in B but outside A

P(A)=3/5P(A)=3/5, P(B)=3/4P(B)=3/4 and P(A∣B′)=1/2P(A\mid B')=1/2.

(i) Find P(A∩B)P(A\cap B). marks

(ii) Find P(A′∣A∪B)P(A'\mid A\cup B). marks

A third event C has P(C)=1/2P(C)=1/2 and is independent of A.

(iii) Find P(A′∩C)P(A'\cap C). mark

(iv) Hence show that 3/40≤P(A′∩B∩C)≤1/53/40\le P(A'\cap B\cap C)\le1/5. marks

[2014/AJC/II/7]

CJC · Review problems · Question 4

Answer

(i) 19/4019/40; (ii) 11/3511/35; (iii) 1/51/5; (iv) 3/40≤P(A′∩B∩C)≤1/53/40\le P(A'\cap B\cap C)\le1/5.

(i) Remove A outside B

P(A∩B′)=P(A∣B′)P(B′)=12(1−34)=1/8P(A\cap B')=P(A\mid B')P(B')=\tfrac12(1-\tfrac34)=1/8. Therefore P(A∩B)=P(A)−P(A∩B′)=3/5−1/8=19/40P(A\cap B)=P(A)-P(A\cap B')=3/5-1/8=19/40.

How did this part go?

(ii) Divide B-only by the union

Within A∪BA\cup B, the part outside A is exactly A′∩BA'\cap B. Its probability is P(B)−P(A∩B)=3/4−19/40=11/40P(B)-P(A\cap B)=3/4-19/40=11/40. The union is 3/5+3/4−19/40=7/83/5+3/4-19/40=7/8. Thus P(A′∣A∪B)=(11/40)/(7/8)=11/35P(A'\mid A\cup B)=(11/40)/(7/8)=11/35.

How did this part go?

(iii) Independence passes to the complement

Since A and C are independent, A′ and C are independent. Hence P(A′∩C)=P(A′)P(C)=(2/5)(1/2)=1/5P(A'\cap C)=P(A')P(C)=(2/5)(1/2)=1/5.

How did this part go?

(iv) Upper bound

The event A′∩B∩CA'\cap B\cap C is contained in A′∩CA'\cap C, so its probability is at most 1/5. Equality holds if all of the latter event lies in B.

How did this part go?

(iv) Lower bound, source method 1: outside-B capacity

P(A′∩B′)=1−P(A∪B)=1/8P(A'\cap B')=1-P(A\cup B)=1/8. Of the mass P(A′∩C)=1/5P(A'\cap C)=1/5, at most 1/8 can lie outside B, so at least 1/5−1/8=3/401/5-1/8=3/40 lies in B. The source’s small-C sketch is schematic and does not itself demonstrate the extremal placement; the probability-capacity argument establishes the bound.

How did this part go?

(iv) Lower bound, source method 2: union cannot exceed 1

P(A∪B∪C)=P(A∪B)+P(A′∩C)−P(A′∩B∩C)≤1P(A\cup B\cup C)=P(A\cup B)+P(A'\cap C)-P(A'\cap B\cap C)\le1. With P(A∪B)=7/8P(A\cup B)=7/8, this gives P(A′∩B∩C)≥7/8+1/5−1=3/40P(A'\cap B\cap C)\ge7/8+1/5-1=3/40. Combining bounds proves 3/40≤P(A′∩B∩C)≤1/53/40\le P(A'\cap B\cap C)\le1/5.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Four panels: A-only shaded; B-only shaded; source Method 1 sketch with a small C near the A/B overlap inside B; source Method 2 three-circle union sketch. The small C sketch is not an extremal lower-bound construction. Bounds follow the retained probability arguments, not scale or area in these schematic circles.
Four panels: A-only shaded; B-only shaded; source Method 1 sketch with a small C near the A/B overlap inside B; source Method 2 three-circle union sketch. The small C sketch is not an extremal lower-bound construction. Bounds follow the retained probability arguments, not scale or area in these schematic circles.
Four panels: A-only shaded; B-only shaded; source Method 1 sketch with a small C near the A/B overlap inside B; source Method 2 three-circle union sketch. The small C sketch is not an extremal lower-bound construction. Bounds follow the retained probability arguments, not scale or area in these schematic circles.

Common mistakes

C is independent of A; no independence with B is given. Use set containment/capacity for the bounds.

Solution notes

All four asks, all source algebra, both lower-bound methods and four diagram panels retained. Source diagrams are not area-proportional; the Method 1 small-C sketch does not demonstrate a minimum-overlap position, so the algebraic bound is explained independently. Original native text unchanged.

Read the hint