10 min readConcept sheet1-mark MCQ or assertion-reason; 2 or 3-mark question; also a JEE favourite

Independent events

E and F are independent if one happening does not change the chance of the other. Then P(E ∩ F) = P(E) · P(F).

Independent eventsP(E ∩ F) = P(E) · P(F)

The same idea with conditional probabilityP(E|F) = P(E) (when P(F) ≠ 0) and P(F|E) = P(F) (when P(E) ≠ 0)

Three events A, B, C (mutually independent)P(A ∩ B) = P(A)P(B), P(B ∩ C) = P(B)P(C), P(A ∩ C) = P(A)P(C) and P(A ∩ B ∩ C) = P(A)P(B)P(C)

If E and F are independentSo are E and F′, E′ and F, and E′ and F′

Independent vs mutually exclusive

IndependentOne event does not affect the chance of the other. Test: P(E ∩ F) = P(E) · P(F). They can happen together.

Mutually exclusiveThe events cannot happen together: E ∩ F = φ, so P(E ∩ F) = 0.

If P(E) and P(F) are both non-zero, mutually exclusive events cannot be independent, because then P(E) · P(F) > 0.

Proof: if E and F are independent, so are E′ and F′

1
P(E′ ∩ F′) = P((E ∪ F)′), by De Morgan's law
2
= 1 − P(E ∪ F)
3
= 1 − [P(E) + P(F) − P(E)P(F)], since E and F are independent
4
= [1 − P(E)][1 − P(F)]
5★
So, P(E′ ∩ F′) = P(E′) · P(F′), and E′ and F′ are independent.

★ key step

Example 1: test for independence

Problem

A number is chosen at random from 1 to 12. E: the number is even. F: the number is a multiple of 3. Are E and F independent? Repeat for the numbers 1 to 10.

Working

1 to 12: P(E) = 6/12 = 1/2 and P(F) = 4/12 = 1/3.
E ∩ F = {6, 12}, so P(E ∩ F) = 2/12 = 1/6 = 1/2 × 1/3. Independent.
1 to 10: P(E) = 5/10 = 1/2 and P(F) = 3/10 (3, 6 and 9).
E ∩ F = {6}, so P(E ∩ F) = 1/10, but P(E) · P(F) = 3/20. Not independent.

Answer

Independent for 1 to 12; not independent for 1 to 10.

Example 2: using independence

Problem

A and B are independent, with P(A) = 0.25 and P(B) = 0.4. Find P(A ∩ B), P(A ∪ B) and P(A′ ∩ B′).

Working

P(A ∩ B) = 0.25 × 0.4 = 0.1
P(A ∪ B) = 0.25 + 0.4 − 0.1 = 0.55
P(A′ ∩ B′) = P(A′) · P(B′) = 0.75 × 0.6 = 0.45
Check: P(A′ ∩ B′) = 1 − P(A ∪ B) = 1 − 0.55 = 0.45

Answer

0.1, 0.55 and 0.45.

Example 3: at least one success

Problem

Asha, Bilal and Chitra each try a puzzle on their own. Their chances of solving it are 1/2, 1/3 and 1/4. Find the probability that (i) the puzzle is solved, (ii) exactly one of them solves it.

Working

P(none solves) = (1/2)(2/3)(3/4) = 6/24 = 1/4
(i) P(solved) = 1 − 1/4 = 3/4
Only Asha: (1/2)(2/3)(3/4) = 6/24
Only Bilal: (1/2)(1/3)(3/4) = 3/24; only Chitra: (1/2)(2/3)(1/4) = 2/24
(ii) P(exactly one) = (6 + 3 + 2)/24 = 11/24

Answer

(i) 3/4 (ii) 11/24

In real life

Two traffic signals on different roads that work independently: the chance that both are green when you arrive is the product of their separate chances.

Exam angle

1 marks: “A and B are independent with P(A) = 0.2 and P(B) = 0.5. Find P(A ∩ B).” P(A ∩ B) = 0.2 × 0.5 = 0.1.

2 marks: “P(A) = 0.4, P(B) = 0.5 and P(A ∪ B) = 0.7. Are A and B independent?” P(A ∩ B) = 0.4 + 0.5 − 0.7 = 0.2, and P(A) · P(B) = 0.4 × 0.5 = 0.2. They are equal, so A and B are independent.

3 marks: “The chances that three students solve a problem on their own are 1/3, 1/4 and 1/5. Find the probability that exactly one of them solves it.” (1/3)(3/4)(4/5) + (2/3)(1/4)(4/5) + (2/3)(3/4)(1/5) = 12/60 + 8/60 + 6/60 = 26/60 = 13/30.

Remember it

Independent: multiply. Exclusive: they never meet.