10 min readConcept sheet1-mark MCQ, 2 or 3-mark question; part of case-based questions; also a JEE favourite

Conditional probability

The probability of E given that F has happened is P(E|F) = P(E ∩ F)/P(F), where P(F) ≠ 0. F becomes the new sample space.

Conditional probabilityP(E|F) = P(E ∩ F)/P(F), where P(F) ≠ 0

Equally likely outcomesP(E|F) = n(E ∩ F)/n(F)

Range0 ≤ P(E|F) ≤ 1

Read 'P(E|F)' asThe probability of E given F (F has already happened)

What 'given' changes

Suppose you already know that event F has happened. Then outcomes outside F are no longer possible. F becomes the reduced sample space. Event E can now happen only through the outcomes it shares with F, that is E ∩ F. So the chance of E is the share of F taken up by E ∩ F. We write it P(E|F) and read it 'the probability of E given F'.

From counting to the formula

Let the sample space S have equally likely outcomes, and let P(F) ≠ 0.

1
Given F, the new sample space is F, with n(F) outcomes.
2
The favourable outcomes are those in E ∩ F.
3
P(E|F) = n(E ∩ F)/n(F)
4
Divide the top and the bottom by n(S): P(E|F) = [n(E ∩ F)/n(S)] / [n(F)/n(S)]
5★
So, P(E|F) = P(E ∩ F)/P(F). This is taken as the definition in all cases.

★ key step

Example 1: two dice

Problem

Two dice are thrown. Find the probability that at least one die shows 3, given that the sum is 8.

Working

F: the sum is 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}, so n(F) = 5.
E ∩ F: sum 8 with a 3 showing = {(3, 5), (5, 3)}, so n(E ∩ F) = 2.
P(E|F) = n(E ∩ F)/n(F) = 2/5

Answer

P(E|F) = 2/5.

Example 2: survey data

Problem

In a class of 60 students, 35 like cricket, 25 like kabaddi and 15 like both. A student who likes kabaddi is chosen. Find the probability that she also likes cricket.

Working

Let C = likes cricket and K = likes kabaddi.
P(K) = 25/60 and P(C ∩ K) = 15/60.
P(C|K) = P(C ∩ K)/P(K) = (15/60)/(25/60) = 15/25 = 3/5

Answer

P(C|K) = 3/5.

Example 3: from given probabilities

Problem

P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.7. Find P(A|B) and P(B|A).

Working

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.4 − 0.7 = 0.2
P(A|B) = 0.2/0.4 = 1/2
P(B|A) = 0.2/0.5 = 2/5

Answer

P(A|B) = 1/2 and P(B|A) = 2/5.

P(E|F) vs P(F|E)

P(E|F)F is given. Divide P(E ∩ F) by P(F).

P(F|E)E is given. Divide P(E ∩ F) by P(E).

The numerator is the same; only the denominator changes. They are equal when P(E) = P(F), or when P(E ∩ F) = 0.

In real life

A doctor reading a test report asks, 'Given this result, how likely is the disease?' That is a conditional probability. A weather forecast of rain given dark clouds is another.

Exam angle

1 marks: “If P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2, find P(A|B).” P(A|B) = P(A ∩ B)/P(B) = 0.2/0.5 = 0.4.

2 marks: “Three coins are tossed. Find the probability that all three show heads, given that at least two show heads.” F = {HHH, HHT, HTH, THH}, so n(F) = 4. E ∩ F = {HHH}. P(E|F) = 1/4.

3 marks: “Two dice are thrown. Find the probability that the sum is 10, given that the first die shows a number greater than 4.” F: first die 5 or 6, so n(F) = 12. E ∩ F = {(5, 5), (6, 4)}, so n(E ∩ F) = 2. P(E|F) = 2/12 = 1/6.

Remember it

Given F? Shrink the world to F. Divide by the probability of what is given.