MHT-CET Maths · Probability Distribution

Conditional Probability, Independence and Bayes' Theorem

Restrict the sample space to compute P(A|B), chain events with the multiplication rule, exploit independence for 'at least one / exactly one' shortcuts, and reverse the conditioning with total probability and Bayes' theorem.

All Probability Distribution notes MHT-CET Maths strategy

Why this matters

This is the densest subtopic in the chapter: 23 PYQs sit here (4 HARD, 15 MODERATE, 4 EASY). MHT-CET tests the whole conditional-probability chain — the definition P(A|B) = P(A∩B)/P(B), sequential draws without replacement, the independence identity P(A∩B) = P(A)P(B), the 1 − P(none) shortcut for 'the target is hit / the problem is solved', and Bayes' theorem for bag/box/disease posteriors. The recurring traps are all here too: confusing 'exactly one' with 'at least one', forgetting P(A'|B) = P(A') only when A and B are independent, and swapping priors with likelihoods in the Bayes ratio.

Concept 1 of 7

Conditional Probability — Restricting the Sample Space

Intuition

Once you are told that event B has happened, B becomes your new, smaller universe. The conditional probability P(A|B) asks: within that shrunken world of B-outcomes, what fraction also lie in A? You are simply re-measuring A's chance against B instead of against the whole sample space.

Definition

For events A and B with $P(B) > 0$ , the conditional probability of A given B is

$P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}$ — the share of B's probability that also lies in A.
Equivalently, rearranged, $P(A\cap B) = P(B)\,P(A\mid B) = P(A)\,P(B\mid A)$ — the multiplication rule.
When outcomes are equally likely, this reduces to counting: $P(A\mid B) = \dfrac{n(A\cap B)}{n(B)}$ .

Definition of conditional probability

P(A\mid B) = \dfrac{P(A\cap B)}{P(B)},\qquad P(B) > 0

P(A\cap B)probability that both A and B occur
P(B)probability of the conditioning (given) event — the new universe
P(A\mid B)probability of A once B is known to have occurred

Diagram · P(A | B) restricts the world to B

Once B is given, only the amber region counts — it's the new whole. P(A | B) is the slice of B that also lies in A: P(A | B) = P(A∩B) / P(B). Dividing by P(B) is exactly "rescale B to be the new 100%".

Worked example

A fair die is rolled. Given that the outcome is even, find the probability that it is greater than 3.

Conditioning event $B = \{2,4,6\}$ , so $P(B) = \tfrac36 = \tfrac12$ .
Event $A = \{4,5,6\}$ ; the overlap $A\cap B = \{4,6\}$ , so $P(A\cap B) = \tfrac26 = \tfrac13$ .
Restrict: $P(A\mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{1/3}{1/2} = \dfrac{2}{3}$ .
Check by counting inside B: of the three even outcomes $\{2,4,6\}$ , two exceed 3, giving $\tfrac23$ .

Answer:

P(A\mid B) = \dfrac{2}{3}

P(A|B) and P(B|A) are not the same number

The condition goes in the denominator:

P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}

divides by

P(B)

, while

P(B\mid A) = \dfrac{P(A\cap B)}{P(A)}

divides by

P(A)

. The numerator

P(A\cap B)

is shared, but swapping which event you condition on gives a different answer unless

P(A)=P(B)

Divide by the GIVEN event's probability, not by 1

A conditional probability is measured against the reduced universe B, so

P(A\mid B)

can be much larger than

P(A)

. Forgetting to divide by

P(B)

— reporting just

P(A\cap B)

— is the most common conditional-probability slip.

Concept 2 of 7

Multiplication Rule and Sequential Draws Without Replacement

Intuition

When events happen in sequence — draw a ticket, then another, then another — chain them with the multiplication rule, updating the pool after every draw. Each factor is the conditional probability given everything drawn so far. If the draws come from separate independent sources (one ball from each of several urns), the factors just multiply and no updating is needed.

Definition

The general multiplication rule for a chain of events:

$P(E_1\cap E_2\cap E_3) = P(E_1)\,P(E_2\mid E_1)\,P(E_3\mid E_1\cap E_2)$ .
Without replacement: after each draw the counts shrink, so denominators drop by 1 and the relevant numerator drops by 1 as well. Drawing tickets one at a time is exactly this.
One item from each of several independent sources: the joint probability is just the product of each source's single-draw probability; add over all the ways a target composition can occur.

Chain rule for a sequence of dependent draws

P(E_1\cap E_2\cap E_3) = P(E_1)\,P(E_2\mid E_1)\,P(E_3\mid E_1\cap E_2)

Visualization · total probability & Bayes tree

P(B₁): 0.6P(A|B₁): 0.2P(A|B₂): 0.5

P(A) = 0.12 + 0.2 = 0.32P(B₁|A) = 0.12/0.32 = 0.375

Each leaf is a route product P(Bᵢ)·P(A|Bᵢ). Total probability adds the two leaves that end in A; Bayes' theorem divides one of those leaves by that total to flip the conditioning.

Worked example

A box has 4 red and 3 green tickets. Three tickets are drawn one at a time without replacement. Find the probability that they come out in the order red, green, red.

First draw red: $\dfrac{4}{7}$ .
Given a red is gone, draw green: $\dfrac{3}{6}$ (6 tickets left, 3 green).
Given red and green gone, draw red: $\dfrac{3}{5}$ (5 left, 3 red remain).
Multiply the chain: $\dfrac{4}{7}\cdot\dfrac{3}{6}\cdot\dfrac{3}{5} = \dfrac{36}{210} = \dfrac{6}{35}$ .

Answer:

P(\text{R, G, R}) = \dfrac{6}{35}

Practice this conceptself-check · 4 quick reps

Try it yourself

A box contains 9 tickets numbered 1 to 9. Three tickets are drawn one at a time without replacement. Find the probability that they come out alternately as odd, even, odd.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
From 5 red and 3 blue balls, two are drawn without replacement. Probability both red?
2.
Urn 1 has 2 white of 5; Urn 2 has 3 white of 5. Draw one from each — probability both white?
3.
9 tickets 1–9, draw 3 without replacement. P(even, odd, even)?
4.
In a chain of dependent draws, what does the third factor condition on?

From the bank · past-year question

Example 2Probability DistributionMODERATE

Three urns respectively contain 2 white and 3 black, 3 white and 2 black and 1 white and 4 black balls. If one ball is drawn from each urn, then the probability that the selection contains 1 black and 2 white balls is

[Q143 · 22 April Shift II · 2025]

Without replacement: shrink BOTH the numerator and the denominator

After drawing an odd ticket from 9, the next 'another odd' probability is

\dfrac{4}{8}

(one fewer odd, one fewer total) — not

\dfrac{5}{8}

. Freezing the count at its original value is the classic sequential-draw error.

Add over all favourable orderings for a composition

For '1 black and 2 white, one ball from each of three urns', the black can come from urn 1, 2, or 3 — so sum the three products

P(B_1W_2W_3)+P(W_1B_2W_3)+P(W_1W_2B_3)

. Computing only one ordering undercounts.

'Alternately O,E,O OR E,O,E' means add both patterns

Two disjoint arrangements satisfy the requirement, so compute each chain separately and add:

P(\text{O,E,O}) + P(\text{E,O,E})

. Treating it as a single pattern halves the answer.

Drill 1 more on multiplication rule and sequential draws without replacement

Concept 3 of 7

Computing P(A|B) by Restriction — Distributions, Counting and Composite Events

Intuition

Many MHT-CET conditional questions are just the definition applied carefully: build the conditioning event B, find the overlap A∩B, and divide. B may be a range of a random variable, a family of tickets sharing a property, or a birth-order event. The whole skill is identifying A∩B correctly and dividing by P(B), never by the full sample space.

Definition

Apply $P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}$ to composite events:

Distribution table: $P(a\le X<b\mid X\le c)$ is the sum of the P(X) values in the numerator range that also satisfy $X\le c$ , divided by $P(X\le c)$ .
'At least one' condition: for family/coin problems, $P(\text{all girls}\mid \text{at least one girl}) = \dfrac{P(\text{all girls})}{P(\text{at least one girl})}$ , and 'at least one' is best found as $1 - P(\text{none})$ .
Counting-based: $P(A\mid B) = \dfrac{n(A\cap B)}{n(B)}$ — list the outcomes in B, then count how many also lie in A.

Restriction form for composite conditioning

P(A\mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{n(A\cap B)}{n(B)}\ \text{(equally likely)}

Worked example

A random variable X takes values 0,1,2,3 with P(X) equal to

2k, 3k, 4k, 6k

respectively. Find

P(X\ge 2\mid X\ge 1)

Total probability is 1: $2k+3k+4k+6k = 15k = 1$ , so $k = \tfrac{1}{15}$ (not needed as k cancels).
Numerator $P(X\ge 2\cap X\ge 1) = P(X\ge 2) = P(2)+P(3) = 4k+6k = 10k$ .
Denominator $P(X\ge 1) = 3k+4k+6k = 13k$ .
Divide: $\dfrac{10k}{13k} = \dfrac{10}{13}$ .

Answer:

P(X\ge 2\mid X\ge 1) = \dfrac{10}{13}

Practice this conceptself-check · 4 quick reps

Try it yourself

A family has 3 children. Find the probability that all three are girls, given that at least one of them is a girl.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
For X with P(X=1)=2k, P(X=2)=4k, P(X=0)=k, find P(1≤X≤2 | X≤2).
2.
A die is rolled. P(prime | odd)?
3.
'At least one girl' in 3 children — probability?
4.
Given digit-sum 8 among tickets 00–49, how many tickets qualify?

From the bank · past-year question

Example 3Probability DistributionMODERATE

A random variable

X

has the following probability distribution ------------------------------------------------------

X:

0 1 2 3 4 ----------- ------- -------- -------- -------- -------

P(X):

k

2k

4k

2k

k

------------------------------------------------------ then the value of

P(1 \leqslant X< 4 \mid X\leqslant 2) =

[Q120 · 20 April Shift I · 2025]

The overlap A∩B is measured inside B, not over the whole space

For

P(1\le X<4\mid X\le 2)

the numerator is only

P(X=1)+P(X=2)

— the values that satisfy BOTH conditions (

X=3

is excluded by

X\le 2

). Summing the full range

1\le X<4

in the numerator over-counts.

Compute 'at least one' as the complement

P(\text{at least one girl})

in 3 children is

1 - P(\text{no girls}) = 1 - \tfrac18 = \tfrac78

, not

\tfrac38

(that is exactly-one) and not

\tfrac12

. The at-least-one event is large; its complement 'none' is the single easy case.

Watch for a 'None of these' answer when your value is not listed

A carefully computed conditional probability may not appear among the four numeric options; MHT-CET occasionally makes the correct choice 'None of these'. Recount the conditioning set before assuming an arithmetic slip.

Drill 2 more on computing p(a|b) by restriction — distributions, counting and composite events

Concept 4 of 7

Independence and Event Algebra with Unions

Intuition

Two events are independent when knowing one occurred does not change the other's probability — formally P(A∩B) = P(A)P(B). This single identity powers a whole family of algebra problems: given a union probability and one event, solve for the other; combine complements; or evaluate P(A'|B), which collapses to P(A') when A and B are independent. Odds 'in favour a:b' convert to probability a/(a+b) first.

Definition

The independence identity and its consequences:

$P(A\cap B) = P(A)\,P(B)$ (definition); then $P(A\cup B) = P(A)+P(B)-P(A)P(B)$ .
If A, B are independent, so are the pairs $(A',B)$ , $(A,B')$ , $(A',B')$ ; hence $P(A'\mid B) = P(A')$ and $P(B'\mid A') = P(B')$ .
Odds: 'odds in favour $a:b$ ' means $P = \dfrac{a}{a+b}$ ; 'odds against $a:b$ ' means $P = \dfrac{b}{a+b}$ .
The complement-conditional sum $P(A'\mid B') + P(B'\mid A') = \dfrac{P(A'\cap B')}{P(B')} + \dfrac{P(A'\cap B')}{P(A')}$ , using $P(A'\cap B') = 1 - P(A\cup B)$ .

Independence and the union it produces

P(A\cap B) = P(A)\,P(B),\qquad P(A\cup B) = P(A)+P(B)-P(A)\,P(B)

P(A)P(B)the product form — holds ONLY for independent A, B
P(A'\mid B)equals P(A') when A, B are independent

Diagram · mutually exclusive ≠ independent

Mutually exclusive events can't both happen, so they don't overlap and P(A∩B) = 0. Independent events do overlap — one happening doesn't change the other, so P(A∩B) = P(A)·P(B). Disjoint events with non-zero probability are therefore never independent.

Worked example

A and B are independent events with

P(A) = \tfrac13

and

P(A\cup B) = \tfrac35

. Find

P(B)

Independence: $P(A\cup B) = P(A) + P(B) - P(A)P(B)$ .
Substitute: $\tfrac35 = \tfrac13 + P(B) - \tfrac13 P(B) = \tfrac13 + \tfrac23 P(B)$ .
So $\tfrac23 P(B) = \tfrac35 - \tfrac13 = \tfrac{9-5}{15} = \tfrac{4}{15}$ .
Hence $P(B) = \tfrac{4}{15}\cdot\tfrac32 = \tfrac{2}{5}$ .

Answer:

P(B) = \dfrac{2}{5}

Practice this conceptself-check · 5 quick reps

Try it yourself

A and B are independent events with

P(A) = \tfrac{3}{10}

and

P(B) = \tfrac{2}{5}

. Find

P(A'\cup B)

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

1.
A, B independent, P(A)=1/4, P(B)=2/5. Find P(A∩B).
2.
Odds in favour of an event are 2:5. Find its probability.
3.
A, B independent. Simplify P(A'|B).
4.
If P(A∪B)=1/3, find P(A'∩B').
5.
P(A')=0.75, so P(A)=?

From the bank · past-year question

Example 4Probability DistributionMODERATE

Let A and B are independent events with

P(B) = \frac{2}{5}

P(A \cup B) = \frac{11}{20}

, then

P(A'|B)

is root of the equation

[Shift || · 2025]

P(A'|B) = P(A') needs INDEPENDENCE

For independent A and B,

P(A'\mid B) = P(A')

— so if

P(A) = \tfrac14

then

P(A'\mid B) = \tfrac34

. This collapse is FALSE for dependent events; there you must use

P(A'\mid B) = \dfrac{P(A'\cap B)}{P(B)}

The union formula loses its cross-term only when independent

P(A\cup B) = P(A) + P(B) - P(A\cap B)

always holds; replacing

P(A\cap B)

P(A)P(B)

is legal ONLY under independence. Do not use the product form for mutually exclusive or unspecified events.

Convert odds to probability before plugging in

'Odds in favour

a:b

' is

P = \dfrac{a}{a+b}

, not

\dfrac{a}{b}

. Odds

2:5

give

P = \tfrac27

; using

\tfrac25

is the standard odds-vs-fraction mistake in ship/selection problems.

Independent is not the same as mutually exclusive

Mutually exclusive means

P(A\cap B) = 0

; independent means

P(A\cap B) = P(A)P(B)

. Two events with nonzero probabilities cannot be both — if they never co-occur, knowing one occurred forces the other out, which is maximal dependence.

Drill 9 more on independence and event algebra with unions

Concept 5 of 7

At Least One and Exactly One for Independent Trials

Intuition

'The problem is solved' or 'the target is hit' means at least one of several independent attempts succeeds — always compute this as 1 minus the probability that they ALL fail. 'Exactly one is selected' is different: sum the products where one succeeds and the rest fail. Getting these two apart is the single most-tested distinction in this subtopic.

Definition

For independent events $A_1,\dots,A_n$ with success probabilities $p_i$ :

At least one succeeds: $P(\text{at least one}) = 1 - \prod_i (1-p_i)$ — one minus 'all fail'.
Exactly one (of two): $P = P(A)P(B') + P(A')P(B)$ .
A composite pattern like 'hit by P or Q but not R' is a sum of the independent triple-products matching that description, e.g. $P Q' R' + P' Q R' + P Q R'$ (all with R failing).

At-least-one and exactly-one

P(\text{at least one}) = 1 - \prod_{i}(1-p_i),\qquad P(\text{exactly one of }A,B) = P(A)P(B') + P(A')P(B)

Diagram · two coin tosses → 2² = 4 outcomes

Each toss branches into H or T with probability ½, and the branches multiply: every leaf is ½ × ½ = ¼. Tossing n coins gives 2ⁿ equally likely outcomes — so "at least one head" is easiest via the complement, 1 − P(all tails).

Worked example

Three shooters hit a target independently with probabilities

\tfrac13, \tfrac14, \tfrac15

. Find the probability that the target is hit.

'Target is hit' means at least one shooter hits — use the complement.
$P(\text{all miss}) = (1-\tfrac13)(1-\tfrac14)(1-\tfrac15) = \tfrac23\cdot\tfrac34\cdot\tfrac45$ .
$= \dfrac{24}{60} = \dfrac25$ .
$P(\text{hit}) = 1 - \tfrac25 = \tfrac35$ .

Answer:

P(\text{target hit}) = \dfrac{3}{5}

Practice this conceptself-check · 4 quick reps

Try it yourself

Two candidates A and B are selected independently with probabilities

\tfrac25

and

\tfrac47

. Find the probability that exactly one of them is selected.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Four people hit independently with p = 1/2, 1/3, 1/4, 1/5. P(target hit)?
2.
A, B, C solve a problem independently with p = 1/2, 1/3, 1/4. P(problem solved)?
3.
Husband selected 1/7, wife 1/5, independent. P(only one selected)?
4.
State the shortcut for 'at least one of many independent successes'.

From the bank · past-year question

Example 5Probability DistributionEASY

A problem in statistics is given to three students A, B and C. Their probabilities of solving the problem are

\frac{1}{2},\frac{1}{3}

and

\frac{1}{4}

respectively. If all of them try independently, then the probability that the problem is solved is

[Q142 · 9th May Shift 2 · 2024]

'At least one' is 1 − P(none), NOT the sum of individual probabilities

Adding

p_1 + p_2 + p_3

over-counts overlaps and can exceed 1. For 'the target is hit / the problem is solved', always use

1 - \prod(1-p_i)

. E.g. shooters

\tfrac12,\tfrac13,\tfrac14

give

1 - \tfrac14 = \tfrac34

, not

\tfrac12+\tfrac13+\tfrac14

Exactly one ≠ at least one

'One of them is selected' asking for EXACTLY one is

P(A)P(B') + P(A')P(B)

. Using

P(A\cup B) = P(A)+P(B)-P(A)P(B)

gives AT LEAST one — a bigger number. Read whether the problem excludes the both-succeed case.

Complement each event correctly inside a composite pattern

'Hit by P or Q but not R' forces R to FAIL in every favourable term, so each product carries

P(R') = 1 - P(R)

. Sum

PQ'R' + P'QR' + PQR'

. Forgetting to put

R'

in each term, or dropping a valid case, is the usual error.

Drill 5 more on at least one and exactly one for independent trials

Concept 6 of 7

Total Probability Theorem

Intuition

When an event can happen through several mutually exclusive, exhaustive 'routes' (which bag, which first draw, which machine), its overall probability is the weighted average of the route-conditional probabilities — each weighted by how likely that route is. Draw a two-stage tree: branch to the routes, then to the event, and add the branch products.

Definition

If $H_1,\dots,H_n$ are mutually exclusive and exhaustive (they partition the sample space) with $P(H_i) > 0$ , then for any event E:

$P(E) = \displaystyle\sum_{i=1}^{n} P(H_i)\,P(E\mid H_i)$ .
Each term is one route's prior times its conditional; the routes' priors sum to 1.
Polya-urn / draw-then-add problems fit here: the first draw's colour defines the partition, and the second-draw probability is conditional on the updated bag.

Total probability theorem

P(E) = \sum_{i=1}^{n} P(H_i)\,P(E\mid H_i)

H_ithe partition (mutually exclusive, exhaustive routes)
P(H_i)prior probability of route i
P(E\mid H_i)probability of E along route i

Visualization · total probability & Bayes tree

P(B₁): 0.6P(A|B₁): 0.2P(A|B₂): 0.5

P(A) = 0.12 + 0.2 = 0.32P(B₁|A) = 0.12/0.32 = 0.375

Each leaf is a route product P(Bᵢ)·P(A|Bᵢ). Total probability adds the two leaves that end in A; Bayes' theorem divides one of those leaves by that total to flip the conditioning.

Worked example

Bag I has 2 red and 3 white balls; Bag II has 4 red and 1 white. A bag is chosen at random and one ball is drawn. Find the probability the ball is red.

Routes: $P(\text{Bag I}) = P(\text{Bag II}) = \tfrac12$ .
Conditionals: $P(R\mid \text{I}) = \tfrac25$ , $P(R\mid \text{II}) = \tfrac45$ .
Total probability: $P(R) = \tfrac12\cdot\tfrac25 + \tfrac12\cdot\tfrac45 = \tfrac{2}{10} + \tfrac{4}{10} = \tfrac{6}{10}$ .
Simplify: $P(R) = \tfrac35$ .

Answer:

P(R) = \dfrac{3}{5}

Practice this conceptself-check · 4 quick reps

Try it yourself

A bag has 4 red and 6 black balls. A ball is drawn, its colour noted, and it plus 3 more of the same colour are returned. A second ball is then drawn. Find the probability the second ball is red.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Two machines make 60% and 40% of items; defect rates 2% and 5%. P(item defective)?
2.
What two properties must the routes H_i have for total probability?
3.
Bag chosen 1/2 each; P(R|I)=2/5, P(R|II)=3/8. P(R)?
4.
Do the priors P(H_i) sum to 1?

From the bank · past-year question

Example 6Probability DistributionHARD

A bag contains 4 Red and 6 Black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with 3 additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red is

[Q119 · 2nd May Shift 1 · 2023]

The routes must partition the space — exclusive AND exhaustive

Total probability

P(E) = \sum P(H_i)P(E\mid H_i)

is valid only when the

H_i

are mutually exclusive (no overlap) and exhaustive (cover everything). If your routes leave a gap or overlap, the weighted sum is wrong.

In draw-then-add problems, update the bag before the conditional

After returning the drawn ball plus 3 of its colour, the bag size grows to

10 + 3 = 13

and the matching colour count grows by 4. Use

P(R_2\mid R_1) = \tfrac{7}{13}

, not the original

\tfrac{4}{10}

Drill 1 more on total probability theorem

Concept 7 of 7

Bayes' Theorem — Reversing the Conditioning

Intuition

Bayes' theorem answers the reverse question: given that the effect E was observed, which cause H_i was most likely responsible? It reweights each route's prior by how strongly that route predicts the observed evidence (its likelihood), then normalises by the total probability of the evidence. It is total probability run backwards.

Definition

For a partition $H_1,\dots,H_n$ and observed event E with $P(E) > 0$ :

$P(H_k\mid E) = \dfrac{P(H_k)\,P(E\mid H_k)}{\displaystyle\sum_{i} P(H_i)\,P(E\mid H_i)}$ .
Numerator = the chosen route's prior × likelihood; denominator = total probability of E (the sum of ALL routes' prior × likelihood).
With equal priors $P(H_i) = \tfrac1n$ , the priors cancel and the posterior is just $\dfrac{P(E\mid H_k)}{\sum_i P(E\mid H_i)}$ — a ratio of likelihoods.

Bayes' theorem (posterior from priors and likelihoods)

P(H_k\mid E) = \dfrac{P(H_k)\,P(E\mid H_k)}{\sum_{i} P(H_i)\,P(E\mid H_i)}

P(H_k)prior — probability of cause k before the evidence
P(E\mid H_k)likelihood — how well cause k predicts the evidence E
P(H_k\mid E)posterior — probability of cause k after seeing E

Visualization · total probability & Bayes tree

P(B₁): 0.6P(A|B₁): 0.2P(A|B₂): 0.5

P(A) = 0.12 + 0.2 = 0.32P(B₁|A) = 0.12/0.32 = 0.375

Each leaf is a route product P(Bᵢ)·P(A|Bᵢ). Total probability adds the two leaves that end in A; Bayes' theorem divides one of those leaves by that total to flip the conditioning.

Worked example

Bag I has 3 red and 1 white ball; Bag II has 1 red and 3 white. A bag is chosen at random and a red ball is drawn. Find the probability it came from Bag I.

Priors: $P(\text{I}) = P(\text{II}) = \tfrac12$ . Likelihoods: $P(R\mid \text{I}) = \tfrac34$ , $P(R\mid \text{II}) = \tfrac14$ .
Total probability of red: $P(R) = \tfrac12\cdot\tfrac34 + \tfrac12\cdot\tfrac14 = \tfrac{3}{8} + \tfrac18 = \tfrac48 = \tfrac12$ .
Bayes: $P(\text{I}\mid R) = \dfrac{\tfrac12\cdot\tfrac34}{\tfrac12} = \dfrac{3/8}{1/2}$ .
$= \dfrac34$ .

Answer:

P(\text{Bag I}\mid R) = \dfrac{3}{4}

Practice this conceptself-check · 4 quick reps

Try it yourself

For

k = 1,2,3

, box

B_k

contains k red and

(k+1)

white balls, with

P(B_1) = \tfrac12,\ P(B_2) = \tfrac13,\ P(B_3) = \tfrac16

. A box is chosen and a red ball drawn. Find the probability it came from

B_2

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Equal priors on 3 diseases; test-positive likelihoods 0.7, 0.5, 0.8. P(disease 2 | positive)?
2.
In Bayes' theorem, what sits in the denominator?
3.
Bag I: 3R 2G, Bag II: 5R 3G, chosen equally; a green is drawn. P(from Bag I)?
4.
When do priors drop out of the Bayes ratio?

From the bank · past-year question

Example 7Probability DistributionHARD

For

k = 1,2,3

the box

B_{k}

contains k red balls and

(k + 1)

white balls. Let

P\left( B_{1} \right)=\frac{1}{2},P\left( B_{2} \right)=\frac{1}{3}

and

P\left( B_{3} \right)=\frac{1}{6}.A

box is selected at random and a ball is drawn from it. If a red ball is drawn from it, then the probability that it comes from box

B_{2}

[Q144 · 20 April Shift II · 2025]

Numerator is ONE route; denominator is ALL routes

P(H_k\mid E)

puts only the chosen route

P(H_k)P(E\mid H_k)

on top, but the FULL total probability

\sum_i P(H_i)P(E\mid H_i)

on the bottom. Using the same single term top and bottom gives 1 — a classic Bayes setup error.

Do not swap priors and likelihoods

The prior

P(H_i)

(which box/bag was chosen) multiplies the likelihood

P(E\mid H_i)

(chance of the observed ball given that box). Interchanging them — using

P(H_i\mid E)

where a likelihood belongs — corrupts every term.

Equal priors cancel — reduce to a likelihood ratio

With equal priors

P(H_i) = \tfrac1n

, the

\tfrac1n

factors out of every term and the posterior becomes

\dfrac{P(E\mid H_k)}{\sum_i P(E\mid H_i)}

. For disease tests with equal prior probability, this shortcut avoids carrying

\tfrac13

through the arithmetic.

Drill 2 more on bayes' theorem — reversing the conditioning

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (7)

Conditional Probability — Restricting the Sample Space
Definition of conditional probability
$P(A\mid B) = \dfrac{P(A\cap B)}{P(B)},\qquad P(B) > 0$
Multiplication Rule and Sequential Draws Without Replacement
Chain rule for a sequence of dependent draws
$P(E_1\cap E_2\cap E_3) = P(E_1)\,P(E_2\mid E_1)\,P(E_3\mid E_1\cap E_2)$
Computing P(A|B) by Restriction — Distributions, Counting and Composite Events
Restriction form for composite conditioning
$P(A\mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{n(A\cap B)}{n(B)}\ \text{(equally likely)}$
Independence and Event Algebra with Unions
Independence and the union it produces
$P(A\cap B) = P(A)\,P(B),\qquad P(A\cup B) = P(A)+P(B)-P(A)\,P(B)$
At Least One and Exactly One for Independent Trials
At-least-one and exactly-one
$P(\text{at least one}) = 1 - \prod_{i}(1-p_i),\qquad P(\text{exactly one of }A,B) = P(A)P(B') + P(A')P(B)$
Total Probability Theorem
Total probability theorem
$P(E) = \sum_{i=1}^{n} P(H_i)\,P(E\mid H_i)$
Bayes' Theorem — Reversing the Conditioning
Bayes' theorem (posterior from priors and likelihoods)
$P(H_k\mid E) = \dfrac{P(H_k)\,P(E\mid H_k)}{\sum_{i} P(H_i)\,P(E\mid H_i)}$

Watch out for (20)

P(A|B) and P(B|A) are not the same number
→ Conditional Probability — Restricting the Sample Space
Divide by the GIVEN event's probability, not by 1
→ Conditional Probability — Restricting the Sample Space
Without replacement: shrink BOTH the numerator and the denominator
→ Multiplication Rule and Sequential Draws Without Replacement
Add over all favourable orderings for a composition
→ Multiplication Rule and Sequential Draws Without Replacement
'Alternately O,E,O OR E,O,E' means add both patterns
→ Multiplication Rule and Sequential Draws Without Replacement
The overlap A∩B is measured inside B, not over the whole space
→ Computing P(A|B) by Restriction — Distributions, Counting and Composite Events
Compute 'at least one' as the complement
→ Computing P(A|B) by Restriction — Distributions, Counting and Composite Events
Watch for a 'None of these' answer when your value is not listed
→ Computing P(A|B) by Restriction — Distributions, Counting and Composite Events
P(A'|B) = P(A') needs INDEPENDENCE
→ Independence and Event Algebra with Unions
The union formula loses its cross-term only when independent
→ Independence and Event Algebra with Unions
Convert odds to probability before plugging in
→ Independence and Event Algebra with Unions
Independent is not the same as mutually exclusive
→ Independence and Event Algebra with Unions
'At least one' is 1 − P(none), NOT the sum of individual probabilities
→ At Least One and Exactly One for Independent Trials
Exactly one ≠ at least one
→ At Least One and Exactly One for Independent Trials
Complement each event correctly inside a composite pattern
→ At Least One and Exactly One for Independent Trials
The routes must partition the space — exclusive AND exhaustive
→ Total Probability Theorem
In draw-then-add problems, update the bag before the conditional
→ Total Probability Theorem
Numerator is ONE route; denominator is ALL routes
→ Bayes' Theorem — Reversing the Conditioning
Do not swap priors and likelihoods
→ Bayes' Theorem — Reversing the Conditioning
Equal priors cancel — reduce to a likelihood ratio
→ Bayes' Theorem — Reversing the Conditioning

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Probability DistributionMODERATE

A box contains 9 tickets numbered 1 to 9 both inclusive. If 3 tickets are drawn from the box one at a time, then the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is

[Q107 · 19 April Shift I · 2025]

Example 2Probability DistributionMODERATE

Out of 50 tickets numbered 00, 01, 02, \ldots, 49, one ticket is drawn randomly. The probability of the ticket having the product of its digits 7, given that the sum of the digits is 8, is

[Q112 · May Shift 1 · 2021]

Example 3Probability DistributionMODERATE

A and B are independent events with

P(A) = \frac{3}{10}

P(B) = \frac{2}{5}

, then

P(A' \cup B)

has the value

[Q108 · 10th May Shift 2 · 2023]

Example 4Probability DistributionEASY

A man and his wife appear for an interview for two posts. The probability of the husband's selection is

\frac{1}{7}

and that of the wife's selection is

\frac{1}{5}

. If they appear for the interview independently, then the probability that only one of them is selected, is

[Q140 · 4th May Shift 2 · 2023]

Example 5Probability DistributionHARD

The probability, that a year selected at random will have 53 Mondays, is

[Q127 · 15th May Shift 1 · 2023]

Drill every past-year question on this subtopic

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Related notes

Probability Distribution — Random Variables & Expectation →