MHT-CET Maths · Probability Distribution

Classical Probability, Addition Theorem and Odds

Count favourable outcomes over equally-likely total outcomes, combine events with the addition theorem P(A∪B) = P(A)+P(B)−P(A∩B), and convert freely between probability and odds — the foundation layer every later probability topic rests on.

All Probability Distribution notes MHT-CET Maths strategy

Why this matters

This is the entry point of the chapter and a near-certain 1–2 marks on every MHT-CET paper: 20 PYQs sit here (6 EASY, 12 MODERATE, 2 HARD). The bank tests three recurring shapes — combinatorial counting (tickets, balls via nCr, word-letter arrangements, dice and 'with replacement' pairs), the addition theorem (often applied to a given probability-distribution table, or as 'exactly one occurs'), and odds ↔ probability (single die, and the 'one of A, B, C must and only one can happen' setup). The classic slips are all here: subtracting P(A∩B) when you should add it, forgetting the complement in 'at least one', and reading 'odds against' backwards.

Concept 1 of 5: Classical Probability — Favourable over Total

Perform a random experiment whose outcomes are all equally likely. The sample space S is the set of every possible outcome; an event E is any subset of S. The probability of E is simply the fraction of outcomes that make E happen — how many favourable, out of how many in all.

Definition

For a finite experiment with equally-likely outcomes:

Sample space $S$ — the set of all possible outcomes; $n(S)$ is the total count.
Event $E$ — a subset of $S$ ; $n(E)$ is the number of favourable outcomes.
Classical probability: $P(E) = \dfrac{n(E)}{n(S)}$ , and always $0 \le P(E) \le 1$ .
Complement: $P(E') = 1 - P(E)$ , where $E'$ is 'E does not happen'.

The whole game is counting $n(E)$ and $n(S)$ correctly — everything below is just smarter counting or smarter combining.

Classical probability of an event

P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{favourable outcomes}}{\text{total equally-likely outcomes}},\qquad P(E') = 1 - P(E)

n(S)size of the sample space (total outcomes)
n(E)number of outcomes favourable to E
E'complement of E — the event 'E does not occur'

Diagram · event = subset of the sample space

P(E) = favourable / total = 3 / 6 = 1/2

The sample space S is all six equally likely outcomes; the event E is the subset {4, 5, 6}. For equally likely outcomes, P(E) is simply the number of favourable outcomes over the total.

Worked example

A fair die is rolled once. Find the probability of getting an even number.

Practice this conceptself-check · 4 quick reps

Probability needs EQUALLY-likely outcomes

P(E)=\dfrac{n(E)}{n(S)}

is only valid when every outcome in

S

is equally likely. Do not count 'sum = 7' as one outcome out of the eleven possible sums 2..12 — the eleven sums are NOT equally likely; count over the 36 equally-likely ordered dice pairs instead.

A probability can never exceed 1 or go below 0

If your count gives

n(E) > n(S)

or a negative value, the counting is wrong. Every valid

P(E)

satisfies

0 \le P(E) \le 1

; an answer of

\tfrac{7}{6}

or a negative fraction is an immediate signal to re-count.

Concept 2 of 5: Counting Probabilities with Combinations and Arrangements

Most probability questions are really counting questions. When order does not matter (drawing balls, choosing vertices, picking numbers) count with combinations

{}^nC_r

; when order matters (arranging letters of a word) count with permutations. Build

n(S)

and

n(E)

with the same tool, then divide.

Definition

The counting tools and standard shapes:

Selection (order irrelevant): choose $r$ from $n$ in ${}^nC_r = \dfrac{n!}{r!\,(n-r)!}$ ways. Draw-without-replacement problems use this for BOTH $n(S)$ and $n(E)$ .
Different-category draws: to draw one of each colour multiply the per-colour combinations, e.g. ${}^3C_1\,{}^4C_1\,{}^2C_1$ over ${}^9C_3$ .
Arrangements of a word: $n$ letters with a letter repeated $k$ times arrange in $\dfrac{n!}{k!}$ ways.
'Two alike together': glue them into one block — the block arrangements count the 'together' case, and 'not together' $= 1 - P(\text{together})$ .
With replacement / independent choices: each of $m$ picks from $k$ options gives $k^m$ equally-likely ordered outcomes.

Combination count and word-arrangement count

{}^nC_r = \dfrac{n!}{r!\,(n-r)!},\qquad \text{(word, one letter } k \text{ times)}\ \#\text{arrangements} = \dfrac{n!}{k!}

Visualization · two-dice sample space

Target sum: 7

favourable = 6P(sum = 7) = 6/36 = 1/6

Each of the 36 cells is one equally-likely ordered outcome (first die, second die). The highlighted anti-diagonal is the event "sum = 7"; its size over 36 is the probability. The count peaks at 6 for a sum of 7 and tapers to 1 at sums 2 and 12.

Worked example

A bag has 5 red and 3 black balls. Two balls are drawn at random without replacement. Find the probability that both are red.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q105Moderate

Example 2 · Probability Distribution · Classical Probability, Addition Theorem and Odds

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is

Use combinations when the draw order does NOT matter

Drawing 3 balls 'at random' from an urn is an unordered selection: use

{}^9C_3

for the total, not

9\cdot 8\cdot 7

. Mixing an ordered numerator with an unordered denominator (or vice-versa) is the commonest counting error — keep both the same.

Multiply combinations for 'one of each category'

For 'one red AND one blue AND one green', count each colour separately and MULTIPLY:

{}^3C_1\times{}^4C_1\times{}^2C_1 = 24

, then divide by

{}^9C_3=84

. Adding the counts instead of multiplying is wrong — the choices are made together, not as alternatives.

'With replacement' means $k^m$ ordered outcomes

Choosing

p

and

q

from

\{1,2,3,4\}

with replacement gives

4\times 4 = 16

equally-likely ordered pairs (not

{}^4C_2

). Order matters and repeats are allowed, so the sample space is

4^2

'Not together' = 1 − 'together' (glue the alike letters)

To count arrangements with two identical letters together, glue them into ONE block and arrange

(n-1)

items. Then P(not together)

= 1 - P(\text{together})

. For UNIVERSITY this is

1 - \tfrac{9!}{10!/2!} = 1 - \tfrac{2}{10} = \tfrac45

Drill 9 more on counting probabilities with combinations and arrangements

Concept 3 of 5: Mutually Exclusive and Exhaustive Events

Two events are mutually exclusive when they cannot both happen (no common outcome), and a set of events is exhaustive when together they cover the whole sample space. If events are BOTH — mutually exclusive and exhaustive — their probabilities are disjoint pieces that fill S completely, so they add up to exactly 1.

Definition

Key facts:

Mutually exclusive: $A \cap B = \varnothing$ , so $P(A \cap B) = 0$ and $P(A \cup B) = P(A) + P(B)$ .
Exhaustive: the events together are the whole space, $A \cup B \cup \cdots = S$ .
Mutually exclusive AND exhaustive: the probabilities partition S, so $P(A) + P(B) + P(C) + \cdots = 1$ .

This last identity is the workhorse: given the probabilities in terms of one unknown, set their sum to 1 and solve.

Mutually exclusive and exhaustive events sum to 1

A_1, A_2, \ldots, A_n \text{ mut. excl. \& exhaustive} \;\Longrightarrow\; \sum_{i=1}^{n} P(A_i) = 1

Diagram · exhaustive events tile the sample space

The three events leave no gap and no overlap — they exhaust S. When events are both exhaustive and mutually exclusive (a partition), their probabilities add to exactly 1: 0.5 + 0.3 + 0.2 = 1. This is the backbone of the total-probability rule.

Worked example

A, B, C are mutually exclusive and exhaustive events with

P(B) = 2P(A)

and

P(C) = 3P(A)

. Find

P(A)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q146Easy

Example 3 · Probability Distribution · Classical Probability, Addition Theorem and Odds

A,B,C

are mutually exclusive and exhaustive events of a sample space

S

such that

P(B) =\frac{3}{2}P(A)

and

P(C) =\frac{1}{2}P(B)

, then

P(A) =

Mutually exclusive is NOT the same as independent

Mutually exclusive means

P(A\cap B)=0

(they cannot co-occur). Independent means

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

. Two events with non-zero probability cannot be both — if they were mutually exclusive AND independent,

P(A)P(B)=0

, forcing one to be impossible.

Only add all the pieces to 1 when the events are BOTH exclusive AND exhaustive

The identity

\sum P(A_i)=1

needs the events to be mutually exclusive (no overlap to double-count) and exhaustive (nothing left uncovered). If they merely partition part of S, or overlap, the sum is not 1.

Concept 4 of 5: The Addition Theorem — P(A∪B), Exactly One, and Complements

For any two events, 'A or B' counts A plus B but double-counts the overlap, so subtract it once: P(A∪B) = P(A)+P(B)−P(A∩B). This single identity, plus the complement rule, answers 'A or B', 'exactly one of A, B', 'neither', and reads events straight off a probability-distribution table (an event is just a set of X-values, so its probability is the sum of those rows).

Definition

The addition theorem and its friends:

Addition theorem: $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ .
Rearranged: $P(A) + P(B) = P(A\cup B) + P(A\cap B)$ ; combined with $P(A')+P(B') = 2 - [P(A)+P(B)]$ .
Exactly one of A, B occurs: $P(A\cup B) - P(A\cap B)$ (the union minus the shared middle).
Reading a distribution table: an event like $E=\{X\text{ is prime}\}$ or $F=\{X<5\}$ is a set of X-values; $P(E)$ is the sum of $P(X=x)$ over those $x$ , and $P(E\cap F)$ sums the rows in BOTH.

Addition theorem and its derived identities

P(A\cup B) = P(A) + P(B) - P(A\cap B),\qquad P(\text{exactly one}) = P(A\cup B) - P(A\cap B)

$P(A\cup B)$ probability that A or B (or both) occurs
$P(A\cap B)$ probability that both occur (the overlap)
P(A')complement, $1 - P(A)$

Visualization · two events in the sample space

P(A): 0.5P(B): 0.4P(A∩B): 0.2

P(A∪B) = 0.7neither = 0.3exactly one = 0.5

P(A∪B) = P(A) + P(B) − P(A∩B): the lens is counted once, not twice. "Neither" is everything outside both circles, 1 − P(A∪B). The overlap is held inside its feasible range, so it never claims more than the smaller event or less than the forced minimum.

Worked example

For two events,

P(A) = 0.5

P(B) = 0.4

and

P(A\cap B) = 0.2

. Find

P(A\cup B)

and the probability that exactly one of A, B occurs.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q137Moderate

Example 4 · Probability Distribution · Classical Probability, Addition Theorem and Odds

A random variable X has the following probability distribution: | X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | | P(X=x) | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 | For the events

E=\{X \text{ is prime number}\}

F=\{X<4\}

. Then

P(E \cup F)=

ADD the intersection back, don't subtract, to get P(A)+P(B)

From

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

we get

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

— you ADD the intersection. So

P(A')+P(B')=2-[P(A\cup B)+P(A\cap B)]

. Subtracting

P(A\cap B)

here is the standard MHT-CET distractor.

'Exactly one' is the union MINUS the intersection

'Exactly one of A, B occurs' excludes the both-happen case, so it is

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

, equivalently

P(A)+P(B)-2P(A\cap B)

. Do not confuse it with

P(A\cup B)

(which INCLUDES both occurring).

On a distribution table, an event's probability is a SUM of rows

For

E=\{X\text{ is prime}\}

add

P(2)+P(3)+P(5)+P(7)

; for the intersection

E\cap F

add only the X-values in BOTH sets. Forgetting a prime (2 is prime; 1 is not) or mis-listing the overlap throws

P(E\cup F)

$P(A')+P(B') = 2 - [P(A)+P(B)]$ , not $1-[P(A)+P(B)]$

Each complement subtracts from 1, and there are TWO of them:

P(A')+P(B') = (1-P(A)) + (1-P(B)) = 2 - [P(A)+P(B)]

. Using a single 1 is a frequent slip on the complement-sum questions.

Drill 5 more on the addition theorem — p(a∪b), exactly one, and complements

Concept 5 of 5: Odds in Favour and Odds Against a Probability

Odds compare favourable outcomes to UNfavourable ones (not to the total). If an event has m favourable and n unfavourable equally-likely outcomes, the odds in favour are m : n and the odds against are n : m. Convert to probability by putting favourable over the total m + n. The 'one of A, B, C must and only one can happen' setup turns three odds into three probabilities that must sum to 1.

Definition

Odds ↔ probability conversions:

Odds in favour $= m : n$ means $P(E) = \dfrac{m}{m+n}$ ; odds against $= n : m$ means $P(E) = \dfrac{m}{m+n}$ (favourable is still the second term).
From a probability: odds in favour $= P(E) : P(E') = P(E) : (1-P(E))$ ; odds against $= P(E') : P(E)$ .
'One of A, B, C must and only one can happen' means A, B, C are mutually exclusive and exhaustive, so $P(A)+P(B)+P(C) = 1$ . Convert each given odds to a probability, use the sum to find the missing one, then convert back to odds.

Odds and probability

\text{odds in favour } m:n \;\Longleftrightarrow\; P(E) = \dfrac{m}{m+n},\qquad \text{odds against} = \dfrac{P(E')}{P(E)} = \dfrac{1-P(E)}{P(E)}

Worked example

In a single throw of a fair die, find the odds against getting a number greater than 4.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q132Moderate

Example 5 · Probability Distribution · Classical Probability, Addition Theorem and Odds

A, B, C are three events, one of which must and only one can happen. The odds in favor of A are

4:6

, the odds against B are

7:3

. Thus, odds against C are

Odds compare favourable to UNfavourable, not to the total

Odds in favour

m:n

means

m

favourable to

n

unfavourable, so

P = \dfrac{m}{m+n}

— NOT

\dfrac{m}{n}

and NOT

\dfrac{m}{\text{total already including } m}

. The denominator of the probability is the SUM of the two odds terms.

'Odds against' puts the unfavourable term first

Odds against an event are

P(E') : P(E)

— unfavourable first. For

P(E)=\tfrac13

the odds against are

\tfrac23:\tfrac13 = 2:1

, while odds IN FAVOUR are

1:2

. Reading the ratio in the wrong order is the most common odds mistake.

'Must and only one can happen' = mutually exclusive and exhaustive

When exactly one of A, B, C happens, they are mutually exclusive and exhaustive, so

P(A)+P(B)+P(C)=1

. Convert each odds to a probability, solve for the missing probability from this sum, THEN convert back to the odds the question asks for.

Drill 2 more on odds in favour and odds against a probability

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 (5)

Classical Probability — Favourable over Total
Classical probability of an event
$P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{favourable outcomes}}{\text{total equally-likely outcomes}},\qquad P(E') = 1 - P(E)$
Counting Probabilities with Combinations and Arrangements
Combination count and word-arrangement count
${}^nC_r = \dfrac{n!}{r!\,(n-r)!},\qquad \text{(word, one letter } k \text{ times)}\ \#\text{arrangements} = \dfrac{n!}{k!}$
Mutually Exclusive and Exhaustive Events
Mutually exclusive and exhaustive events sum to 1
$A_1, A_2, \ldots, A_n \text{ mut. excl. \& exhaustive} \;\Longrightarrow\; \sum_{i=1}^{n} P(A_i) = 1$
The Addition Theorem — P(A∪B), Exactly One, and Complements
Addition theorem and its derived identities
$P(A\cup B) = P(A) + P(B) - P(A\cap B),\qquad P(\text{exactly one}) = P(A\cup B) - P(A\cap B)$
Odds in Favour and Odds Against a Probability
Odds and probability
$\text{odds in favour } m:n \;\Longleftrightarrow\; P(E) = \dfrac{m}{m+n},\qquad \text{odds against} = \dfrac{P(E')}{P(E)} = \dfrac{1-P(E)}{P(E)}$

Watch out for (15)

Probability needs EQUALLY-likely outcomes
→ Classical Probability — Favourable over Total
A probability can never exceed 1 or go below 0
→ Classical Probability — Favourable over Total
Use combinations when the draw order does NOT matter
→ Counting Probabilities with Combinations and Arrangements
Multiply combinations for 'one of each category'
→ Counting Probabilities with Combinations and Arrangements
'With replacement' means $k^m$ ordered outcomes
→ Counting Probabilities with Combinations and Arrangements
'Not together' = 1 − 'together' (glue the alike letters)
→ Counting Probabilities with Combinations and Arrangements
Mutually exclusive is NOT the same as independent
→ Mutually Exclusive and Exhaustive Events
Only add all the pieces to 1 when the events are BOTH exclusive AND exhaustive
→ Mutually Exclusive and Exhaustive Events
ADD the intersection back, don't subtract, to get P(A)+P(B)
→ The Addition Theorem — P(A∪B), Exactly One, and Complements
'Exactly one' is the union MINUS the intersection
→ The Addition Theorem — P(A∪B), Exactly One, and Complements
On a distribution table, an event's probability is a SUM of rows
→ The Addition Theorem — P(A∪B), Exactly One, and Complements
$P(A')+P(B') = 2 - [P(A)+P(B)]$ , not $1-[P(A)+P(B)]$
→ The Addition Theorem — P(A∪B), Exactly One, and Complements
Odds compare favourable to UNfavourable, not to the total
→ Odds in Favour and Odds Against a Probability
'Odds against' puts the unfavourable term first
→ Odds in Favour and Odds Against a Probability
'Must and only one can happen' = mutually exclusive and exhaustive
→ Odds in Favour and Odds Against a Probability

Test yourself on Probability Distribution

20 past MHT-CET questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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