MHT-CET Maths · Formula sheet

Probability Distribution formulas

28 formulas and 77 common traps for MHT-CET Maths Probability Distribution, grouped by subtopic.

Classical Probability, Addition Theorem and Odds

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)

$n(S)$ 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'

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)

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

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)}

Common traps

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.

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

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.

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.

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.

Conditional Probability, Independence and Bayes' Theorem

Learn this subtopic in the notes

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

$P(A\cap B)$ 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

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)

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

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

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

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

$P(H_k)$ 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

Common traps

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.

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.

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.

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.

'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.

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}

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.

Discrete Random Variables, PMF and CDF

Learn this subtopic in the notes

Discrete Random Variable and Its Probability Mass Function

The two pmf axioms

0 \le P(X=x_i) \le 1, \qquad \sum_{i} P(X=x_i) = 1

$X$ the discrete random variable
$x_i$ each value X can take
$P(X=x_i)$ the probability mass at that value

Finding the Constant k from a Linear Probability Table

Linear normalisation

\sum_x P(X=x) = 1 \;\Longrightarrow\; (\text{multiple of } k) = 1 \;\Longrightarrow\; k = \frac{1}{\text{that multiple}}

Reading a Range Probability from the pmf Table

Complement for a tail probability

P(X \ge a) = 1 - P(X < a) = 1 - \sum_{x < a} P(X=x)

Finding k from a Quadratic Probability Table

The two recurring MHT-CET quadratics

6k^2+5k-1=(6k-1)(k+1),\qquad 10k^2+9k-1=(10k-1)(k+1)

Finding k for an Exponential pmf on a Finite Range

Finite geometric normalisation

\sum_{x=0}^{n} k\,r^{x} = k\cdot\frac{r^{\,n+1}-1}{r-1} = 1

Finding k for an Infinite pmf k(x+1)rˣ

AGP normalisation for k(x+1)rˣ

\sum_{x=0}^{\infty}(x+1)r^{x} = \frac{1}{(1-r)^{2}} \;\Longrightarrow\; k = (1-r)^{2}

$r$ the common ratio, $|r|<1$
$(x+1)$ the arithmetic factor
$k$ the normalising constant $(1-r)^2$

Constructing a Probability Distribution from an Experiment

Binomial and hypergeometric building blocks

P(X=r)=\binom{n}{r}p^{r}(1-p)^{n-r}, \qquad P(X=r)=\frac{\binom{D}{r}\binom{N-D}{n-r}}{\binom{N}{n}}

Cumulative Distribution Function and pmf ↔ CDF Differencing

CDF definition and differencing

F(x)=P(X\le x)=\sum_{x_i\le x}P(X=x_i), \qquad P(X=x_i)=F(x_i)-F(x_{i-1})

Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)

Continuous normalisation, CDF, interval

\int_{-\infty}^{\infty} f(x)\,dx = 1,\quad F(x)=\int_{-\infty}^{x} f(t)\,dt,\quad P(a<X<b)=\int_a^b f(x)\,dx

$f(x)$ probability density function
$F(x)$ cumulative distribution function, $F'=f$

Common traps

A pmf must sum to exactly 1, over ALL values

The single most common error is summing over only some of the listed values (forgetting the last row, or a

P=0

row). Every value the variable can take contributes to

\sum P(X=x)=1

. If a row shows

P=0

it still belongs in the table — it just contributes nothing to the sum.

Probabilities can never exceed 1 or go negative

When you solve

\sum P = 1

for a constant and get two roots (say from a quadratic), reject any root that makes some

P(X=x)

negative or bigger than 1. Only the root keeping every entry in

[0,1]

is admissible.

Include every row — even a fixed number — in ΣP = 1

P(0)=0.1,\ P(1)=k,\dots

the fixed

0.1

is part of the total:

0.1 + 6k = 1

, not

6k = 1

. Omitting the constant term gives

k = \tfrac16

instead of the correct

0.15

Match the range operator exactly: strict vs inclusive

P(3 < X \le 6)

means

P(4)+P(5)+P(6)

— it EXCLUDES

x=3

and INCLUDES

x=6

. Reading it as

P(3)+P(4)+P(5)+P(6)

or dropping

x=6

is the classic off-by-one range slip.

$P(X > a)$ and $P(X \ge a)$ are not the same

P(X>2)

excludes

x=2

;

P(X\ge 2)

includes it. For entries

k^2,2k,k,2k,5k^2

P(X>2)=k+2k+5k^2

but

P(X\ge 2)=1-k^2

. Mixing the two is the top range error in this subtopic.

Use the complement only when it has fewer terms

P(X\ge a)=1-P(X<a)

is a shortcut, not a rule to apply blindly. On a table running

0\ldots 7

, the tail

P(X\ge 6)=\{6,7\}

is SHORT — add it directly. Switch to the complement only when the 'below' side has fewer cells than the tail; count both sides before choosing.

Reject the negative root of the k-quadratic

Both

6k^2+5k-1=0

and

10k^2+9k-1=0

have

k=-1

as a root. A negative

k

forces negative probabilities, so it is inadmissible — always keep the positive root (

\tfrac16

\tfrac1{10}

Don't drop the $k^2$ terms when evaluating a range

For

P(X\ge 6)=2k^2+(7k^2+k)=9k^2+k

with

k=\tfrac1{10}

, you must square:

9k^2=\tfrac{9}{100}

, giving

\tfrac{19}{100}

. Treating

k^2

k

gives the wrong tail probability.

$P(X\ge 2)=1-P(X=1)$ , not $1-P(X\le 2)$

For entries

k^2,2k,k,2k,5k^2

with

k=\tfrac16

P(X\ge 2)=1-P(1)=1-k^2=\tfrac{35}{36}

. Subtracting

P(X\le 2)

instead would wrongly remove

x=2

, which the '

\ge 2

' event includes.

Sum a FINITE range fully — don't stop early

For

P=k\cdot 2^x

x=0,1,2,3,4

, the sum is

1+2+4+8+16=31

, so

k=\tfrac1{31}

. Stopping at

x=3

(

=15

) gives

k=\tfrac1{15}

— a designed distractor. Count every listed value.

A finite exponential pmf is NOT the infinite geometric sum

\sum_{x=0}^{4}2^x = 31

, not

\dfrac{1}{1-2}

(which diverges anyway for

r>1

). Use the finite formula

\dfrac{r^{n+1}-1}{r-1}

; the infinite

\dfrac{1}{1-r}

only applies when

|r|<1

over an infinite range.

$\sum(x+1)r^x = \dfrac{1}{(1-r)^2}$ , NOT $\dfrac{1}{1-r}$

The arithmetic factor

(x+1)

squares the denominator. For

r=\tfrac15

\sum(x+1)r^x = \dfrac{1}{(4/5)^2}=\dfrac{25}{16}

, giving

k=\tfrac{16}{25}

. Using the plain geometric

\dfrac{1}{1-r}=\tfrac54

gives the wrong

k=\tfrac45

The range is INFINITE here — use $|r|<1$

P(X=x)=k(x+1)r^x

runs over all

x=0,1,2,\ldots

, so the infinite AGP sum applies (it converges because

r=\tfrac12

\tfrac15

satisfies

|r|<1

). Don't confuse it with the finite

k\cdot 2^x

case, whose ratio exceeds 1.

With replacement is binomial; without replacement is hypergeometric

'Successively WITH replacement' means each draw is independent with fixed

p

— use

\binom{n}{r}p^r(1-p)^{n-r}

. 'Drawn from the lot' WITHOUT replacement changes the pool each draw — use the

\dfrac{\binom{D}{r}\binom{N-D}{n-r}}{\binom{N}{n}}

ratio. Choosing the wrong model is the top construction error.

$P(X=1)$ counts BOTH orders — include the factor of 2

For two independent draws,

P(\text{exactly one success}) = 2\,p(1-p)

(success-then-fail OR fail-then-success), e.g.

2\cdot\tfrac4{52}\cdot\tfrac{48}{52}=\tfrac{24}{169}

. Forgetting the 2 halves the middle probability.

In a bounded 'until' experiment, the last cell POOLS two branches

A coin tossed until a head OR 4 tails:

P(X=4)=P(\text{TTTH})+P(\text{TTTT})=\tfrac1{16}+\tfrac1{16}=\tfrac18

. The forced stop at 4 means the experiment ends whether the 4th toss is H or T, so both outcomes count toward

X=4

Read $P(X=x_i)$ as a CDF DIFFERENCE, not the CDF value

Except for the smallest value,

P(X=x_i)=F(x_i)-F(x_{i-1})

, not

F(x_i)

. Only

P(X=x_1)=F(x_1)

(nothing accumulated before it). Reading

P(X=0)=F(0)=0.5

directly (when earlier values exist) double-counts.

$P(X\le 0)=F(0)$ ; $P(X>0)=1-F(0)$

From a CDF,

P(X\le 0)

is exactly

F(0)

, and

P(X>0)=1-F(0)

. If

F(0)=0.5

the ratio

\dfrac{P(X\le0)}{P(X>0)}=\dfrac{0.5}{0.5}=1

. Watch

\le

<

P(X<0)

excludes the mass exactly at 0.

Integrate over the SUPPORT only

For a continuous pdf

f(x)=kx^2

0\le x\le 6

, normalise

\int_0^6 kx^2\,dx = \left[\tfrac{kx^3}{3}\right]_0^6 = 72k = 1

, giving

k=\tfrac1{72}

— integrate only where

f\ne 0

(

f=0

outside

[0,6]

). Applying

\int_{-\infty}^{\infty}

blindly, or a discrete SUM

\sum_{x=0}^{6} kx^2

(which would give

91k=1

), is the standard continuous-vs-discrete mix-up.

A two-unknown pdf needs TWO equations

f(x)=\tfrac{ax^2}{2}+bx

[1,3]

has unknowns

a,b

, one equation is

\int_1^3 f\,dx=1

; the second is a given value like

f(2)=2\Rightarrow 2a+2b=2

. Solve the pair — normalisation alone is not enough.

The CDF is the running integral, and $F'(x)=f(x)$

F(x)=\int_{-\infty}^{x} f

. For

f(x)=12x^2(1-x)

(0,1)

F(x)=\int_0^x 12t^2(1-t)\,dt = 4x^3-3x^4

. Differentiating back must return

f

— a quick check that catches sign or coefficient errors.

$P(|X|<2)$ is a symmetric integral $\int_{-2}^{2}$

For

f(x)=\tfrac{x^2}{18}

(-3,3)

P(|X|<2)=\int_{-2}^{2}\tfrac{x^2}{18}\,dx=\tfrac1{18}\cdot\tfrac{16}{3}=\tfrac{8}{27}

. Integrate from

-2

2

, not

0

2

—

|X|<2

means

-2<X<2

Expectation, Variance and Standard Deviation

Learn this subtopic in the notes

Expectation as the Long-Run Average

Expected value of a discrete random variable

E(X) = \mu = \sum_{i} x_i\,P(X=x_i)

$x_i$ the values X can take
$P(X=x_i)$ the probability of each value (the pmf)
$\mu$ the mean / expected value — a weighted average, not always an attainable value

Computing the Mean E(X) from a Probability Distribution

Mean of a listed distribution

E(X) = \sum_{i} x_i\,P(X=x_i),\qquad E(X+Y) = E(X)+E(Y)

Variance and Standard Deviation: Var(X) = E(X²) − [E(X)]²

Variance and standard deviation

\mathrm{Var}(X) = E(X^2) - [E(X)]^2,\qquad E(X^2) = \sum x^2 P(x),\qquad \sigma = \sqrt{\mathrm{Var}(X)}

$E(X^2)$ average of the SQUARES: $\sum x^2 P(x)$
$[E(X)]^2$ the SQUARE of the mean — a different, smaller number in general
$\sigma$ standard deviation = $\sqrt{\text{Var}}$ , in the same units as X

Expected Winnings of a Game: E(g(X)) = Σ g(x)·P(x)

Expected value of a payoff (function of X)

E\big(g(X)\big) = \sum g(x)\,P(x)

$g(x)$ the cash payoff for outcome x — positive for a gain, NEGATIVE for a loss
$P(x)$ the probability of that outcome

Uniform Distribution on 1 to n: E(X) = (n+1)/2, Var(X) = (n²−1)/12

Discrete uniform on 1..n

E(X) = \dfrac{n+1}{2},\qquad \mathrm{Var}(X) = \dfrac{n^2-1}{12},\qquad \dfrac{\mathrm{Var}(X)}{E(X)} = \dfrac{n-1}{6}

$n$ the number of equally-likely integer values 1, 2, …, n

Finding Unknown Probabilities from the Mean and ΣP = 1

The determining system

\sum P(x) = 1 \quad\text{and}\quad E(X) = \sum x\,P(x) = \mu_{\text{given}}

Expectation of Standard Distributions: Geometric and Hypergeometric

Means of named distributions

\text{Geometric: } E(X) = \dfrac{1}{p},\qquad \text{Hypergeometric: } E(X) = \dfrac{nK}{N}

$p$ success probability of one trial (geometric)
$N$ total items in the lot (hypergeometric)
$K$ number of successes in the lot
$n$ number of items drawn without replacement

Common traps

The mean is a weighted average, not a plain average of the values

E(X) = \sum x_i p_i

— each value is weighted by its OWN probability. Averaging the values while ignoring the probabilities (e.g. reporting

\tfrac{1+2+3+4}{4}

for a non-uniform pmf) is the most common beginner slip.

Always verify $\sum p_i = 1$ before computing anything

If a table contains an unknown like

k

2k

, the normalization

\sum p_i = 1

is the equation that determines it. Compute the mean only after fixing every probability, or the whole answer is off.

Use linearity for the sum on dice, don't build all 36 outcomes

The expected sum on two dice is

E(X_1)+E(X_2) = 3.5+3.5 = 7

. You never need the full 2..12 distribution — but if you build it,

\sum x P(x) = \tfrac{252}{36} = 7

confirms the same answer.

Solve for the unknown probability before taking the mean

A row like

P(X=5) = 2k

is meaningless until

\sum P = 1

fixes

k

. Find

k

first, substitute back, THEN compute

E(X) = \sum xP(x)

E(X²) is NOT [E(X)]²

E(X^2) = \sum x^2 P(x)

averages the squares;

[E(X)]^2

squares the average. They are equal only when the variance is zero. The whole variance formula is the gap between them:

\mathrm{Var}(X) = E(X^2) - [E(X)]^2

Convert a CDF to a pmf before computing an expectation

If the question gives

F(x)

(cumulative), you must first difference it:

P(x_i) = F(x_i) - F(x_{i-1})

. Plugging the CDF values straight into

\sum x^2 F(x)

is a classic wrong answer.

Standard deviation vs variance — don't hand back the wrong one

= \sqrt{\mathrm{Var}}

. If the variance works out to 1, the SD is also 1; if the variance is 2, the SD is

\sqrt2

, not 2. Options are deliberately built to punish reporting the variance when the SD is asked (and vice versa).

Variance is never negative

\mathrm{Var}(X) = E(X^2) - [E(X)]^2 \ge 0

always. If you get a negative variance, you have squared the mean wrong or mixed up E(X²) and [E(X)]² — recheck before choosing an option.

A loss is a negative payoff — carry the minus sign

In a game that pays ₹150 on all-heads/all-tails and requires paying ₹50 otherwise, the second term is

-50

, not

+50

E = \tfrac14(150) + \tfrac34(-50) = 37.5 - 37.5 = 0

. Dropping the minus turns a fair game into a phantom win.

Get the all-heads/all-tails probability right

With 3 coins there are 8 equally-likely outcomes. All heads OR all tails is 2 of them, so

P = \tfrac{2}{8} = \tfrac14

; exactly one or two heads is the remaining 6, so

P = \tfrac{6}{8} = \tfrac34

. Using

\tfrac18

(only all heads) mis-weights the whole expectation.

Variance of a winning amount is still E(X²) − [E(X)]²

Treat the cash winnings as the values of X, list them with their probabilities, and apply the ordinary variance formula. Don't confuse the variance of the payoff with the expected payoff itself.

Memorise both uniform formulas — mean (n+1)/2 AND variance (n²−1)/12

The variance is

\dfrac{n^2-1}{12}

, not

\dfrac{n^2-1}{6}

\dfrac{n^2+1}{12}

— those are the standard distractors. And E(X) is

\dfrac{n+1}{2}

, the midpoint, not

\dfrac{n}{2}

Cancel the (n+1) factor for 'find n' questions

Equations like

\dfrac{n^2-1}{12} = \dfrac{n+1}{2}

collapse instantly once you write

n^2-1 = (n-1)(n+1)

and cancel the shared

(n+1)

. This avoids solving a full quadratic.

P(x) = 2x/[n(n+1)] is NOT the uniform distribution

When probability grows with x, you cannot use

E = (n+1)/2

. Compute

E(X) = \sum x\cdot\dfrac{2x}{n(n+1)}

using

\sum x^2 = \tfrac{n(n+1)(2n+1)}{6}

, which gives

\dfrac{2n+1}{3}

Watch the sign in the E(X) equation

For X = 30, 10, −10 with P = 1/5, A, B and E(X) = 4:

6 + 10A - 10B = 4

, so

A - B = -\tfrac15

(a MINUS). A sign slip here flips A and B and gives the wrong product AB.

Use the extra stated relation as your second equation

A clue like

P(X=3) = 2P(X=1)

is not decoration — it is one of the two equations you need. Combined with the mean and

\sum P = 1

, it pins every probability down.

For range problems, apply non-negativity to EVERY row

When probabilities like

\tfrac{1+p}{5}, \tfrac{2-2p}{5}, \tfrac{2-p}{5}, \tfrac{2p}{5}

contain p, each must lie in

[0,1]

. The tightest of those inequalities gives p's actual range; the mean's min/max occur at the ends of that range, not by guessing.

'Until success' means geometric, mean = 1/p

The expected number of trials to the first success is

\dfrac1p

. Rolling an n-faced die until a number

< n

appears has

p = \dfrac{n-1}{n}

, so the mean is

\dfrac{n}{n-1}

— don't confuse this with the value of a single roll.

Hypergeometric mean is nK/N — no replacement needed for the mean

Even though sampling is without replacement, the expected count of successes is simply

\dfrac{nK}{N}

(the same as the with-replacement binomial mean). You do NOT need to build the full pmf just to get the mean.

For E(X²) build the small combination pmf first

A question asking for

2E(X)+3E(X^2)

(e.g. queens in 2 cards) needs both moments: build

P(X=0),P(X=1),P(X=2)

from

\binom{}{}

-ratios, then compute

E(X)=\sum kP

and

E(X^2)=\sum k^2 P

with the SAME probabilities.

Probability Distribution formulas

Classical Probability, Addition Theorem and Odds

Classical Probability — Favourable over Total

Counting Probabilities with Combinations and Arrangements

Mutually Exclusive and Exhaustive Events

The Addition Theorem — P(A∪B), Exactly One, and Complements

Odds in Favour and Odds Against a Probability

Common traps

Conditional Probability, Independence and Bayes' Theorem

Conditional Probability — Restricting the Sample Space

Multiplication Rule and Sequential Draws Without Replacement

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

Independence and Event Algebra with Unions

At Least One and Exactly One for Independent Trials

Bayes' Theorem — Reversing the Conditioning

Common traps

Discrete Random Variables, PMF and CDF

Discrete Random Variable and Its Probability Mass Function

Finding the Constant k from a Linear Probability Table

Reading a Range Probability from the pmf Table

Finding k from a Quadratic Probability Table

Finding k for an Exponential pmf on a Finite Range

Finding k for an Infinite pmf k(x+1)rˣ

Constructing a Probability Distribution from an Experiment

Cumulative Distribution Function and pmf ↔ CDF Differencing

Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)

Common traps

Expectation, Variance and Standard Deviation

Expectation as the Long-Run Average

Computing the Mean E(X) from a Probability Distribution

Variance and Standard Deviation: Var(X) = E(X²) − [E(X)]²

Expected Winnings of a Game: E(g(X)) = Σ g(x)·P(x)

Uniform Distribution on 1 to n: E(X) = (n+1)/2, Var(X) = (n²−1)/12

Finding Unknown Probabilities from the Mean and ΣP = 1

Expectation of Standard Distributions: Geometric and Hypergeometric

Common traps

More MHT-CET Maths formula sheets