MHT-CET Maths · Probability Distribution

Discrete Random Variables, PMF and CDF

A random variable assigns a number to each outcome; its probability mass function lists P(X=x) for every value, obeys 0 ≤ P ≤ 1 and ΣP = 1, and its cumulative distribution function F(x) = P(X ≤ x) accumulates those probabilities.

All Probability Distribution notes MHT-CET Maths strategy

Why this matters

This is the technique-richest subtopic of the chapter: 30 PYQs (4 EASY, 21 MODERATE, 5 HARD). The bank tests four separate skills that all begin from ΣP = 1 — solving a linear-k table, a quadratic-in-k table (the 6k²+5k−1 and 10k²+9k−1 factorings recur almost every year), an exponential pmf, and an infinite arithmetico-geometric pmf — plus building a distribution from a coin/card/draw experiment, reading a CDF, and normalising a continuous pdf. Expectation and variance are taught separately; here the whole game is finding the constant, reading a range probability, and constructing the table correctly.

Concept 1 of 9: Discrete Random Variable and Its Probability Mass Function

A random variable X is a rule that turns each outcome of an experiment into a number (heads count, number drawn, tosses needed). It is DISCRETE when it takes isolated values — 0, 1, 2, … The probability mass function (pmf) is simply the table that lists P(X = x) against each value x.

Definition

A probability mass function $P(X=x)$ of a discrete random variable must satisfy TWO axioms:

Each probability is valid: $0 \le P(X=x_i) \le 1$ for every value $x_i$ .
The total mass is one: $\displaystyle\sum_i P(X=x_i) = 1$ — summed over ALL values the variable can take.

These two rules are the engine of the whole subtopic: every 'find the constant' question is just ΣP = 1 solved for the unknown, and every 'is this a valid distribution?' check is these two axioms.

The two pmf axioms

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

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

Worked example

P(X=x) = \dfrac{x}{6}

for

x = 1,2,3

a valid pmf?

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.

Concept 2 of 9: Finding the Constant k from a Linear Probability Table

The most direct 'find k' shape: every entry is a whole-number multiple of k (or a simple piecewise rule in k). Add them, set the total to 1, and one division gives k. Once k is known, any range probability is just adding the right cells.

Definition

When the pmf entries are linear in $k$ (e.g. $2k, k, 2k, 4k, k$ , or a piecewise rule $P=kx$ / $P=k(5-x)$ ):

Sum all entries and set the total equal to 1: $\sum P(X=x)=1$ .
Solve the resulting LINEAR equation for $k$ — a single step.
Substitute back to read off any required probability or range.

If a fixed number appears (e.g. $P(0)=0.1$ with the rest in $k$ ), include it in the sum: $0.1 + (\text{terms in }k) = 1$ .

Linear normalisation

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

Worked example

For the distribution

P(X=x)

x=1,2,3,4

with

P = k, 3k, 3k, k

, find

k

and

P(X \le 2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q109Moderate

Example 2 · Probability Distribution · Discrete Random Variables, PMF and CDF

A random variable X has the following probability distribution. Then value of k is___ and

P(3<x\leq6)

has the value. \begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline X=x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ \hline P(x) & k & 2k & 3k & 4k & 4k & 3k & 2k & k & k\\ \hline\end{array}

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.

Drill 1 more on finding the constant k from a linear probability table

Concept 3 of 9: Reading a Range Probability from the pmf Table

Once the pmf is known, a range probability is a sum of cells. The only skill is decoding the inequality correctly — which endpoints are included — and using the complement

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

when it is shorter.

Definition

Translate the inequality into exactly which values to add:

$P(X < a)$ : all values strictly below $a$ .
$P(a \le X < b)$ : from $a$ up to but NOT including $b$ .
$P(X \ge a) = 1 - P(X < a)$ — use the complement to avoid adding a long tail.
$P(X > a) = 1 - P(X \le a)$ .

The complement rule is the workhorse whenever the 'up to' side has fewer cells than the 'from' side.

Complement for a tail probability

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

Worked example

A pmf on

x=0,1,2,3,4,5

P = 0.05, 0.15, 0.30, 0.25, 0.15, 0.10

. Find

P(2 \le X < 5)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q115Moderate

Example 3 · Probability Distribution · Discrete Random Variables, PMF and CDF

The probability distribution of a discrete random variable X is

$X$	0	1	2	3	4
$P(X = x)$	$2k$	$k$	$2k$	$4k$	$k$

a = P(x< 3)

and

b = P(2 \leq X < 4)

, then

$P(X > a)$ 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.

Concept 4 of 9: Finding k from a Quadratic Probability Table

When some pmf entries carry

k^2

(like

k^2, 2k^2, 7k^2+k

), summing to 1 gives a QUADRATIC in k. Two recurring MHT-CET quadratics factor cleanly:

6k^2+5k-1=(6k-1)(k+1)

and

10k^2+9k-1=(10k-1)(k+1)

. Take the positive root (the negative one makes probabilities invalid).

Definition

The two standard quadratics and their admissible roots:

$6k^2 + 5k - 1 = 0 \Rightarrow (6k-1)(k+1)=0 \Rightarrow k = \tfrac16$ (reject $k=-1$ ).
$10k^2 + 9k - 1 = 0 \Rightarrow (10k-1)(k+1)=0 \Rightarrow k = \tfrac1{10}$ (reject $k=-1$ ).

Always reject the negative root — a negative $k$ would make some $P(X=x)$ negative. After finding k, evaluate the required range, remembering the $k^2$ terms: e.g. $P(X\ge 6)=2k^2+(7k^2+k)=9k^2+k$ .

The two recurring MHT-CET quadratics

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

Worked example

A pmf on

x=1,2,3

has

P = k, 3k, 6k^2

. Find

k

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift I · Q146Moderate

Example 4 · Probability Distribution · Discrete Random Variables, PMF and CDF

If a random variable

X

has the following probability distribution of X

$X = x$	$P(X = x)$
0	0
1	$k$
2	$2k$
3	$2k$
4	$3k$
5	$k^{2}$
6	$2k^{2}$
7	$7k^{2} + k$

Then

P(X\geq 6) =

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.

Drill 5 more on finding k from a quadratic probability table

Concept 5 of 9: Finding k for an Exponential pmf on a Finite Range

When

P(X=x) = k\cdot 2^x

over a FINITE set

x=0,1,\ldots,n

, summing to 1 needs a finite geometric series, not an infinite one. The trap is stopping the sum early or using an infinite-sum formula.

Definition

For $P(X=x)=k\,r^x$ on $x=0,1,\ldots,n$ :

Finite geometric sum: $\displaystyle\sum_{x=0}^{n} r^x = \dfrac{r^{n+1}-1}{r-1}$ (for $r\ne 1$ ).
Set $k \cdot (\text{that sum}) = 1$ and solve for $k$ .

For $r=2,\ n=4$ : $2^0+2^1+2^2+2^3+2^4 = 1+2+4+8+16 = 31$ , so $31k=1$ and $k=\tfrac1{31}$ .

Finite geometric normalisation

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

Worked example

P(X=x) = k\cdot 3^x

for

x=0,1,2

, find

k

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q129Moderate

Example 5 · Probability Distribution · Discrete Random Variables, PMF and CDF

f(x) = k\cdot 2^x

is a probability distribution of a random variable

X

that can take on the values

x = 0, 1, 2, 3, 4

, then

k

is equal to

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.

Concept 6 of 9: Finding k for an Infinite pmf k(x+1)rˣ

When

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

over ALL

x=0,1,2,\ldots

, the normalisation sum is an arithmetico-geometric progression (AGP). Its closed form

\sum_{x\ge 0}(x+1)r^x = \dfrac{1}{(1-r)^2}

turns the whole problem into one substitution.

Definition

For the infinite pmf $P(X=x)=k(x+1)r^x,\ x=0,1,2,\ldots$ :

Key AGP sum (memorise): $\displaystyle\sum_{x=0}^{\infty}(x+1)r^{x} = \frac{1}{(1-r)^{2}}$ for $|r|<1$ .
Set $k\cdot\dfrac{1}{(1-r)^2} = 1$ , so $k = (1-r)^2$ .
For $r=\tfrac12$ : $k=(1-\tfrac12)^2=\tfrac14$ . For $r=\tfrac15$ : $k=(1-\tfrac15)^2=(\tfrac45)^2=\tfrac{16}{25}$ .

Then any $P(X=x)$ follows by direct substitution — e.g. $P(X=0)=k(1)(r^0)=k$ .

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}

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

Worked example

P(X=x) = k(x+1)\left(\tfrac13\right)^x

for

x=0,1,2,\ldots

. Find

k

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift II · Q145Hard

Example 6 · Probability Distribution · Discrete Random Variables, PMF and CDF

A random variable X takes the values 0, 1, 2, 3,

\_\_\_\_

with probability

P(X =x) = k(x+ 1)\left( \frac{1}{5} \right)^{x}

, where k is a constant.Then

P(X = 0)

$\sum(x+1)r^x = \dfrac{1}{(1-r)^2}$ , 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.

Drill 4 more on finding k for an infinite pmf k(x+1)rˣ

Concept 7 of 9: Constructing a Probability Distribution from an Experiment

Instead of a table with an unknown k, you are handed an experiment (toss coins, draw cards, throw a die) and must BUILD the pmf. List the possible values of X, compute each probability from the experiment, and check the total is 1.

Definition

Identify the values $X$ takes, then find each $P(X=x)$ by the right counting rule:

With-replacement draws (independent trials): binomial — $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$ . Two cards with replacement, jack has $p=\tfrac4{52}=\tfrac1{13}$ : $P(0)=(\tfrac{12}{13})^2=\tfrac{144}{169},\ P(1)=2\cdot\tfrac1{13}\cdot\tfrac{12}{13}=\tfrac{24}{169},\ P(2)=\tfrac1{169}$ .
Without-replacement draws: hypergeometric — $P(X=r)=\dfrac{\binom{D}{r}\binom{N-D}{n-r}}{\binom{N}{n}}$ . 4 defective + 16 good, draw 3: $P(0)=\tfrac{\binom{16}{3}}{\binom{20}{3}}=\tfrac{28}{57}$ , and so on.
Counting outcomes (equally likely): three fair coins, $X=$ heads: $P(X=k)=\binom{3}{k}(\tfrac12)^3$ , giving $\tfrac18,\tfrac38,\tfrac38,\tfrac18$ .
Sequential 'until' experiments: multiply along each branch — a coin tossed until a head or 4 tails gives $\tfrac12,\tfrac14,\tfrac18,\tfrac18$ for $X=1,2,3,4$ (the last cell pools TTTH and TTTT).

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}}

Worked example

Two fair coins are tossed. Let

X

be the number of heads. Build the probability distribution.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift II · Q148Moderate

Example 7 · Probability Distribution · Discrete Random Variables, PMF and CDF

Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is

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

Drill 5 more on constructing a probability distribution from an experiment

Concept 8 of 9: Cumulative Distribution Function and pmf ↔ CDF Differencing

The CDF

F(x)=P(X\le x)

accumulates the pmf from the left. Because it is a running total, you recover any single mass by DIFFERENCING consecutive CDF values:

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

. This is how you answer 'find P(X=a)' or a probability ratio when only the CDF is given.

Definition

For a discrete random variable with values $x_1<x_2<\cdots$ :

Definition: $F(x)=P(X\le x)=\displaystyle\sum_{x_i \le x}P(X=x_i)$ ; it is non-decreasing and reaches 1 at the top value.
Recover the pmf (differencing): $P(X=x_i)=F(x_i)-F(x_{i-1})$ , with $P(X=x_1)=F(x_1)$ .
Tail from the CDF: $P(X>a)=1-F(a)$ ; $P(X\le a)=F(a)$ .

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})

Worked example

A discrete CDF has

F(1)=0.2,\ F(2)=0.5,\ F(3)=1

at values

x=1,2,3

. Find

P(X=2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q120Easy

Example 8 · Probability Distribution · Discrete Random Variables, PMF and CDF

The c.d.f. of a discrete random variable

X

$X$	$F(X = x)$
$-3$	0.1
$-1$	0.3
0	0.5
1	0.65
3	0.75
5	0.85
7	0.90
9	1

Then

\frac{P\lbrack X = - 3\rbrack}{P\lbrack X < 0\rbrack}=

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.

Drill 1 more on cumulative distribution function and pmf ↔ cdf differencing

Concept 9 of 9: Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)

For a continuous variable, probability lives in a density

f(x)

, not at single points. Every discrete 'sum' becomes an INTEGRAL: normalise with

\int f\,dx=1

to find a constant, get the CDF by

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

, and read an interval probability as

\int_a^b f\,dx

Definition

The continuous analogues of the pmf rules:

Normalisation (find the constant): $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$ , integrated over the support only.
CDF: $F(x)=\displaystyle\int_{-\infty}^{x} f(t)\,dt$ ; $F$ rises from 0 to 1, and $F'(x)=f(x)$ .
Interval probability: $P(a<X<b)=\displaystyle\int_a^b f(x)\,dx = F(b)-F(a)$ . For continuous $X$ , $<$ and $\le$ give the same value.
Two-condition pdf: if the pdf has TWO unknowns, use $\int f=1$ AND a given point value (like $f(2)=2$ ) to solve the pair.

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, $F'=f$

Worked example

A pdf is

f(x)=kx

for

0<x<2

(and 0 otherwise). Find

k

and

P(X<1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q134Moderate

Example 9 · Probability Distribution · Discrete Random Variables, PMF and CDF

The following is p.d.f. of continuous random variable X

f(x) =\left\{ \begin{matrix} \frac{x}{8} & ,\text{~if~}0 <x< 4 \\ 0 & ,\text{~otherwise~} \end{matrix} \right.

.Then

F(0.5),F(1.7)

and

F(5)

is respectively

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

Drill 6 more on continuous random variables — pdf, normalisation, cdf and p(a < x < b)

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 (9)

Discrete Random Variable and Its Probability Mass Function
The two pmf axioms
$0 \le P(X=x_i) \le 1, \qquad \sum_{i} P(X=x_i) = 1$
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
$\sum_{x=0}^{\infty}(x+1)r^{x} = \frac{1}{(1-r)^{2}} \;\Longrightarrow\; k = (1-r)^{2}$
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
$\int_{-\infty}^{\infty} f(x)\,dx = 1,\quad F(x)=\int_{-\infty}^{x} f(t)\,dt,\quad P(a<X<b)=\int_a^b f(x)\,dx$

Watch out for (22)

A pmf must sum to exactly 1, over ALL values
→ Discrete Random Variable and Its Probability Mass Function
Probabilities can never exceed 1 or go negative
→ Discrete Random Variable and Its Probability Mass Function
Include every row — even a fixed number — in ΣP = 1
→ Finding the Constant k from a Linear Probability Table
Match the range operator exactly: strict vs inclusive
→ Finding the Constant k from a Linear Probability Table
$P(X > a)$ and $P(X \ge a)$ are not the same
→ Reading a Range Probability from the pmf Table
Use the complement only when it has fewer terms
→ Reading a Range Probability from the pmf Table
Reject the negative root of the k-quadratic
→ Finding k from a Quadratic Probability Table
Don't drop the $k^2$ terms when evaluating a range
→ Finding k from a Quadratic Probability Table
$P(X\ge 2)=1-P(X=1)$ , not $1-P(X\le 2)$
→ Finding k from a Quadratic Probability Table
Sum a FINITE range fully — don't stop early
→ Finding k for an Exponential pmf on a Finite Range
A finite exponential pmf is NOT the infinite geometric sum
→ Finding k for an Exponential pmf on a Finite Range
$\sum(x+1)r^x = \dfrac{1}{(1-r)^2}$ , NOT $\dfrac{1}{1-r}$
→ Finding k for an Infinite pmf k(x+1)rˣ
The range is INFINITE here — use $|r|<1$
→ Finding k for an Infinite pmf k(x+1)rˣ
With replacement is binomial; without replacement is hypergeometric
→ Constructing a Probability Distribution from an Experiment
$P(X=1)$ counts BOTH orders — include the factor of 2
→ Constructing a Probability Distribution from an Experiment
In a bounded 'until' experiment, the last cell POOLS two branches
→ Constructing a Probability Distribution from an Experiment
Read $P(X=x_i)$ as a CDF DIFFERENCE, not the CDF value
→ Cumulative Distribution Function and pmf ↔ CDF Differencing
$P(X\le 0)=F(0)$ ; $P(X>0)=1-F(0)$
→ Cumulative Distribution Function and pmf ↔ CDF Differencing
Integrate over the SUPPORT only
→ Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)
A two-unknown pdf needs TWO equations
→ Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)
The CDF is the running integral, and $F'(x)=f(x)$
→ Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)
$P(|X|<2)$ is a symmetric integral $\int_{-2}^{2}$
→ Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)

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.

Take the Probability Distribution chapter test Drill all 30 questions

Related notes

NDA Maths — Binomial Distribution notes →