Mathematics 9709 · Probability & Statistics 1

Building a probability distribution

A discrete random variable is a number decided by chance.
Examples: a die score, the number of heads, or how many red counters are drawn.
A probability distribution lists every value the variable can take.
Under each value, it gives the probability of that value.
The whole table must add to $1$ .

Build the table; make it sum to 1

Use $X$ for the random variable and $x$ for one value it can take.
List all possible values of $X$ .
Find each $P(X=x)$ and put it directly under the matching value.
Check $\sum p=1$ .
If a probability contains an unknown, the $\sum p=1$ equation usually solves it.
After solving, reject any value that makes a probability negative or bigger than 1.

Values across the top, probabilities under them, summing to 1. That Σp = 1 line both checks the table and solves an unknown.

Two coins, X = number of heads. Each value's height is its probability, and the three heights add to 1.

Worked example

Start here: read a value straight off a finished table

For $x=1,2,3,4$ the probabilities are $0.1,\ 0.2,\ 0.4,\ 0.3$ . The table is complete (they sum to 1), so just add the right cells: $P(X\ge3)=0.4+0.3$ .

Answer

P(X\ge3)=0.7

Worked example

Next rung: use Σp = 1 to find a constant

For $x=1,2,3,4$ the probabilities are $0.1,\ k,\ 2k,\ 0.3$ . The one missing fact is $\sum p=1$ : $0.1+k+2k+0.3=1\Rightarrow 3k=0.6$ , so $k=0.2$ (all probabilities valid). Then $P(X\ge3)=2k+0.3$ .

Answer

k=0.2,\quad P(X\ge3)=0.7

Build the table from the outcomes themselves

Sometimes the probabilities are not handed to you.
List the equally likely outcomes first.
Map each outcome to its value of $x$ .
If several outcomes give the same $x$ , pool them into one probability.

Worked example

Two spinners: build from the grid

A fair 4-sided spinner (labelled 1, 2, 3, 4) is spun twice and the scores are added to give $X$ . A grid of the $4\times4=16$ equally likely outcomes shows $X\in\{2,3,4,5,6,7,8\}$ .

Count how many of the 16 cells give each total: $1,2,3,4,3,2,1$ for $x=2,\dots,8$ .
Each probability is that count over 16:

$P(X{=}x)=\tfrac{1}{16},\tfrac{2}{16},\tfrac{3}{16},\tfrac{4}{16},\tfrac{3}{16},\tfrac{2}{16},\tfrac{1}{16}$

Answer

\tfrac{1+2+3+4+3+2+1}{16}=1\ \checkmark

The total 5 sits in the middle because it has the most ways to happen, the classic triangular shape of a sum of two dice or spinners.

Worked example

Selection without replacement: count with combinations

Eight youths, one man and one woman want to board a bus with room for three. The driver picks 3 of the 10 at random; let $Y$ be the number of youths chosen. Selecting (order does not matter) means combinations, over ${}^{10}C_3=120$ .

Only two people are non-youths, so $Y\in\{1,2,3\}$ (you cannot avoid youths).
Choose the youths and the rest separately, then divide by 120:

$P(Y{=}1)=\tfrac{{}^8C_1\cdot{}^2C_2}{120}=\tfrac{1}{15},\quad P(Y{=}2)=\tfrac{{}^8C_2\cdot{}^2C_1}{120}=\tfrac{7}{15},\quad P(Y{=}3)=\tfrac{{}^8C_3}{120}=\tfrac{7}{15}$

Answer

\tfrac{1}{15}+\tfrac{7}{15}+\tfrac{7}{15}=1\ \checkmark

Worked example

Another selection table (smaller numbers)

A box has 3 red, 5 blue (8 total); draw 2 without replacement. With $R=$ reds drawn and ${}^8C_2=28$ :

$P(R{=}0)=\tfrac{{}^5C_2}{28}=\tfrac{10}{28},\ \ P(R{=}1)=\tfrac{3\cdot5}{28}=\tfrac{15}{28},\ \ P(R{=}2)=\tfrac{3}{28}$

Answer

\tfrac{10}{28}+\tfrac{15}{28}+\tfrac{3}{28}=1\ \checkmark

When a formula gives each probability

Worked example

Formula table: substitute, then Σp = 1 for the constant

$X\in\{1,2,3\}$ , $P(X=x)=c\,x^2$ . $c(1+4+9)=1\Rightarrow c=\tfrac{1}{14}$ , so the table is $\tfrac{1}{14},\tfrac{4}{14},\tfrac{9}{14}$ — give the exact fractions asked. (9709/52/M/J/23 Q1)

Answer

c=\tfrac{1}{14}

Worked example

Bigger formula table (exact fractions)

$Q\in\{3,4,5,6\}$ , $P(Q=q)=c\,q^2$ . Sum the squares: $c(9+16+25+36)=1\Rightarrow 86c=1$ .

$c=\tfrac{1}{86}$ ; the table is $\tfrac{9}{86},\tfrac{16}{86},\tfrac{25}{86},\tfrac{36}{86}$ .
Then $P(Q>4)=\tfrac{25}{86}+\tfrac{36}{86}=\tfrac{61}{86}$ .

Answer

c=\tfrac{1}{86},\quad P(Q>4)=\tfrac{61}{86}

A common denominator (here 86) keeps every cell exact, which is what a “leave in fraction form” question wants.

When the constant comes out of a quadratic: reject invalid roots

Worked example

Two roots, one is impossible

$V\in\{2,3,4,5,6\}$ with probabilities $0.05,\ c^2,\ c+0.1,\ 2c+0.05,\ 0.16$ . Find $c$ , then $P(V>4)$ .

$\sum p=1$ : $0.05+c^2+c+0.1+2c+0.05+0.16=1$ , so $c^2+3c-0.64=0$ .
Factorise: $(c-0.2)(c+3.2)=0\Rightarrow c=0.2$ or $c=-3.2$ .
Reject $c=-3.2$ : it makes $P(V{=}4)=c+0.1=-3.1<0$ . So $c=0.2$ .
$P(V>4)=P(V{=}5)+P(V{=}6)=(2(0.2)+0.05)+0.16$ .

Answer

c=0.2,\quad P(V>4)=0.61

List every value $X$ can take.
Each $P(X=x)$ — count outcomes (combinations if selecting), or plug $x$ into the formula.
Lay them under their $x$ ; pool repeated x values.
Write the $\sum p=1$ line (it scores); it also solves a single unknown.

The question gives	Each P(X=x) by	Watch
a formula f(x)	substitute x; Σp=1 fixes the constant	give exact fractions
'select n at random'	combinations ÷ ⁿCᵣ	without replacement → denominators shrink
a verbal die/coin rule	list outcomes, map to x, pool equal x's	don't drop a branch
table, 1 unknown	Σp = 1	reject p<0 or p>1
table, 2 unknowns	Σp = 1 AND E(X)	one equation can't fix two

Examiner note

Write the

\sum p=1

line explicitly (a scored step) and draw a two-row table with each probability directly under its x. A “show that” needs the calculation, not the bare value. (9709/52/M/J/24 Q5)

Common mistake

Always write the

\sum p=1

check line (it scores a mark). If solving for an unknown gives two roots, reject the one that makes a probability negative or above 1, and say why. For selection tables, count with combinations (order doesn't matter).

Common mistake

In build-type tables several outcomes can give the same

x

(a coin and die rule sending many results to one total) — pool them, then re-check

\sum p=1

Two unknowns? You need two facts

Probs $0.3,a,b,0.15$ on $X\in\{0,1,2,3\}$ with $E(X)=1.35$ : $\sum p=1\Rightarrow a+b=0.55$ , and $E(X)=a+2b+0.45=1.35\Rightarrow a+2b=0.9$ ; solve $b=0.35,\ a=0.20$ . One equation can't pin two letters. (9709/52/O/N/23 Q1)

PreviousCombinations in probability NextExpectation & variance

Building a probability distribution

A discrete random variable is a number decided by chance.
Examples: a die score, the number of heads, or how many red counters are drawn.
A probability distribution lists every value the variable can take.
Under each value, it gives the probability of that value.
The whole table must add to $1$ .

Build the table; make it sum to 1

Use $X$ for the random variable and $x$ for one value it can take.
List all possible values of $X$ .
Find each $P(X=x)$ and put it directly under the matching value.
Check $\sum p=1$ .
If a probability contains an unknown, the $\sum p=1$ equation usually solves it.
After solving, reject any value that makes a probability negative or bigger than 1.

Values across the top, probabilities under them, summing to 1. That Σp = 1 line both checks the table and solves an unknown.

Two coins, X = number of heads. Each value's height is its probability, and the three heights add to 1.

Worked example

Start here: read a value straight off a finished table

For $x=1,2,3,4$ the probabilities are $0.1,\ 0.2,\ 0.4,\ 0.3$ . The table is complete (they sum to 1), so just add the right cells: $P(X\ge3)=0.4+0.3$ .

Answer

P(X\ge3)=0.7

Worked example

Next rung: use Σp = 1 to find a constant

For $x=1,2,3,4$ the probabilities are $0.1,\ k,\ 2k,\ 0.3$ . The one missing fact is $\sum p=1$ : $0.1+k+2k+0.3=1\Rightarrow 3k=0.6$ , so $k=0.2$ (all probabilities valid). Then $P(X\ge3)=2k+0.3$ .

Answer

k=0.2,\quad P(X\ge3)=0.7

Build the table from the outcomes themselves

Sometimes the probabilities are not handed to you.
List the equally likely outcomes first.
Map each outcome to its value of $x$ .
If several outcomes give the same $x$ , pool them into one probability.

Worked example

Two spinners: build from the grid

A fair 4-sided spinner (labelled 1, 2, 3, 4) is spun twice and the scores are added to give $X$ . A grid of the $4\times4=16$ equally likely outcomes shows $X\in\{2,3,4,5,6,7,8\}$ .

Count how many of the 16 cells give each total: $1,2,3,4,3,2,1$ for $x=2,\dots,8$ .
Each probability is that count over 16:

$P(X{=}x)=\tfrac{1}{16},\tfrac{2}{16},\tfrac{3}{16},\tfrac{4}{16},\tfrac{3}{16},\tfrac{2}{16},\tfrac{1}{16}$

Answer

\tfrac{1+2+3+4+3+2+1}{16}=1\ \checkmark

The total 5 sits in the middle because it has the most ways to happen, the classic triangular shape of a sum of two dice or spinners.

Worked example

Selection without replacement: count with combinations

Only two people are non-youths, so $Y\in\{1,2,3\}$ (you cannot avoid youths).
Choose the youths and the rest separately, then divide by 120:

$P(Y{=}1)=\tfrac{{}^8C_1\cdot{}^2C_2}{120}=\tfrac{1}{15},\quad P(Y{=}2)=\tfrac{{}^8C_2\cdot{}^2C_1}{120}=\tfrac{7}{15},\quad P(Y{=}3)=\tfrac{{}^8C_3}{120}=\tfrac{7}{15}$

Answer

\tfrac{1}{15}+\tfrac{7}{15}+\tfrac{7}{15}=1\ \checkmark

Worked example

Another selection table (smaller numbers)

A box has 3 red, 5 blue (8 total); draw 2 without replacement. With $R=$ reds drawn and ${}^8C_2=28$ :

$P(R{=}0)=\tfrac{{}^5C_2}{28}=\tfrac{10}{28},\ \ P(R{=}1)=\tfrac{3\cdot5}{28}=\tfrac{15}{28},\ \ P(R{=}2)=\tfrac{3}{28}$

Answer

\tfrac{10}{28}+\tfrac{15}{28}+\tfrac{3}{28}=1\ \checkmark

When a formula gives each probability

Worked example

Formula table: substitute, then Σp = 1 for the constant

$X\in\{1,2,3\}$ , $P(X=x)=c\,x^2$ . $c(1+4+9)=1\Rightarrow c=\tfrac{1}{14}$ , so the table is $\tfrac{1}{14},\tfrac{4}{14},\tfrac{9}{14}$ — give the exact fractions asked. (9709/52/M/J/23 Q1)

Answer

c=\tfrac{1}{14}

Worked example

Bigger formula table (exact fractions)

$Q\in\{3,4,5,6\}$ , $P(Q=q)=c\,q^2$ . Sum the squares: $c(9+16+25+36)=1\Rightarrow 86c=1$ .

$c=\tfrac{1}{86}$ ; the table is $\tfrac{9}{86},\tfrac{16}{86},\tfrac{25}{86},\tfrac{36}{86}$ .
Then $P(Q>4)=\tfrac{25}{86}+\tfrac{36}{86}=\tfrac{61}{86}$ .

Answer

c=\tfrac{1}{86},\quad P(Q>4)=\tfrac{61}{86}

A common denominator (here 86) keeps every cell exact, which is what a “leave in fraction form” question wants.

When the constant comes out of a quadratic: reject invalid roots

Worked example

Two roots, one is impossible

$V\in\{2,3,4,5,6\}$ with probabilities $0.05,\ c^2,\ c+0.1,\ 2c+0.05,\ 0.16$ . Find $c$ , then $P(V>4)$ .

$\sum p=1$ : $0.05+c^2+c+0.1+2c+0.05+0.16=1$ , so $c^2+3c-0.64=0$ .
Factorise: $(c-0.2)(c+3.2)=0\Rightarrow c=0.2$ or $c=-3.2$ .
Reject $c=-3.2$ : it makes $P(V{=}4)=c+0.1=-3.1<0$ . So $c=0.2$ .
$P(V>4)=P(V{=}5)+P(V{=}6)=(2(0.2)+0.05)+0.16$ .

Answer

c=0.2,\quad P(V>4)=0.61

List every value $X$ can take.
Each $P(X=x)$ — count outcomes (combinations if selecting), or plug $x$ into the formula.
Lay them under their $x$ ; pool repeated x values.
Write the $\sum p=1$ line (it scores); it also solves a single unknown.

The question gives	Each P(X=x) by	Watch
a formula f(x)	substitute x; Σp=1 fixes the constant	give exact fractions
'select n at random'	combinations ÷ ⁿCᵣ	without replacement → denominators shrink
a verbal die/coin rule	list outcomes, map to x, pool equal x's	don't drop a branch
table, 1 unknown	Σp = 1	reject p<0 or p>1
table, 2 unknowns	Σp = 1 AND E(X)	one equation can't fix two

Examiner note

Write the

\sum p=1

line explicitly (a scored step) and draw a two-row table with each probability directly under its x. A “show that” needs the calculation, not the bare value. (9709/52/M/J/24 Q5)

Common mistake

Always write the

\sum p=1

Common mistake

In build-type tables several outcomes can give the same

x

(a coin and die rule sending many results to one total) — pool them, then re-check

\sum p=1

Two unknowns? You need two facts