Completing and using a distribution | CAIE 9709 Probability & Statistics 1 Notes

Probability & Statistics 1

Mathematics9709

Substitute every possible value first

If $P(X=x)=k(x^2+2)$ for $x\in\{-2,-1,0,2,3\}$ , then $k$ is one fixed constant used in every cell. (9709/51/O/N/24 Q2)

Substitute the five values to get $6k,3k,2k,6k,11k$ .
Use $\sum p=1$ : $28k=1$ , so $k=1/28$ .
Put the numerical probabilities into the table.

x	P(X = x)
−2	6/28
−1	3/28
0	2/28
2	6/28
3	11/28

x	−2	−1	0	2	3
P(X = x)	6/28	3/28	2/28	6/28	11/28

If a question says probabilities are proportional to $f(x)$ , write $P(X=x)=k f(x)$ and find $k$ in the same way.

A probability test chooses the valid root

Worked example

When the total gives a quadratic

For $V=2,3,4,5,6$ , suppose the probabilities are $0.05,c^2,c+0.1,2c+0.05,0.16$ .

c^2+3c+0.36=1

c^2+3c-0.64=0

The equation solver is the quickest method here. It gives $c=0.2$ or $c=-3.2$ . The second root makes some probabilities negative or greater than 1, so reject it.

Answer

c=0.2

If the roots are convenient, write the factorised line. If they are awkward, write the equation and both calculator roots, then test each root. The mathematical decision is the validity test.

Key idea

A valid distribution needs every probability in

0\le P(X=x)\le1

and a total of 1.

Translate the event, then add its cells

Use a table with probabilities 0.1, 0.2, 0.4 and 0.3 for $X=0,1,2,3$ .

$P(X\ge2)=P(X=2)+P(X=3)=0.7$ .
$P(0<X<3)=P(X=1)+P(X=2)=0.6$ .
The modal value is 2 because it has the largest probability.

A conditional probability restricts the table first. For example,

P(X\ge1)=0.9

P(X=3\mid X\ge1)=\frac{0.3}{0.9}=\frac13

Common mistake

“At least 2” includes 2. “More than 2” does not. Mark the accepted values before adding probabilities.

Your turnIndependent check [4 marks]

For $X=0,1,2$ , $P(X=x)=c(x+1)$ . Find $c$ , $P(X\ge1)$ and $P(X=2\mid X\ge1)$ .

Show worked answer

The probabilities are $c,2c,3c$ , so $6c=1$ .

Within $X\ge1$ , the total probability is $5/6$ , while $X=2$ has probability $3/6$ .

Answer

c=\frac16

Answer

P(X\ge1)=\frac56

Answer

P(X=2\mid X\ge1)=\frac35

Substitute every possible value first

If $P(X=x)=k(x^2+2)$ for $x\in\{-2,-1,0,2,3\}$ , then $k$ is one fixed constant used in every cell. (9709/51/O/N/24 Q2)

Substitute the five values to get $6k,3k,2k,6k,11k$ .
Use $\sum p=1$ : $28k=1$ , so $k=1/28$ .
Put the numerical probabilities into the table.

x	P(X = x)
−2	6/28
−1	3/28
0	2/28
2	6/28
3	11/28

x	−2	−1	0	2	3
P(X = x)	6/28	3/28	2/28	6/28	11/28

If a question says probabilities are proportional to $f(x)$ , write $P(X=x)=k f(x)$ and find $k$ in the same way.

A probability test chooses the valid root

Worked example

When the total gives a quadratic

For $V=2,3,4,5,6$ , suppose the probabilities are $0.05,c^2,c+0.1,2c+0.05,0.16$ .

c^2+3c+0.36=1

c^2+3c-0.64=0

The equation solver is the quickest method here. It gives $c=0.2$ or $c=-3.2$ . The second root makes some probabilities negative or greater than 1, so reject it.

Answer

c=0.2

If the roots are convenient, write the factorised line. If they are awkward, write the equation and both calculator roots, then test each root. The mathematical decision is the validity test.

Key idea

A valid distribution needs every probability in

0\le P(X=x)\le1

and a total of 1.

Translate the event, then add its cells

Use a table with probabilities 0.1, 0.2, 0.4 and 0.3 for $X=0,1,2,3$ .

$P(X\ge2)=P(X=2)+P(X=3)=0.7$ .
$P(0<X<3)=P(X=1)+P(X=2)=0.6$ .
The modal value is 2 because it has the largest probability.

A conditional probability restricts the table first. For example,

P(X\ge1)=0.9

P(X=3\mid X\ge1)=\frac{0.3}{0.9}=\frac13

Common mistake

“At least 2” includes 2. “More than 2” does not. Mark the accepted values before adding probabilities.

Your turnIndependent check [4 marks]

For $X=0,1,2$ , $P(X=x)=c(x+1)$ . Find $c$ , $P(X\ge1)$ and $P(X=2\mid X\ge1)$ .

Show worked answer

The probabilities are $c,2c,3c$ , so $6c=1$ .

Within $X\ge1$ , the total probability is $5/6$ , while $X=2$ has probability $3/6$ .

Answer

c=\frac16

Answer

P(X\ge1)=\frac56

Answer

P(X=2\mid X\ge1)=\frac35