In a mock driving test, the random variable X represents the number of minor faults recorded for a randomly chosen candidate, where X can take the values 0, 1, 2 or 3. The probability distribution of X is shown in the table below, where p is a constant. x 0 1 2 3 P(X=x) 0.5 3p 2…

A Level Maths 9709: Distribution Table with Two Unknown Probabilities

$x$	0	1	2	3
$P(X=x)$	$0.5$	$3p$	$2p$	$0.1$

x

P(X=x)

0.5

3p

2p

0.1

Part (a): Finding the unknown constant

Since $X$ is a discrete random variable, all of its probabilities must sum to $1$ : $P(X=0)+P(X=1)+P(X=2)+P(X=3)=1$ $0.5+3p+2p+0.1=1$

Combine the known probabilities and the $p$ -terms: $0.6+5p=1$

Solve for $p$ : $5p=1-0.6=0.4 \implies p=\frac{0.4}{5}=0.08$

So $P(X=1)=3p=3(0.08)=0.24$ and $P(X=2)=2p=2(0.08)=0.16$ .

Check: $0.5+0.24+0.16+0.1=1.00$ ✓, and every value lies between $0$ and $1$ , so this is a valid probability distribution.

Part (b): Finding E(X)

$E(X)=\sum xP(X=x)=0(0.5)+1(0.24)+2(0.16)+3(0.1)$ $=0+0.24+0.32+0.3=0.86$

Part (c): Finding Var(X)

First find $E(X^2)=\sum x^2P(X=x)$ : $E(X^2)=0^2(0.5)+1^2(0.24)+2^2(0.16)+3^2(0.1)$ $=0(0.5)+1(0.24)+4(0.16)+9(0.1)$ $=0+0.24+0.64+0.9=1.78$

Then apply $\mathrm{Var}(X)=E(X^2)-\bigl[E(X)\bigr]^2$ : $\mathrm{Var}(X)=1.78-(0.86)^2=1.78-0.7396=1.0404$

Check by the direct definition: $\mathrm{Var}(X)=\sum(x-E(X))^2P(X=x)$ $=(0-0.86)^2(0.5)+(1-0.86)^2(0.24)+(2-0.86)^2(0.16)+(3-0.86)^2(0.1)$ $=(0.7396)(0.5)+(0.0196)(0.24)+(1.2996)(0.16)+(4.5796)(0.1)$ $=0.3698+0.004704+0.207936+0.45796=1.0404$ which agrees exactly with the value found above.

Final answers

(a) $p=\boxed{0.08}$
(b) $E(X)=\boxed{0.86}$
(c) $\mathrm{Var}(X)=\boxed{1.0404}$