A fair spinner is divided into four sectors labelled 1, 2, 3 and 4. The random variable X denotes the score obtained on one spin, and has the probability distribution shown in the table below, where k is a constant. x 1 2 3 4 P(X=x) 0.1 0.3 k 0.2 What is the value of k?

A Level Maths 9709: Finding an Unknown Probability in a Distribution Table (MCQ)

$x$	1	2	3	4
$P(X=x)$	$0.1$	$0.3$	$k$	$0.2$

x

P(X=x)

0.1

0.3

k

0.2

For any discrete random variable, the probabilities in the distribution table must satisfy $\sum P(X=x) = 1$

So here, $P(X=1)+P(X=2)+P(X=3)+P(X=4) = 1$ $0.1 + 0.3 + k + 0.2 = 1$

Add the three known probabilities first: $0.1+0.3+0.2 = 0.6$

So the equation becomes $0.6+k=1$ $k = 1-0.6 = 0.4$

Check: substituting back, $0.1+0.3+0.4+0.2=1.00$ ✓, and every probability lies between $0$ and $1$ , so this is a valid distribution.

A ( $0.6$ ): this is the sum $0.1+0.3+0.2$ itself. The value of $k$ still needs to be found by subtracting this sum from $1$ , not by stopping at the sum.
B ( $0.3$ ): this simply repeats $P(X=2)$ ; there is no reason $k$ should equal another probability already in the table.
D ( $0.5$ ): this could come from an arithmetic slip, e.g. computing $1-0.1-0.3-0.2$ incorrectly as $0.5$ instead of $0.4$ .