Mathematics 9709/62 — October/November 2010

Cambridge AS Level · Probability & Statistics 1 · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Discrete Random Variables · Representation of Data · Probability · The Normal Distribution · Permutations and Combinations

Q13MDiscrete Random VariablesFree sample

The discrete random variable $X$ takes the values 1, 4, 5, 7 and 9 only. The probability distribution of $X$ is shown in the table.

$x$	1	4	5	7	9
$P(X = x)$	$4p$	$5p^2$	$1.5p$	$2.5p$	$1.5p$

Find $p$ .

DifficultyMedium-Easy

Worked solution

Approach

For a probability distribution, the probabilities of all the possible values of $X$ must sum to $1$ . Use this to form an equation in $p$ , solve the quadratic, then reject any value that gives an impossible probability.

Working

Since the probabilities sum to $1$ ,

4p + 5p^2 + 1.5p + 2.5p + 1.5p = 1

Collect the $p$ -terms:

5p^2 + (4 + 1.5 + 2.5 + 1.5)p = 1

5p^2 + 9.5p = 1

Multiply by 2 to clear the decimal:

10p^2 + 19p - 2 = 0

Factorise:

(10p - 1)(p + 2) = 0

p = \frac{1}{10} \quad \text{or} \quad p = -2

A probability cannot be negative, so reject $p = -2$ . Therefore

p = 0.1

Answer

p = 0.1

Final answer

0.1

Detailed explanation

Walkthrough

Start with the key property of any probability distribution: the sum of the probabilities for all possible outcomes is exactly 1. The table gives five probabilities, so add them together and set the total equal to 1.

This gives

4p + 5p^2 + 1.5p + 2.5p + 1.5p = 1

Next, collect all the terms involving $p$ . The terms $4p$ , $1.5p$ , $2.5p$ and $1.5p$ combine to $9.5p$ , so the equation becomes

5p^2 + 9.5p = 1

It is easier to work with whole numbers, so multiply the whole equation by 2:

10p^2 + 19p - 2 = 0

This quadratic factorises:

(10p - 1)(p + 2) = 0

so the two possible solutions are

p = \frac{1}{10} \quad \text{or} \quad p = -2

The second value, $p = -2$ , cannot be correct because it would make terms such as $4p$ and $1.5p$ negative. For example, $P(X = 1) = 4p = -8$ , which is impossible since probabilities must be between 0 and 1. Therefore reject $p = -2$ and keep

p = 0.1

Key Takeaways

The probabilities in a discrete probability distribution must always sum to 1.
After forming the equation, solving the resulting quadratic is an algebraic step that must be completed carefully.
Not every algebraic solution is a valid probability: all probabilities must be non-negative and no greater than 1.
A probability value must be checked against the actual probabilities it produces, not just accepted because it solves the equation.

Common Mistakes

Forgetting that the probabilities must sum to 1, so no equation is formed.
Making arithmetic errors when adding $4p + 1.5p + 2.5p + 1.5p$ ; the total is $9.5p$ , not $9p$ .
Writing the quadratic incorrectly; it is $10p^2 + 19p - 2 = 0$ after multiplying by 2.
Accepting $p = -2$ without rejecting it. The mark scheme explicitly requires the negative value to be rejected to gain the final mark.
Factorising the quadratic incorrectly. Check: $(10p - 1)(p + 2) = 10p^2 + 20p - p - 2 = 10p^2 + 19p - 2$ .

Things to Be Careful About

Probabilities must lie in the interval $0 \leq P(X = x) \leq 1$ . Since some probabilities contain $p$ with a positive coefficient, $p$ must be positive.
If $p = -2$ , then $P(X = 1) = -8$ , $P(X = 5) = -3$ , $P(X = 7) = -5$ and $P(X = 9) = -3$ , all impossible.
The mark scheme awards a mark for choosing $p = 0.1$ only when the negative solution is explicitly rejected.
It is good practice to verify the final probabilities: for $p = 0.1$ , they are $0.4$ , $0.05$ , $0.15$ , $0.25$ and $0.15$ , which sum to 1.

Techniques used

sum probabilities in a distribution table to 1form and solve a quadratic equationreject an invalid probability value

The rest of this paper

6 more questions

Q2Representation of Data6M
Q3Probability6M
Q4Representation of Data7M
Q5The Normal Distribution7M
Q6Discrete Random Variables · The Normal Distribution10M
Q7Permutations and Combinations · Probability11M

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