9709/51

Mathematics 9709/51October/November 2022

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

6
questions
50
marks
75
minutes

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

Q14MDiscrete Random VariablesFree sample

The probability distribution table for a random variable XX is shown below.

xx2-21-10.50.512
P(X=x)P(X = x)0.12ppqq0.160.3

Given that E(X)=0.28E(X) = 0.28, find the value of pp and the value of qq.

DifficultyMedium-Easy
Worked solution

Approach

Since XX is a discrete random variable, the probabilities must sum to 11, and the expectation is the sum of each value multiplied by its probability. This gives two linear equations in pp and qq, which can be solved simultaneously.

Working

Sum of probabilities:

0.12+p+q+0.16+0.3=10.12 + p + q + 0.16 + 0.3 = 1 p+q=10.120.160.3p + q = 1 - 0.12 - 0.16 - 0.3 p+q=0.42p + q = 0.42

Expectation:

E(X)=(2)(0.12)+(1)(p)+(0.5)(q)+(1)(0.16)+(2)(0.3)E(X) = (-2)(0.12) + (-1)(p) + (0.5)(q) + (1)(0.16) + (2)(0.3) 0.28=0.24p+0.5q+0.16+0.60.28 = -0.24 - p + 0.5q + 0.16 + 0.6

Simplify the constant terms:

0.28=0.52p+0.5q0.28 = 0.52 - p + 0.5q p+0.5q=0.24-p + 0.5q = -0.24

Now solve the two equations:

p+q=0.42p + q = 0.42 p+0.5q=0.24-p + 0.5q = -0.24

From the first equation, q=0.42pq = 0.42 - p. Substitute into the second equation:

p+0.5(0.42p)=0.24-p + 0.5(0.42 - p) = -0.24 p+0.210.5p=0.24-p + 0.21 - 0.5p = -0.24 1.5p+0.21=0.24-1.5p + 0.21 = -0.24 1.5p=0.45-1.5p = -0.45 p=0.3p = 0.3

Then

q=0.420.3=0.12q = 0.42 - 0.3 = 0.12

Answer

p=0.3,q=0.12p = 0.3, \quad q = 0.12
Final answer

p = 0.3, q = 0.12

Detailed explanation

Walkthrough

We are given a probability distribution table for a discrete random variable XX, with two unknown probabilities pp and qq. There are two standard facts we can use to find them.

First, the probabilities in any probability distribution must add up to 11. This gives the equation

0.12+p+q+0.16+0.3=1,0.12 + p + q + 0.16 + 0.3 = 1,

which simplifies to p+q=0.42p + q = 0.42.

Second, the expectation is defined as

E(X)=xP(X=x).E(X) = \sum x \, P(X = x).

We substitute each value of xx and its probability, using pp and qq for the unknown probabilities. This gives

0.28=(2)(0.12)+(1)p+(0.5)q+(1)(0.16)+(2)(0.3).0.28 = (-2)(0.12) + (-1)p + (0.5)q + (1)(0.16) + (2)(0.3).

After simplifying the constant terms, we obtain

p+0.5q=0.24.-p + 0.5q = -0.24.

Now we have two linear equations in pp and qq. We solve them simultaneously. From p+q=0.42p + q = 0.42, we can write q=0.42pq = 0.42 - p and substitute into the second equation. This gives a single equation in pp:

p+0.5(0.42p)=0.24,-p + 0.5(0.42 - p) = -0.24,

which simplifies to 1.5p+0.21=0.24-1.5p + 0.21 = -0.24, so p=0.3p = 0.3. Substituting back gives q=0.12q = 0.12.

Key Takeaways

  • For any discrete probability distribution, the sum of all probabilities is always 11.
  • The expectation formula E(X)=xP(X=x)E(X) = \sum x \, P(X = x) is the key tool for using a given mean to find unknown probabilities.
  • Two unknown probabilities require two independent equations: one from the total probability and one from the expectation.
  • Solving simultaneous linear equations is a core algebraic skill needed in probability and statistics.

Common Mistakes

  • Forgetting to include all probabilities in the sum-to-1 equation, or incorrectly adding the constants.
  • Making sign errors when computing E(X)E(X), especially with negative values of xx.
  • Using the wrong signs when simplifying p+0.5q=0.24-p + 0.5q = -0.24.
  • Not simplifying the expectation equation before attempting to solve, leading to arithmetic errors.
  • Stopping after finding only one of the two unknowns.

Things to Be Careful About

  • The mark scheme awards marks for forming both equations: the sum-of-probabilities equation and the expectation equation. You must show both clearly.
  • The expectation equation may be left unsimplified and still earn the method mark, but simplifying helps avoid mistakes.
  • When solving, be careful with decimal arithmetic. Multiplying the equations by 10 or 2 can make the algebra easier.
  • Always check that the final values are sensible: probabilities must lie between 00 and 11, and they should satisfy both original equations.
Techniques used
apply the sum of probabilities conditionform the expectation equationsolve simultaneous linear equations

The rest of this paper

5 more questions
  • Q2Discrete Random Variables · The Normal Distribution8M
  • Q3Representation of Data9M
  • Q4The Normal Distribution9M
  • Q5Probability10M
  • Q6Permutations and Combinations10M
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