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How do you calculate probabilities and work with discrete random variables and the binomial distribution?

Probability of combined events, mutually exclusive and independent events, conditional probability, discrete random variables with their expectation and variance, and the binomial distribution as a model.

A CCEA AS Further Maths Statistics answer on probability of combined events, mutually exclusive and independent events, conditional probability, discrete random variables with expectation and variance, and the binomial distribution as a model.

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What this dot point is asking

CCEA wants you to calculate probabilities of combined events, distinguish mutually exclusive from independent events, use conditional probability, work with discrete random variables (their distribution, expectation and variance), and recognise and apply the binomial distribution as a model.

The answer

Probability of combined events

Independence and conditional probability

Discrete random variables

The binomial distribution

Examples in context

Example 1. Reliability of a delivery service. If each parcel arrives on time with probability pp independently, the number of on-time deliveries in a batch of nn follows B(n,p)B(n, p). The expected number npnp tells a manager the typical service level, and the tail probabilities flag unusually bad days.

Example 2. Medical screening. Conditional probability underlies test interpretation: the chance a patient actually has a condition given a positive test, P(ill+)P(\text{ill} \mid +), depends on how common the condition is. Confusing this with P(+ill)P(+ \mid \text{ill}) is a classic real-world error the conditional formula corrects.

Try this

Q1. Events AA and BB are mutually exclusive with P(A)=0.3P(A) = 0.3, P(B)=0.4P(B) = 0.4. Find P(AB)P(A \cup B). [1 mark]

  • Cue. 0.3+0.4=0.70.3 + 0.4 = 0.7 (since P(AB)=0P(A \cap B) = 0).

Q2. For XB(10,0.3)X \sim B(10, 0.3), write down E(X)E(X). [1 mark]

  • Cue. E(X)=np=10(0.3)=3E(X) = np = 10(0.3) = 3.

Q3. A discrete variable has P(X=0)=0.6P(X = 0) = 0.6, P(X=1)=0.4P(X = 1) = 0.4. Find E(X)E(X). [1 mark]

  • Cue. 0(0.6)+1(0.4)=0.40(0.6) + 1(0.4) = 0.4.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA AS 20206 marksA discrete random variable XX has probability distribution P(X=1)=0.2P(X = 1) = 0.2, P(X=2)=0.5P(X = 2) = 0.5, P(X=3)=kP(X = 3) = k. Find kk, then find E(X)E(X) and Var(X)\operatorname{Var}(X).
Show worked answer →

Probabilities sum to 11: 0.2+0.5+k=10.2 + 0.5 + k = 1, so k=0.3k = 0.3.

The expectation is E(X)=xP(X=x)E(X) = \sum x\,P(X = x):

E(X)=1(0.2)+2(0.5)+3(0.3)=0.2+1.0+0.9=2.1.E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1.

For the variance, find E(X2)=x2P(X=x)E(X^2) = \sum x^2 P(X = x):

E(X2)=1(0.2)+4(0.5)+9(0.3)=0.2+2.0+2.7=4.9.E(X^2) = 1(0.2) + 4(0.5) + 9(0.3) = 0.2 + 2.0 + 2.7 = 4.9.

Var(X)=E(X2)[E(X)]2=4.92.12=4.94.41=0.49.\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 4.9 - 2.1^2 = 4.9 - 4.41 = 0.49.

Markers reward finding kk, the expectation, and the variance using E(X2)[E(X)]2E(X^2) - [E(X)]^2.

CCEA AS 20186 marksA fair coin is tossed 8 times. Using the binomial distribution, find the probability of exactly 5 heads, and the probability of at least 7 heads.
Show worked answer →

Let XX be the number of heads, XB(8,0.5)X \sim B(8, 0.5).

Exactly 55 heads: P(X=5)=(85)(0.5)5(0.5)3=56×(0.5)8=56256=0.21875.P(X = 5) = \binom{8}{5}(0.5)^5(0.5)^3 = 56 \times (0.5)^8 = \dfrac{56}{256} = 0.21875.

At least 77 heads means X=7X = 7 or X=8X = 8:

P(X=7)=(87)(0.5)8=8256,P(X=8)=(88)(0.5)8=1256.P(X = 7) = \binom{8}{7}(0.5)^8 = \dfrac{8}{256}, \qquad P(X = 8) = \binom{8}{8}(0.5)^8 = \dfrac{1}{256}.

P(X7)=8+1256=9256=0.0352.P(X \geq 7) = \dfrac{8 + 1}{256} = \dfrac{9}{256} = 0.0352.

Markers reward identifying the binomial model, the correct binomial coefficient and powers, and adding the two tail probabilities.

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