S3 Probability — printable coverage pack

S3.1 · Understand and use mutually exclusive and independent events when calculating probabilities; link to discrete and continuous distributions.

Mutually exclusive events cannot occur together, so $P(A\cap B)=0$ and $P(A\cup B)=P(A)+P(B)$ .
Independent events satisfy $P(A\cap B)=P(A)P(B)$ , equivalently $P(A\mid B)=P(A)$ when $P(B)>0$ .
Use $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ for any two events, subtracting the overlap to avoid double-counting.
Events of positive probability cannot be both mutually exclusive and independent; the same rules apply to events defined from discrete or continuous random variables.

1. Events $A$ and $B$ are mutually exclusive, with $P(A)=0.38$ and $P(B)=0.27$ . Find $P(A\cup B)$ .[2 marks]
Answer
- $0.65$
Method: Mutually exclusive events have no overlap, so $P(A\cup B)=P(A)+P(B)=0.38+0.27=0.65$ .

1. Events $A$ and $B$ are independent. Given $P(A)=0.45$ and $P(A\cap B)=0.18$ , find $P(B)$ and $P(A\cup B)$ .[4 marks]
Answer
- $P(B)=0.40$ .
- $P(A\cup B)=0.67$ .
Method: Independence gives $0.18=P(A)P(B)=0.45P(B)$ , so $P(B)=0.18/0.45=0.40$ . Then $P(A\cup B)=0.45+0.40-0.18=0.67$ .

1. A discrete random variable $X$ has $P(X=0)=0.2$ , $P(X=1)=0.5$ and $P(X=2)=0.3$ . Independently, $Y$ is uniformly distributed on $0\leq y\leq5$ . Let $A$ be the event $X\geq1$ and $B$ the event $Y<2$ . Find $P(A\cap B)$ and $P(A\cup B)$ .[5 marks]
Answer
- $P(A\cap B)=0.32$ .
- $P(A\cup B)=0.88$ .
Method: $P(A)=0.5+0.3=0.8$ . Since $Y$ is uniform, $P(B)=2/5=0.4$ . The variables are independent, so $P(A\cap B)=0.8(0.4)=0.32$ . Therefore $P(A\cup B)=0.8+0.4-0.32=0.88$ .

Conditional probability restricts the sample space: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ for $P(B)>0$ .
On a tree diagram, multiply probabilities along a path and add the probabilities of mutually exclusive paths that satisfy the event.
Without replacement, later branch probabilities change because both the total and the relevant category count may have changed.
In a Venn diagram or two-way table, use the condition as the denominator; reversing $P(A\mid B)$ to $P(B\mid A)$ is a common error.

1. Of $60$ students, $24$ travel by bus and $15$ of those bus travellers arrive before $8{:}30$ . Find the probability that a randomly chosen bus traveller arrives before $8{:}30$ .[2 marks]
Answer
- $\frac{15}{24}=\frac58$
Method: The condition restricts the denominator to the $24$ bus travellers. Of these, $15$ arrived before $8{:}30$ , so the probability is $15/24=5/8$ .

1. A bag contains $5$ amber counters and $3$ blue counters. Two counters are taken without replacement. Find the probability that exactly one counter of each colour is taken.[4 marks]
Answer
- $\frac{15}{28}$
Method: There are two valid tree paths. Amber then blue has probability $\frac58\cdot\frac37=\frac{15}{56}$ . Blue then amber has probability $\frac38\cdot\frac57=\frac{15}{56}$ . Adding gives $\frac{30}{56}=\frac{15}{28}$ .

1. For events $A$ and $B$ , $P(A)=0.60$ , $P(B)=0.50$ and $P(A\mid B)=0.70$ . Find $P(A'\mid B')$ .[5 marks]
Answer
- $0.50$
Method: From the conditional probability formula, $P(A\cap B)=P(A\mid B)P(B)=0.70(0.50)=0.35$ . Hence $P(A\cup B)=0.60+0.50-0.35=0.75$ , so $P(A'\cap B')=1-0.75=0.25$ . Also $P(B')=0.50$ . Therefore $P(A'\mid B')=0.25/0.50=0.50$ .

A probability model simplifies a real process by specifying possible outcomes and assigning probabilities to them.
State assumptions explicitly, such as independence, constant probabilities, identical trials or equally likely outcomes, and judge them in context.
More realistic dependence or changing probabilities can alter both central probabilities and tail risks, so identify the likely direction of the effect where possible.
Validate a model by comparing its predictions with observed data; a close fit in one sample does not prove that its assumptions are true.

1. A model assigns probability $1/200$ to each ticket winning a draw. State one assumption behind this model and one reason it might fail.[2 marks]
Answer
- Assumption: all $200$ tickets are equally likely to be selected.
- It could fail if the mixing or selection mechanism favours some tickets.
Method: Equal probabilities require a fair randomising process. Any systematic difference in ticket placement, shape or handling would undermine that assumption.

1. A factory model treats defects in two items from the same batch as independent, each with probability $0.03$ . It therefore predicts probability $0.03^2$ that both are defective. Explain how batch-to-batch variation is likely to affect this prediction.[3 marks]
Answer
- Items from the same poor-quality batch are positively associated rather than independent.
- Therefore two defects together are likely to occur more often than the modelled probability $0.0009$ .
Method: The model gives $P(\text{both})=0.03^2=0.0009$ . If an unobserved batch condition raises the defect probability for both items, learning that one is defective increases the chance that the other is defective. This positive dependence makes the simple independent model likely to underestimate the joint probability.

1. A transport model assumes that each of $5$ commuters independently chooses route A with probability $0.4$ . Calculate the modelled probability that all $5$ choose route A. A closure on route B can influence every commuter on the same day. Critique the independence assumption and state the likely effect on the probability just calculated.[5 marks]
Answer
- Modelled probability $=0.01024$ .
- A shared closure creates positive dependence, so the model is likely to underestimate the probability that all $5$ choose route A.
Method: Under the model, $P(\text{all choose A})=0.4^5=0.01024$ . A route-B closure is a common cause affecting all five choices, so the choices are not independent on that day. It makes simultaneous choices of route A more likely, so a model including road conditions would usually give a larger probability for this event.