Mu1–Mu5 - Events, independence and probability diagrams | OCR A-Level Mathematics B (MEI)

Use the event rules in Mu1–Mu4 to distinguish mutual exclusivity from independence and calculate combined probabilities. You will organise outcomes in grids and trees, then use the overlapping Venn regions of Mu5, including diagrams with three events.

Complements complete the probability model

Is at least one 6 in five fair die throws more likely than no 6? Each throw has only a one-in-six chance, but there are five opportunities. Adding five copies of $1/6$ feels tempting; it counts some outcomes more than once. We will find the answer without that trap.

First, name outcomes carefully. A fair spinner with eight equal sectors labelled 1 to 8 gives a label greater than 6 with probability $2/8=1/4$ . If the model were biased, the eight labels would need their actual probability weights instead.

You already know

For finitely many equally likely outcomes, use favourable outcomes divided by all outcomes. The complement $A'$ means “not $A$ ”. In $n$ comparable trials, an event with probability $p$ has expected frequency $np$ ; this is not a promised count.

For the spinner, 60 spins have expected frequency $60(1/4)=15$ for a label greater than 6. Different observed counts are compatible with this model.

Event and complement

An event is a set of outcomes in a sample space. Its complement contains every outcome in that sample space that is not in the event. The event and its complement are disjoint and together exhaustive. For $A=\{7,8\}$ on this spinner, $A'=\{1,2,3,4,5,6\}$ , so $\mathrm P(A')=3/4$ .

The complement rule $\mathrm P(A')=1-\mathrm P(A)$ and the expected-frequency rule $np$ are prerequisite knowledge, not separately printed general formulas on the supplied sheet. All probabilities lie between 0 and 1 and the complete outcome probabilities sum to 1.

There is a useful warning in two fair independent coin tosses. HH, HT, TH and TT are equally likely, but the head counts 0, 1 and 2 are not. One head has probability $1/2$ , while zero and two each have probability $1/4$ . Grouping outcomes can destroy equal likelihood.

Key idea

Count equally likely underlying outcomes, rather than assuming that every named category has the same probability.

Keep the model and the expected frequency distinct in this check.

Add disjoint events and multiply independent events

Mutual exclusivity concerns shared outcomes; independence concerns unchanged probabilities.

Use $A\cap B$ for “ $A$ and $B$ ”, and $A\cup B$ for “ $A$ or $B$ ”, including both. Mutually exclusive events cannot occur together. Independent events satisfy the product condition below; they may overlap.

Two different event conditions

Mutually exclusive events satisfy

\mathrm P(A\cap B)=0,\qquad \mathrm P(A\cup B)=\mathrm P(A)+\mathrm P(B).

Independent events satisfy

\mathrm P(A\cap B)=\mathrm P(A)\mathrm P(B).

The general addition and conditional multiplication formulas are printed on the sheet; these conditions and special cases are not separately printed.

Not in the formula sheet: learn it.

For a uniformly selected integer from 1 to 8, the events $E=\{1,2\}$ and $F=\{7,8\}$ are exclusive: their union probability is $1/2$ . They are not independent because their intersection is zero but their product is $1/16$ . Two exclusive events with positive probabilities cannot be independent; an event of probability zero is a boundary case for the product test.

Worked example 1: independent events in one selection

Question. Select an integer uniformly from 1 to 8. Let $C$ mean even and $D$ mean greater than 4. Determine whether the events are independent.

Plan. Compare the probability of their intersection with the product of their probabilities.

\begin{aligned} \mathrm P(C)&=4/8=1/2,\quad \mathrm P(D)=4/8=1/2 &&\text{count each event}\\ \mathrm P(C\cap D)&=2/8=1/4 &&\text{shared labels }6,8\\ \mathrm P(C)\mathrm P(D)&=(1/2)(1/2)=1/4 &&\text{test the product condition} \end{aligned}

Why this step? Sharing outcomes does not settle independence; the numerical equality does.

Answer. The events are independent, because the intersection probability equals the product.

Check. Half the labels above 4 are even, matching the overall even proportion. Their positive overlap also shows they are not mutually exclusive.

Change only the threshold in $D$ , keeping the same selection and even event.

Worked example 2: a changed threshold

Question. Select an integer uniformly from 1 to 8. Let $C$ mean even and $D$ mean greater than 5. Determine whether the events are independent.

Plan. Recount $D$ and the intersection before testing independence.

\begin{aligned} \mathrm P(D)&=3/8 &&\text{labels }6,7,8\\ \mathrm P(C\cap D)&=2/8=1/4 &&\text{labels }6,8\\ \mathrm P(C)\mathrm P(D)&=(1/2)(3/8)=3/16 &&\text{independent benchmark} \end{aligned}

Why this step? The new threshold changes the event probability even though the intersection happens to be unchanged.

Answer. They are not independent, since $1/4\ne3/16$ .

Check. Two of the three labels above 5 are even, which differs from $1/2$ .

For repeated independent trials, “at least one” is often simpler through “none”. The next pair changes only the number of throws. The complement-and-product expression is derived from the supplied rules; it is not separately printed in the formula sheet.

Worked example 3: the five-throw puzzle

Question. Find the probability of at least one 6 in five independent throws of a fair die.

Plan. Find the probability of no 6, then subtract from 1.

\begin{aligned} \mathrm P(\text{no 6})&=(5/6)^5 &&\text{multiply independent failures}\\ \mathrm P(\text{at least one 6})&=1-(5/6)^5\\ &=\frac{4651}{7776}\approx0.60 &&\text{take the complement} \end{aligned}

Why this step? “None” is one repeated-product event; “at least one” includes many overlapping possibilities.

Answer. The probability is approximately $0.60$ , which exceeds $1/2$ .

Check. The probability is below 1 and above the single-throw $1/6$ , as extra opportunities suggest.

Four throws give one fewer opportunity; nothing else in the model changes.

Worked example 4: one fewer throw

Question. Find the probability of at least one 6 in four independent throws of a fair die.

Plan. Use the same complement method with four failure factors.

\begin{aligned} \mathrm P(\text{at least one 6})&=1-(5/6)^4 &&\text{exclude four failures}\\ &=\frac{671}{1296}\approx0.52 &&\text{evaluate last} \end{aligned}

Why this step? Independence allows multiplication, not addition, of the four failure probabilities.

Answer. The probability is approximately $0.52$ .

Check. This is less than the five-throw probability but still exceeds $1/2$ . Four opportunities already make a 6 more likely than none.

The word “exhaustive” describes a third idea: the union covers the sample space. It is not a synonym for mutually exclusive.

Watch out

Wrong: add $4(1/6)$ for at least one 6. Outcomes with several sixes are counted repeatedly. Right: subtract the no-six probability from 1. When explaining exclusivity, refer to a zero or non-zero intersection; a statement about the union does not settle it.

Keep the fair-die model and change the throw count to three.

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Mu1–Mu5 - Events, independence and probability diagrams

Complements complete the probability model

Add disjoint events and multiply independent events

Continue this lesson.

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