E8.1, E8.3 - Probability notation and Venn regions

E8.1, E8.3 - Probability notation and Venn regions

Probability notation makes a long event description compact without changing its meaning. In this Extended lesson, E8.1 introduces event and complement notation, while E8.3 combines it with union and intersection in Venn diagrams. You will translate the symbols into regions and calculate probabilities from tables and up to three sets.

Event and complement notation

If AA means “a chosen card is an ace”, what does P(A′)\mathrm{P}(A') mean? The prime mark changes the event before you calculate its probability. It does not ask for a reciprocal, a different ace, or a second selection.

Probability notation

P(A)\mathrm{P}(A) means “the probability of event AA”. A′A' is its complementary event, so P(A′)\mathrm{P}(A') means “the probability of not AA”.

P(A′)=1−P(A).\mathrm{P}(A')=1-\mathrm{P}(A).

The value of P(A)\mathrm{P}(A) is a number between zero and one; AA itself is the event, not its probability.

For an ordinary 5252-card pack chosen uniformly, the four aces form event AA, giving P(A)=452\mathrm{P}(A)=\frac4{52}. The other 4848 cards form A′A'. The notation keeps the distinction between a collection of outcomes and its chance.

Worked example 1: reading and calculating a complement

Question. Event WW is winning a game, and P(W)=310\mathrm{P}(W)=\frac3{10}. State what P(W′)\mathrm{P}(W') represents and find its value.

Plan. Translate the prime mark as “not winning”, then subtract from one.

P(W′)=1−310all non-winning outcomes=710remaining seven tenths\begin{aligned} \mathrm{P}(W')&=1-\frac3{10} &&\text{all non-winning outcomes}\\ &=\frac7{10} &&\text{remaining seven tenths} \end{aligned}

Answer. It is the probability of not winning, 710\frac7{10}. If draws are possible, not winning includes both a draw and a loss.

Check. 310+710=1\frac3{10}+\frac7{10}=1, accounting for every game result.

Reverse only which probability is supplied. Taking a complement twice returns to the original event.

Worked example 2: the complement is supplied

Question. Event WW is winning a game, and P(W′)=710\mathrm{P}(W')=\frac7{10}. Find P(W)\mathrm{P}(W).

Plan. Remove the non-winning chance from the whole.

P(W)=1−P(W′)winning and not winning exhaust the results=1−710=310subtract the supplied complement\begin{aligned} \mathrm{P}(W)&=1-\mathrm{P}(W') &&\text{winning and not winning exhaust the results}\\ &=1-\frac7{10}=\frac3{10} &&\text{subtract the supplied complement} \end{aligned}

Answer. P(W)=310\mathrm{P}(W)=\frac3{10}.

Check. Taking the complement of this answer gives 1−310=7101-\frac3{10}=\frac7{10}, the value in the question.

In particular, an impossible event has probability zero and its complement is certain. A certain event has probability one and its complement is impossible. These boundary cases follow the same subtraction.

Watch out

P(W′)=1/P(W)\mathrm{P}(W')=1/\mathrm{P}(W) is wrong. The prime means complement, not reciprocal; a reciprocal can even exceed one and so cannot be a probability here.

Translate the supplied event before doing the subtraction.

Union and intersection probabilities

The set symbols still describe the event first. P(A∩B)\mathrm{P}(A\cap B) asks for the chance that both events hold. P(A∪B)\mathrm{P}(A\cup B) asks for the chance that at least one holds, including both. They do not mean the same thing as “exactly one”.

For example, 2020 students are classified by whether they attend art and badminton. The entries below count different students, each appearing in one cell.

Badminton BBNot badminton B′B'Total
Art AA44661010
Not art A′A'33771010
Total7713132020

The cell where the art row meets the badminton column is the intersection. The union contains that cell plus the art-only and badminton-only cells. The outside cell contains neither.

Worked example 3: both events

Question. One of these students is chosen with equal chance. Find P(A∩B)\mathrm{P}(A\cap B).

Plan. Use the cell satisfying both conditions over the whole-group total.

n(A∩B)=4art row and badminton columnP(A∩B)=420=15four favourable students\begin{aligned} n(A\cap B)&=4 &&\text{art row and badminton column}\\ \mathrm{P}(A\cap B)&=\frac4{20}=\frac15 &&\text{four favourable students} \end{aligned}

Answer. 15\frac15.

Check. The four cells total 4+6+3+7=204+6+3+7=20; only the first cell satisfies both conditions.

Change only the event from intersection to union. The overlap remains included once.

Worked example 4: at least one event

Question. For the same selection, find P(A∪B)\mathrm{P}(A\cup B).

Plan. Add the three cells representing art, badminton or both, without repeating a cell.

n(A∪B)=6+4+3=13art only, both, badminton onlyP(A∪B)=1320divide by all students\begin{aligned} n(A\cup B)&=6+4+3=13 &&\text{art only, both, badminton only}\\ \mathrm{P}(A\cup B)&=\frac{13}{20} &&\text{divide by all students} \end{aligned}

Answer. 1320\frac{13}{20}.

Check. The complementary event is neither, with probability 720\frac7{20}. Thus 1−720=13201-\frac7{20}=\frac{13}{20}.

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