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ProbabilityEvent

Reading time: ~45 min

Where a mathematical reasoning can be had, it is as great a folly to make use of any other, as to grope for a thing in the dark, when you have a candle in your hand. — John Arbuthnot

Riya runs the coin stall at the school mela. A player tosses a coin twice and wins a kulfi if exactly one head shows. Before the gates open, Riya wants to be sure what she is promising: which results win the kulfi, and which don't?

You have already met random experiments and their sample spaces. Riya's game has the sample space S = \{HH, HT, TH, TT\} — every possible result, listed once. Nothing that can happen lies outside S, so S plays the part of the universal set for every question we ask about the game.

Riya's prize is a collection of outcomes. Tap the outcomes that win the kulfi, then build the other two events in the list.

Each "happening" you built picked out some outcomes of S and left out the rest. The prize event is the set E = \{HT, TH\}, a subset of S. That is no accident: every description of a result picks out a subset of S, and every subset of S describes a result. Here are a few more for the same game.

Description of the eventSubset of S
Number of tails is exactly 2A = \{TT\}
Number of tails is at least oneB = \{HT, TH, TT\}
Number of heads is at most oneC = \{HT, TH, TT\}
Second toss is not a headD = \{HT, TT\}
Number of tails is at most twoS = \{HH, HT, TH, TT\}
Number of tails is more than two\phi

Two rows deserve a second look. Every outcome has at most two tails, so that event is . No outcome has more than two tails, so that event is . So the two ideas — a description of a result, and a subset of S — are really one idea.

Definition. Any subset E of a sample space S is called an event.

Did It Happen?

An event is a set of outcomes, but one throw produces just one outcome. So when do we say that an event has happened? Throw a , and let E be the event "a number less than 4", so that E = \{1, 2, 3\}.

Roll the die. After each roll, every event whose set contains the number you rolled lights up. Complete the three missions.

So an event happens exactly when the outcome lands inside its set. If the outcome \omega of the experiment satisfies \omega \in E, we say that the event E has occurred; if \omega \notin E, then E has not occurred. A roll of 2 makes E occur; a roll of 5 does not.

Did you watch G and H? G stayed dark whatever you rolled, and H lit up . They are the two extreme kinds of event.

The Two Extremes

Events are sorted into types by how many outcomes their sets hold. Stay with one die, S = \{1, 2, 3, 4, 5, 6\}.

Let G be the event "the number shown is a multiple of 7". No face of a die is a multiple of 7, so no outcome belongs to G, and its set is . An event whose set is \phi can never occur. It is called an impossible event.

Now let H be "the number shown is odd or even". Every face is odd or even, so H = \{1, 2, 3, 4, 5, 6\}, which is . H occurs on every throw: it is the sure event. Every experiment has exactly one of each — the empty set \phi is impossible, and the sample space S itself is sure.

One Outcome or Many?

Between the two extremes lie events with one or more outcomes.

An event with exactly one sample point, like "the die shows 6" = \{6\}, is a simple (or elementary) event. A sample space with n outcomes has exactly n simple events: tossing two coins gives the simple events \{HH\}, \{HT\}, \{TH\} and \{TT\}.

An event with more than one sample point is a compound event. Toss a coin three times: "exactly one head" = \{HTT, THT, TTH\} has sample points, so it is compound. So are "at least one head", with 7 sample points, and "at most one head", with 4.

A die is thrown once. Turn each event into a set in your head, count its outcomes, and drop it into the right bucket.

an odd number
a number greater than 6
an even prime
a number less than 7
a 6
a multiple of 3
a negative number
a number less than 2
No outcome: impossible
One outcome: simple
Several outcomes: compound

Did "a number less than 7" surprise you? It is the sure event S, with six sample points — so the sure event is a compound event too. "A number greater than 6" and "a negative number" both have the empty set: they are the same impossible event, \phi.

Combining Events

In the chapter on Sets you combined sets using union, intersection, difference and complement. Every event is a subset of S, so the same operations combine events — and each of them has an everyday name. Let A and B be events of an experiment with sample space S.

1. "Not A", the complementary event. Every event A has a partner A', the complementary event of A, which occurs exactly when A does not. Toss three coins, so that

S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}

and let A be "only one tail appears", that is A = \{HHT, HTH, THH\}. If the outcome is HTT, then A has not occurred — so "not A" has. Collecting every outcome outside A gives A' = .

In set language, A' = \{\omega : \omega \in S \text{ and } \omega \notin A\} = S - A.

2. The event "A or B". The union A \cup B holds everything that is in A, in B, or in . So the event "A or B" is A \cup B = \{\omega : \omega \in A \text{ or } \omega \in B\}: it occurs when at least one of A and B occurs.

3. The event "A and B". The intersection A \cap B keeps only the elements common to A and B, so the event "A and B" is A \cap B = \{\omega : \omega \in A \text{ and } \omega \in B\}: it occurs when both occur. Throw a die twice, let A be "the first throw shows 6" and B be "the sum of the two scores is at least 11". Then B = \{(5, 6), (6, 5), (6, 6)\}, and the event "first throw 6 and total at least 11" is A \cap B = .

4. The event "A but not B". The difference A - B keeps what is in A but not in B. Removing B from A is the same as keeping the part of A that lies in "not B", so A - B = A \cap B'.

Tip

Read the joining word. Or means union, and means intersection, but not means difference, and not means complement.

Example 1 : A die is rolled. Let A be the event "getting a prime number" and B the event "getting an odd number". Write the sets representing the events (i) A or B, (ii) A and B, (iii) A but not B, (iv) not A.

Here A = \{2, 3, 5\} and B = \{1, 3, 5\}. In the diagram, circle A holds the primes and circle B the odd numbers: 3 and 5 sit in the overlap, while 4 and 6 lie outside both.

Shade the region for each event in turn. The readout shows the set you have shaded.

Solution :

(i) "A or B" = A \cup B = \{1, 2, 3, 5\}

(ii) "A and B" = A \cap B = \{3, 5\}

(iii) "A but not B" = A - B = \{2\}

(iv) "not A" = A' = \{1, 4, 6\}

Events That Never Meet

Roll a die. Let A be "an odd number appears" = \{1, 3, 5\} and B be "an even number appears" = \{2, 4, 6\}. Can a single roll make both of them occur?

Yes, a 3 does it
No, no number is both odd and even
Yes, every roll does it

Right: A \cap B = \phi. Two events are called mutually exclusive when the occurrence of either one rules out the other — they can never happen on the same trial. As sets, A and B are then disjoint: A \cap B = \phi.

Now keep A as "an odd number" but let B be "a number less than 4", so B = \{1, 2, 3\}. A roll of lies in both sets, so this A and B are not mutually exclusive.

Note

The simple events of a sample space are always mutually exclusive: two different one-outcome sets, such as \{2\} and \{5\}, can never share an element.

Leaving Nothing Out

Throw a die and take three events: A "a number less than 4" = \{1, 2, 3\}, B "a number greater than 2 but less than 5" = \{3, 4\}, and C "a number greater than 4" = \{5, 6\}. Together they cover the whole sample space:

A \cup B \cup C = \{1, 2, 3, 4, 5, 6\} = S

So whatever you roll, at least of A, B and C must occur. Events like these are called exhaustive events. In general, events E_1, E_2, \ldots, E_n of a sample space S are exhaustive if

E_1 \cup E_2 \cup \ldots \cup E_n = \bigcup_{i=1}^{n} E_i = S

If, in addition, no two of them overlap — E_i \cap E_j = \phi whenever i \neq j — the events are called mutually exclusive and exhaustive: every outcome falls in exactly one of them. A, B and C above are exhaustive, but they are , because 3 is in both A and B.

Common trap

"Mutually exclusive" and "exhaustive" are two separate demands: mutually exclusive means no overlaps, exhaustive means no gaps. On a die, \{1, 2\} and \{3, 4\} have no overlap but leave out 5 and 6; \{1, 2, 3, 4\} and \{3, 4, 5, 6\} cover S but overlap in 3 and 4.

Example 2 : Two dice are thrown and the sum of the numbers on them is noted. Consider the events A: "the sum is even", B: "the sum is a multiple of 3", C: "the sum is less than 4" and D: "the sum is greater than 11". Which pairs of these events are mutually exclusive?

The sample space has 36 outcomes (x, y) with x, y = 1, 2, \ldots, 6; the grid shows the sum for each (first die down the side, second die along the top). Pick any two events. The first one's outcomes turn blue, the second one's green, and outcomes in both turn purple. Test all six pairs.

Solution : Written out, C = \{(1, 1), (1, 2), (2, 1)\} and D = \{(6, 6)\}, while A has 18 outcomes and B has 12. A and B share six outcomes:

\begin{aligned} A \cap B = \{ & (1, 5), (2, 4), (3, 3), \\ & (4, 2), (5, 1), (6, 6)\} \neq \phi \end{aligned}

so A and B are . In the same way A \cap C, A \cap D, B \cap C and B \cap D are all non-empty. Only C \cap D = \phi, so C and D are the only mutually exclusive pair.

Example 3 : A coin is tossed three times. Consider the events A: "no head appears", B: "exactly one head appears" and C: "at least two heads appear". Do they form a set of mutually exclusive and exhaustive events?

If they do, each of the 8 outcomes belongs to exactly one of A, B and C. Test it: drop each outcome into its event.

HHT
THT
TTT
HHH
HTT
THH
TTH
HTH
A: no head
B: exactly one head
C: at least two heads

Solution : The sample space is S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}, and

A = \{TTT\}, B = \{HTT, THT, TTH\}, C = \{HHT, HTH, THH, HHH\}

Since A \cup B \cup C = S, the events A, B and C are . Also A \cap B = A \cap C = B \cap C = \phi, so they are pairwise disjoint, that is, . Hence A, B and C form a set of mutually exclusive and exhaustive events.

Speed Round: Overlaps and Gaps

One die, two events. Look for an overlap, then look for a gap, and make your call.

Get five right to move on.

Sina