5.3 - Combining events

5.3 - Combining events

Two fair dice have eleven possible totals, but those totals are not equally likely. In this first part of 5.3, you will count equally likely elementary outcomes, combine alternatives without double counting, and distinguish independent events from mutually exclusive events.

Equally likely outcomes

Is a total of 88 as likely as a total of 22 when two fair dice are thrown independently? A total of 22 requires (1,1)(1,1), while a total of 88 can be made in several ways. Listing totals alone hides this difference.

Record an elementary outcome as (a,b)(a,b), where aa is the first die's score and bb the second die's score. There are 6×6=366\times6=36 equally likely ordered pairs. For example, (2,6)(2,6) and (6,2)(6,2) are different outcomes, although both have total 88.

Sample space and event

The sample space is the set of all possible elementary outcomes of an experiment. An event is a set of these outcomes. The probability of an event lies between 00 and 11, inclusive: 00 for an impossible event and 11 for a certain event.

The grid keeps the two scores separate. Its cells are equally likely; the totals written inside them are not.

[DIAGRAM: asset_name: Two-dice elementary outcomes; asset_slug: probability-dice-outcomes; caption: Five of the 36 equally likely ordered pairs give a total of 8; description: A six by six grid records first die score on the vertical axis and second die score on the horizontal axis. Each cell shows the sum of its two scores. The five cells with total 8 are highlighted: first and second scores 2 and 6, 3 and 5, 4 and 4, 5 and 3, 6 and 2. A note distinguishes the 36 equally likely pairs from the eleven unequal-probability totals.]
Five of the 36 equally likely ordered pairs give a total of 8

Counting equally likely outcomes

If a finite sample space contains NN equally likely elementary outcomes, and mm of them belong to event AA, then

P(A)=mN.\mathrm{P}(A)=\frac{m}{N}.

The equal-likelihood condition is essential.
Not in the formula booklet: learn it.

For a biased die, counting favourable faces and dividing by six would generally be wrong. The probabilities of the faces would have to be supplied or modelled.

Worked example 1: a total of eight

Question. Two fair dice are thrown independently. Find the probability that their total is 88.

Plan. Count the ordered pairs with the required total out of all 3636 pairs.

The favourable pairs are

(2,6), (3,5), (4,4), (5,3), (6,2).(2,6),\ (3,5),\ (4,4),\ (5,3),\ (6,2).

Answer. 5/365/36.

Check. The first score can be any of 2,3,4,5,62,3,4,5,6, and each has exactly one matching second score. This confirms five favourable cells.

Change only the required total. Keeping the same sample space makes the probability comparison immediate.

Worked example 2: a total of nine

Question. The same two dice are thrown. Find the probability that their total is 99.

Plan. Find the ordered pairs with sum 99 in the same 3636-outcome space.

The favourable pairs are

(3,6), (4,5), (5,4), (6,3).(3,6),\ (4,5),\ (5,4),\ (6,3).

Answer. 4/36=1/94/36=1/9.

Check. Reflecting the five sum-88 pairs by replacing each score dd with 7−d7-d gives the five sum-66 pairs. Sum 99 instead matches sum 55, which has four pairs: (1,4),(2,3),(3,2),(4,1)(1,4),(2,3),(3,2),(4,1). The different counts confirm that totals cannot all have probability 1/111/11.

For a random selection of objects without replacement, unordered selections can give an equally likely sample space too. Distinguish the physical objects even when some share a colour: two different red balls are still two different balls.

Worked example 3: an unordered ball selection

Question. A bag contains three red and two blue balls. Two balls are selected at random without replacement. Find the probability of selecting one of each colour.

Plan. Count equally likely pairs of distinct balls, then count the pairs with one red and one blue.

all pairs=(52)=10,one of each=(31)(21)=6.\text{all pairs}=\binom52=10,\qquad \text{one of each}=\binom31\binom21=6.

Answer. 6/10=3/56/10=3/5.

Check. Label the balls R1,R2,R3,B1,B2R_1,R_2,R_3,B_1,B_2. Each of the three red balls makes two favourable pairs, giving six. The other four pairs are three red-red pairs and one blue-blue pair, accounting for all ten.

Change only the replacement rule. If the first ball is returned and the bag mixed before the second draw, the same physical ball can appear twice. Ordered draw pairs are now a convenient equally likely sample space.

Worked example 4: returning the first ball

Question. From the same bag, select a ball at random, replace it, mix the bag and select again. Find the probability of one ball of each colour.

Plan. Count the 5×55\times5 ordered physical-ball pairs, including repeats.

There are 3×2=63\times2=6 red-blue pairs and 2×3=62\times3=6 blue-red pairs. These are disjoint outcomes.

P(one of each)=6+625=1225.\mathrm{P}(\text{one of each})=\frac{6+6}{25}=\frac{12}{25}.

Answer. 12/2512/25.

Check. The remaining ordered pairs are 32=93^2=9 red-red and 22=42^2=4 blue-blue. The counts 12+9+4=2512+9+4=25 exhaust the sample space.

With replacement, an unordered pair containing one particular ball twice is not as likely as an unordered pair of two different balls: the latter can occur in two orders. Using ordered elementary outcomes avoids that trap.

Watch out

Red-red, red-blue and blue-blue are possible colour descriptions without replacement, but they are not three equally likely elementary outcomes. Count labelled objects first, or justify the probabilities by another valid method.

Use combinations for a new selection without replacement.

Adding alternatives

“Event A or event B” normally includes outcomes where both occur. You can add the probabilities directly only when the two events have no outcomes in common. Otherwise separate the shared outcomes so that they are counted once.

Union, intersection and complement

A∪BA\cup B, the union, means A or B or both. A∩BA\cap B, the intersection, means both A and B. A′A', the complement of A, means A does not occur. Events are mutually exclusive if they have no outcomes in common.

For example, one card cannot simultaneously show an even number and an odd number. Those events are mutually exclusive. A card can show both an even number and a multiple of three, such as 66; those events overlap.

Disjoint alternatives and complements

For mutually exclusive events A and B,

P(A∪B)=P(A)+P(B).\mathrm{P}(A\cup B)=\mathrm{P}(A)+\mathrm{P}(B).

For any event A,

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

Not in the formula booklet: learn the counting reasoning.

The same addition principle applies to any finite list of disjoint alternatives. The complement is useful for “at least one”: remove the case “none”, rather than adding many possibilities.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account