A. Counting and Probability · Probability

Mutually Exclusive and Non-Mutually Exclusive Events

A1.5About 65 minutes

Learning goals

I can decide whether two or more events are mutually exclusive.
I can use the additive principle P(A ∪ B) = P(A) + P(B) for mutually exclusive events.
I can use P(A ∪ B) = P(A) + P(B) − P(A ∩ B) for events that are not mutually exclusive.
I can choose a strategy (Venn diagram, list, formula) to solve probability problems involving "or", "and" and "not".

Words to know

mutually exclusive events
Events that cannot happen at the same time: A ∩ B = ∅, so P(A ∩ B) = 0. Also called disjoint events.

non-mutually exclusive events
Events that can happen at the same time: they share at least one outcome.

additive principle for probability
For mutually exclusive events, P(A ∪ B) = P(A) + P(B).

general addition rule
For any events, P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Can they happen together?

Two events are mutually exclusive if they cannot both happen on the same trial. Their circles in a Venn diagram do not overlap: A ∩ B = ∅, so P(A ∩ B) = 0.

Rolling a 2 and rolling a 5 on one die: mutually exclusive.
Drawing a red card and drawing a club: mutually exclusive (clubs are black).
Drawing a heart and drawing a face card: not mutually exclusive (the J♥, Q♥ and K♥ are both).

The curriculum gives a nice example: toss a coin 5 times. "An even number of heads" and "an odd number of heads" are mutually exclusive, because the number of heads cannot be both even and odd. (They are also complements, since every count is one or the other.)

Left: a sum of 7 and doubles can never happen on the same roll of two dice, so the circles do not overlap. Right: a sum of 8 and doubles share the outcome (4, 4), so the circles overlap; 4 outcomes give a sum of 8 that is not a double, and 5 doubles do not sum to 8.

Adding probabilities

Mutually exclusive events (additive principle): P(A ∪ B) = P(A) + P(B). This extends to any number of mutually exclusive events: P(A ∪ B ∪ C) = P(A) + P(B) + P(C).

Non-mutually exclusive events (general addition rule): P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

The general rule always works. When the events are mutually exclusive, P(A ∩ B) = 0 and it becomes the simple version. The danger is using the simple version when the events overlap: you count the overlap twice and get an answer that is too big (sometimes bigger than 1!).

Worked example

Example 1: Mutually exclusive or not?

Decide whether the events are mutually exclusive.

(a) Two dice: A = sum is 7, B = doubles. (b) One card: A = heart, B = face card. (c) Five coin tosses: A = even number of heads, B = odd number of heads. (d) A student chosen at random: A = takes MDM4U, B = takes MHF4U.

(a) Doubles have even sums (2, 4, …, 12), so a double can never sum to 7. Mutually exclusive.
(b) The J♥, Q♥ and K♥ are in both. Not mutually exclusive.
(c) A count of heads cannot be both even and odd. Mutually exclusive.
(d) Many students take both courses. Not mutually exclusive.

(a) yes, (b) no, (c) yes, (d) no.

Worked example

Example 2: A card that is a heart or a face card

One card is drawn from a standard deck. Find P(heart or face card).

P(heart) = 13/52 and P(face card) = 12/52.
The events overlap: J♥, Q♥, K♥, so P(heart ∩ face) = 3/52.
P(heart ∪ face) = 13/52 + 12/52 − 3/52 = 22/52.
Check by counting: 13 hearts plus the 9 face cards in other suits = 22. ✓

P(heart or face card) = 22/52 = 11/26 ≈ 0.423.

Worked example

Example 3: Two dice, two ways

Two dice are rolled. Find (a) P(sum is 7 or doubles), (b) P(sum is 8 or doubles).

(a) Mutually exclusive. P(sum 7) = 6/36 and P(doubles) = 6/36. P = 6/36 + 6/36 = 12/36 = 1/3.
(b) Not mutually exclusive: (4, 4) is in both. P(sum 8) = 5/36, P(doubles) = 6/36, P(both) = 1/36.
P = 5/36 + 6/36 − 1/36 = 10/36 = 5/18.

(a) 1/3, (b) 5/18.

Worked example

Example 4: Even or odd number of heads

A coin is tossed 5 times. Find P(an even number of heads), counting 0 as even. Then find P(0 heads or 5 heads).

n(S) = 2⁵ = 32 equally likely sequences.
Number of sequences with k heads: 0 heads: 1; 1: 5; 2: 10; 3: 10; 4: 5; 5: 1. (Check: total 32.) You can find these by listing, and the unit Permutations and Combinations gives a fast way: they are the numbers in row 5 of Pascal's triangle.
Even (0, 2 or 4 heads): these are mutually exclusive, so add: 1 + 10 + 5 = 16. P = 16/32 = 1/2.
0 heads or 5 heads: mutually exclusive, so P = 1/32 + 1/32 = 2/32 = 1/16.

P(even number of heads) = 1/2 and P(0 or 5 heads) = 1/16.

The number of ways to get each number of heads in 5 coin tosses. The even counts (0, 2, 4) and odd counts (1, 3, 5) are mutually exclusive events, and each has 16 of the 32 outcomes.

Mutually exclusive is not the same as independent

Mutually exclusive means "cannot happen together". Independent (lesson 10) means "one does not change the probability of the other". In fact, if A and B are mutually exclusive and both have non-zero probability, knowing A happened tells you B definitely did not, so they are dependent. Keep the two ideas separate.

Just for fun

A pun for the road: two mutually exclusive events walk into a bar. Only one of them gets served, because they never show up together.

Hands-on activity

Deck sort: overlaps and gaps

You need: a standard deck of cards and a table.

Make two piles: event A = "red card", event B = "face card". Some cards belong in both. Lay out two overlapping loops of string (or two sheets of paper that overlap) and sort all 52 cards into the regions: A only, both, B only, neither.
Count each region and check P(A ∪ B) with the general addition rule.
Now use A = "club" and B = "diamond". Sort again. What do you notice about the overlap?
Make up a pair of events that are mutually exclusive and a pair that are not, and test them by sorting.

What to look for: red only 20, both 6 (the red J, Q, K), face only 6 (the black J, Q, K), neither 20. P(red ∪ face) = 26/52 + 12/52 − 6/52 = 32/52 = 8/13. For clubs and diamonds the overlap is empty: they are mutually exclusive, and P = 13/52 + 13/52 = 1/2.

Your turn

One die is rolled. Are "roll an even number" and "roll a 5" mutually exclusive? Find P(even or 5).

One die is rolled. Find P(even or greater than 3).

One card is drawn. Find P(ace or king).

One card is drawn. Find P(spade or ace).

P(A) = 0.25, P(B) = 0.4, A and B mutually exclusive. Find P(A ∪ B) and P(A ∩ B).

P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Find P(A ∪ B). Are A and B mutually exclusive?

Two dice are rolled. Find P(sum is 2, 3 or 12).

More practice: warm-up

Are "heads" and "tails" on one coin toss mutually exclusive?

If A and B are mutually exclusive, what is P(A ∩ B)?

Mutually exclusive, P(A) = 1/5, P(B) = 1/3. P(A or B)?

Are "red card" and "queen" mutually exclusive?

Write the general addition rule.

More practice: on your own

Two dice are rolled. Find P(sum is at least 10 or doubles).

In a group of students, P(plays an instrument) = 0.35, P(plays a sport) = 0.55, P(both) = 0.2. Find P(neither).

A student says P(A) = 0.7 and P(B) = 0.6, and that A and B are mutually exclusive. Explain why this is impossible.

Show that "an even number of heads" and "an odd number of heads" in 3 tosses each have probability 1/2.

One card is drawn. Find P(a red card or a card with rank 2 to 5).

Challenge

Prove that if A and B are mutually exclusive with P(A) > 0 and P(B) > 0, then P(A ∩ B) ≠ P(A) × P(B).

Write a formula for P(A ∪ B ∪ C) for three events that may overlap, and use it: one card is drawn; A = heart, B = face card, C = ace. Find P(A ∪ B ∪ C).

A coin is tossed 6 times. Show that P(even number of heads) = 1/2. (Counts of heads 0 to 6: 1, 6, 15, 20, 15, 6, 1.)

Check your understanding

Quiz

Which pair of events is mutually exclusive when one card is drawn?

One card is drawn. What is P(diamond or queen)?

P(A) = 0.42, P(B) = 0.33 and A and B are mutually exclusive. What is P(A ∪ B)?

Two dice are rolled. What is P(sum is 6 or doubles)?

A coin is tossed 5 times. Which pair of events is mutually exclusive?

P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.9. What can you conclude?

For parents and teachers

Big idea (A1.5, mutually exclusive events): events that cannot happen together are mutually exclusive, and their probabilities simply add. For events that can overlap, subtract the overlap: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

How to help: The card-sorting activity makes the overlap visible. Ask "Can both happen at once?" for everyday pairs (raining and snowing at the same time: yes, it can!).

Common difficulties:

Adding probabilities of overlapping events without subtracting the overlap.
Confusing mutually exclusive with independent (lesson 10).
Forgetting that 0 is an even number in problems about even counts.
Not checking that the answer is at most 1.