Two events, two questions: can they happen together, and does one change the other?
Find the probability
One card is drawn from a standard 52-card deck. What is the probability it is a heart or a face card?
There are 13 hearts and 12 face cards, but 3 cards are both — the jack, queen and king of hearts. So the count is 13 + 12 − 3 = 22, giving 2252 = 1126. A is 2552, which adds 13 and 12 and counts those three cards twice. Whenever two events can happen together, the overlap has to come back out once.
1
"Or" means union. "And" means intersection.
Words
Symbol
Means
A or B
A ∪ B
at least one happens — the union
A and B
A ∩ B
both happen — the intersection
not A
A′
the complement
The Venn diagram shows why: 10 + 3 + 9 = 22 cards are in the union, and only 3 are in both.
With listed outcomes: roll a die, so the sample space is {1, 2, 3, 4, 5, 6}. Let A = {2, 4, 6} and B = {4, 5, 6}.
A ∪ B = {2, 4, 5, 6}
every outcome in A or B, each listed once
A ∩ B = {4, 6}
only the outcomes in both
A′ = {1, 3, 5}
everything in the sample space that is not in A
Two events with no outcomes in common, such as {1, 3} and {2, 4}, are called mutually exclusive: their intersection is empty.
2
The addition rule subtracts the overlap.
P(A or B) = P(A) + P(B) − P(A and B)
P(heart) = 1352
one suit
P(face) = 1252
J, Q, K in four suits
P(both) = 352
J, Q, K of hearts
1352 + 1252 − 352 = 2252
= 1126
Only skip the subtraction when the overlap is empty
Events that cannot happen together are mutually exclusive — a card
cannot be a heart and a spade. Then P(A and B) = 0 and the rule simplifies to plain
addition. Assuming that without checking is how 2552 happens.
3
Complementary events fill the gap to 1.
P(not A) = 1 − P(A)
P(at least one head in 3 tosses)
seven cases to count
P(no heads) = 18
only TTT
= 1 − 18 = 78
one subtraction instead
"At least one" is the signal to use the complement. Counting the opposite is nearly always the shorter route.
4
Independent: the first event changes nothing.
P(A and B) = P(A) × P(B), when they are independent
a die shows 6 and a coin shows heads
the die tells the coin nothing
(16)(12) = 112
multiply
Tossing a coin, rolling a die, spinning a spinner, or drawing a card and putting
it back — all independent. Nothing carries over.
5
Dependent: the first event changes the second.
Once something is removed and not replaced, the pool has shrunk.
bag: 5 red, 3 blue
draw two, no replacement
P(first red) = 58
eight marbles
P(second red) = 47
one red gone, one marble gone
(58)(47) = 2056 = 514
still multiply
With replacement(58)(58) = 2564independent
Without replacement(58)(47) = 514dependent
Both denominators change, not just the numerator. Forgetting to drop 8 to 7 is the
classic error here.
Remember the flow
"Or" → add and subtract the overlap · "And" → multiply · "At least one" → use the complement · Ask whether the pool shrank
Replacement is the word that decides between independent and dependent.
6
Mutually exclusive is not the same as independent.
Mutually exclusive: A and B cannot both happen, so P(A and B) = 0 and P(A or B) = P(A) + P(B).
One die is rolled. What is P(2 or 5)?
one roll cannot show both 2 and 5
mutually exclusive: no overlap
16 + 16 − 0 = 13
the addition rule, overlap 0
A learner takes the bus with probability 0.45 and walks with probability 0.3 (never both). What is P(bus or walk)?
0.45 + 0.3 = 0.75
no overlap to subtract
Multiplying here gives a probability for something impossible
Treating 2 and 5 as independent and multiplying gives (16)(16) = 136 for "2 and 5 on one roll" — but that cannot happen at all, so its probability is 0. Mutually exclusive events are as far from independent as they can be: if one happens, the other certainly did not.
7
Variations: the overlap as the unknown, three draws, "at least one" with unequal odds.
The same three rules, used backwards or stretched.
The overlap is unknown: P(A) = 0.6, P(B) = 0.5, P(A or B) = 0.8. Find P(A and B).
0.8 = 0.6 + 0.5 − P(A and B)
the addition rule
P(A and B) = 1.1 − 0.8 = 0.3
and 0.6 × 0.5 = 0.3, so A and B are independent too
Different: P(A) + P(B) is more than 1, which already tells us the events must overlap.
Three draws: 3 learners are picked from 4 girls and 6 boys. P(all three girls)?
(410)(39)(28)
the pool shrinks each time
= 24720 = 130
simplify at the end
Different: with three draws both the top and the bottom drop by one at every step.
Unequal odds: a seed sprouts with probability 0.8. Two seeds are planted. P(at least one sprouts)?
P(neither) = 0.2 × 0.2 = 0.04
each fails with 1 − 0.8 = 0.2
1 − 0.04 = 0.96
the complement
Different: the outcomes are not equally likely, but "at least one = 1 − none" still works.
On the exam
Expect an "or" item with an overlap (cards are the favourite), an "and" item with or without replacement, an "at least one" item, and a complement. The fastest route is to read the connective: "or" adds and removes the overlap, "and" multiplies, "at least one" uses 1 − none. The 10-second check: every probability lies between 0 and 1, and an "or" answer is at least as big as each event alone.
Now you try
Practice problem
Find the probability
A box holds 4 green and 6 yellow marbles. Two are drawn without replacement. What is the probability that both are yellow?
P(first yellow) = 610. After one yellow is removed, 5 yellow remain out of 9, so P(second yellow) = 59. Multiplying gives 3090 = 13. B is 610 × 610, which is the answer with replacement — it forgets that both counts drop. D is above 1, which is impossible for any probability and comes from adding instead of multiplying.
Hint: "without replacement" means both the top and the bottom of the
second fraction go down by one. Write the second fraction out before multiplying.
