"Or" adds. "And" multiplies. The rest is checking whether the events interfere.
Find the probability
A bag holds 4 red and 6 blue marbles. Two are drawn without replacement. What is P(both red)?
P(first red) = 410. With one red gone and one marble fewer, P(second red) = 39. Multiplying: (410)(39) = 1290 = 215. A is (410)², the answer with replacement — it forgets that both the numerator and the denominator drop. B is just P(first red), which answers a one-draw question.
1
"Or" means add — minus any overlap.
P(A or B) = P(A) + P(B) − P(A and B)
a card: queen or red card?
they overlap
452 + 2652 − 252
two red queens are both
= 2852 = 713
simplify
Only skip the subtraction if the overlap is empty
Mutually exclusive events cannot both happen — a card is not a heart and a spade —
so P(A and B) = 0 and plain addition works. Assuming that without checking is where
3052 comes from.
The deck: 52 cards in 4 suits (hearts, diamonds, clubs, spades) of 13 each. Every suit has an ace, nine number cards 2 to 10, and three face cards (jack, queen, king) — so 36 number cards and 12 face cards in all. Hearts and diamonds are red; clubs and spades are black.
2
"And" means multiply.
P(A and B) = P(A) × P(B), if they are independent
a coin and a die
P(head and a 6)
(12)(16) = 112
independent
Multiplying two probabilities always gives something smaller, which makes sense —
demanding two things at once is harder than demanding one.
3
Independent or dependent: did the pool change?
With replacement(410)(410) = 425independent
Without replacement(410)(39) = 215dependent
Coins, dice, spinnersalways independentnothing is used up
Both numbers change
Going from 410 to 39, the top drops because a red marble was removed and the
bottom drops because a marble was removed. Changing only the numerator is the
standard error.
4
"At least one" — use the complement.
P(at least one) = 1 − P(none)
roll a die three times
P(at least one six)?
P(no six each roll) = 56
the opposite
P(no sixes at all) = (56)³ = 125216
multiply
= 1 − 125216 = 91216
≈ 0.42
Counting "at least one" directly means adding the cases for exactly one, exactly two and exactly three. Counting the single opposite case is far shorter, and the phrase "at least" is the signal to do it.
5
Deciding which rule applies.
The wording
Do this
"A or B"
add, then subtract the overlap
"A and B", or "both"
multiply
"at least one"
1 − P(none)
"neither"
P(not A) × P(not B)
"without replacement"
adjust the second fraction
Final checkis the answer between 0 and 1?if not, something is wrong
"And" answerssmaller than either alonetwo conditions is harder
"Or" answersbigger than either alonemore ways to succeed
Remember the flow
Read for "and", "or" or "at least" · Check for overlap · Check for replacement · Multiply or add · Confirm it sits between 0 and 1
"Without replacement" changes both the top and the bottom of the second fraction.
6
Two dice: list the 36 outcomes, not the 11 sums.
Two dice give 6 × 6 = 36 equally likely ordered pairs. Count the pairs that work.
Two fair dice are rolled. Find P(the sum is 8).
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
first die, second die
5 favourable out of 36
a 1 cannot reach 8
P(sum 8) = 536
check: between 0 and 1 ✓
Treating (3, 5) and (5, 3) as one outcome
Listing only {2, 6}, {3, 5}, {4, 4} gives 336. But a 3 then a 5 and a 5 then a 3 are different rolls of two separate dice. Only a double like (4, 4) is counted once. And "11 possible sums, so 111" fails too: the sums are not equally likely.
7
Variations: "neither", one of each colour, an overlap that is a whole event.
The same add-and-multiply rules where the wording changes the setup.
Independent events with P(A) = 0.3 and P(B) = 0.5. Find P(neither).
A bag has 6 red and 4 blue marbles. Two are drawn without replacement. Find P(one of each colour).
red then blue: (610)(49) = 2490
one order
blue then red: (410)(69) = 2490
the other order
4890 = 815
the two orders cannot both happen, so add
"One of each" hides two orders. Counting only red-then-blue gives 415, half the answer.
A die is rolled. Find P(an even number or a 2).
36 + 16 − 16
the overlap is the 2 itself
= 36 = 12
same as P(even)
When one event sits entirely inside the other, "or" adds nothing new: the answer is the larger event.
On the exam
Expect two draws with or without replacement, an "or" question with an overlap such as a card that is both, an "at least one" question, and two dice with a sum. The fastest route is to read the connective — "and" multiplies, "or" adds then subtracts the overlap, "at least one" is 1 − P(none) — and to adjust both top and bottom for a draw without replacement. The 10-second check: an "and" answer is smaller than either part, an "or" answer is larger, and every answer lies between 0 and 1.
Now you try
Practice problem
Find the probability
A coin is tossed four times. What is the probability of getting at least one head?
Use the complement. The only outcome with no heads is TTTT, with probability (12)⁴ = 116. So P(at least one head) = 1 − 116 = 1516. C is the complement itself — the probability of no heads, which is what you subtract rather than the answer. Sensible: with four tosses, getting at least one head is very likely.
Hint: "at least one" is the signal to count the opposite. There is
exactly one way to get no heads at all.
