Probability and Counting

Arithmetic Reasoning asks probability inside a story. Mathematics Knowledge asks it directly, adds the vocabulary, and pushes a little further into compound events.

Single events

P(event) = favorable outcomes / total outcomes.

A standard die has 6 faces. The probability of rolling a 4 is 1/6. The probability of rolling an even number is 3/6, which reduces to 1/2.

Three habits:

Count the whole sample space. The denominator is every possible outcome, not just the ones you are not interested in.

Reduce. Answer choices are in lowest terms.

Check the range. Every probability lies between 0 and 1. An impossible event has probability 0; a certain one has 1. A negative probability, or one above 1, means the setup is wrong.

The complement

P(not A) = 1 - P(A).

If the chance of rain is 0.3, the chance of no rain is 0.7. This is the single most useful identity in the topic, and the "at least one" section below is why.

Compound events: and

Multiply the probabilities in sequence.

Two dice are rolled. What is the probability both show a 6?

Each die is independent - the first roll does not change the second - so 1/6 times 1/6 is 1/36.

A coin is flipped three times. What is the probability of three heads?

1/2 times 1/2 times 1/2 is 1/8.

An "and" answer is always smaller than either individual probability, because you are multiplying by a number less than 1. If your answer came out larger, you added.

Compound events: or

Add the probabilities, then subtract the overlap.

P(A or B) = P(A) + P(B) - P(both)

One card is drawn from 52. What is the probability it is a king or a heart?

• Kings: 4/52.
• Hearts: 13/52.
• Both - the king of hearts: 1/52.

4 plus 13 minus 1 is 16, so 16/52, which reduces to 4/13.

When events cannot both happen they are mutually exclusive, the overlap is zero, and you just add. Rolling a 2 or a 5 on one die is 1/6 plus 1/6, which is 1/3, because a single roll cannot be both.

The overlap term exists precisely because some pairs of events can co-occur, and forgetting it double-counts.

Independent against dependent

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Independent: the first outcome does not change the second. Coin flips, dice rolls, and any draw where the item is replaced.

Dependent: the first outcome changes the pool. Any draw where the item is kept.

A bag has 5 red and 3 blue marbles. Two are drawn without replacement. What is the probability both are red?

First: 5/8. Second: only 4 reds remain among 7 marbles, so 4/7.

Both: 5/8 times 4/7 is 20/56, which reduces to 5/14.

With replacement it would have been 5/8 times 5/8, which is 25/64. Both values appear as choices, and the words "without replacement" or "and keeps it" are what separate them.

A coin has no memory. After five heads in a row, the next flip is still 1/2. This is asked directly and the intuition that a tail is "due" is the thing being tested.

At least one

Use the complement. The opposite of "at least one" is "none", and "none" is a single multiplication.

Three coins are flipped. What is the probability of at least one head?

The long way is four separate cases. The short way:

• P(no heads) is 1/2 cubed, which is 1/8.
• P(at least one head) is 1 minus 1/8, which is 7/8.

Whenever you see "at least one", reach for this.

Counting

Multiply the number of options at each stage.

Four shirts, three ties and two jackets give 4 times 3 times 2, which is 24 outfits.

When choices come from one shrinking pool and order matters, the pool decreases at each stage:

How many ways can 3 people be arranged from a group of 6 into first, second and third place?

6 times 5 times 4 is 120.

When order does not matter - a committee rather than a ranking - each group has been counted once for every way of ordering it, so divide by the number of those orderings.

How many committees of 3 can be formed from 6 people?

The 120 ordered arrangements each represent the same committee counted 3 times 2 times 1, which is 6 ways. So 120 divided by 6 is 20.

Order matters, divide by nothing. Order does not matter, divide by the arrangements of the group you chose. That is the distinction, and it is the only place this topic gets harder than Arithmetic Reasoning.

What you can skip

Across the 60 questions on this topic:

• Permutation and combination formulas. Neither word appears as a method, and factorial notation never appears. The counting questions are the multiplication principle.
• Expected value and odds never appear.

Where people lose points

Adding for "and" or multiplying for "or".

Forgetting the overlap on an "or" question with events that can co-occur.

Not shrinking the pool on a draw without replacement.

Believing a run of results changes the next independent trial.

Listing every case for "at least one" instead of using the complement, then running out of time or missing a case.

Treating an unordered selection as ordered, which overcounts by a factor.

Work one in under a minute

A drawer holds 4 black and 6 brown socks. Two are drawn at random without replacement. What is the probability they match?

Two ways to match, and they are mutually exclusive, so add.

• Both black: 4/10 times 3/9, which is 12/90.
• Both brown: 6/10 times 5/9, which is 30/90.

Total: 42/90, which reduces to 7/15.

Check: it is under 1 and over the chance of either single case, as an "or" of two cases must be.

Where this leads

The counting rule and the complement are the two ideas that make an unfamiliar probability question tractable, and both carry straight back into the word problems on Arithmetic Reasoning.