Precalculus Counting Probability Lesson

Contents

Key takeaway: This precalculus counting probability lesson explains how to count outcomes before you measure chance. The fundamental counting principle multiplies the choices at each step, permutations count ordered arrangements, and combinations count unordered groups. Probability then becomes favorable outcomes divided by all outcomes. Standards S-CP.B.9 and S-MD.A.2 sit behind the examples, which end with expected value for a game of chance.

The fundamental counting principle multiplies the number of choices at each step. Permutations count ordered arrangements, combinations count unordered selections, and probability is favorable outcomes over total outcomes. The general multiplication rule is P(A and B) = P(A) × P(B given A), and expected value is the sum of each value times its probability.

PrecalculusCommon Core: S-CP.B.8 · S-CP.B.9 · S-MD.A.1 · S-MD.A.2

Precalculus Counting Probability Lesson: key facts

Standards S-CP.B.8, S-CP.B.9, S-MD.A.1 and S-MD.A.2: use the multiplication rule, permutations and combinations, and expected value.
Permutations $_{n}P_{r}=\frac{n!}{(n-r)!}$; combinations $_{n}C_{r}=\frac{n!}{r!\,(n-r)!}$.
General multiplication rule: $P(A\text{ and }B)=P(A)\cdot P(B\mid A)$.
Expected value: $E(X)=\sum x\cdot P(X=x)$; a game is fair when the expected net result is $0.
Two dice have 36 equally likely outcomes, and a standard deck has 52 cards.

What is the fundamental counting principle?

If one task can be done in $m$ ways and a second task in $n$ ways, the two tasks together can be done in $m\times n$ ways. A tree diagram makes the count visible: Maya has 3 shirts and 2 pairs of pants, so she has $3\times 2=6$ outfits.

3 shirts (S1 to S3) and 2 pairs of pants (J, K) give 3 × 2 = 6 outfits.

The same idea works with several decisions. A 4-digit PIN with no repeated digit has $10\cdot 9\cdot 8\cdot 7=5040$ possibilities, because every digit used removes one choice.

One box per decision; multiply the number of choices.

How do permutations and combinations differ?

Permutation and combination

A permutation is an arrangement where order matters: $_{n}P_{r}=\frac{n!}{(n-r)!}$. A combination is a selection where order does not matter: $_{n}C_{r}=\frac{n!}{r!\,(n-r)!}$. Letters repeated $a$, $b$ times in a word of $n$ letters give $\frac{n!}{a!\,b!}$ distinct arrangements.

Gold, silver and bronze among 8 runners is a permutation: $_{8}P_{3}=336$. Choosing 3 of the 8 runners for a relay team is a combination: $_{8}C_{3}=\frac{336}{6}=56$, because each team of 3 was counted $3!=6$ times in the permutations.

Six permutations of the same three letters form one combination.

How do you find a probability by counting?

When all outcomes are equally likely, $P(E)=\frac{\text{favorable outcomes}}{\text{total outcomes}}$, and counting techniques supply both numbers. Two dice have $6\times 6=36$ outcomes, and six of them add to 7, so $P(\text{sum }7)=\frac{6}{36}=\frac{1}{6}$. For two cards drawn from a standard deck, $P(\text{both hearts})=\frac{_{13}C_{2}}{_{52}C_{2}}=\frac{78}{1326}=\frac{1}{17}$.

Each cell is one equally likely outcome.

What is the general multiplication rule?

For any two events, $P(A\text{ and }B)=P(A)\cdot P(B\mid A)$, where $P(B\mid A)$ is the probability of $B$ given that $A$ has happened. Drawing two marbles without replacement from a bag with 5 red and 3 blue marbles changes the second probability, because the bag has changed.

Multiply along each path: the probability of two reds is 5/8 × 4/7 = 5/14.

What is a random variable and its expected value?

A random variable $X$ assigns a number to each outcome. Its expected value is the long-run average, $E(X)=\sum x\cdot P(X=x)$. For the number of heads in two coin flips, $E(X)=0\cdot\frac{1}{4}+1\cdot\frac{1}{2}+2\cdot\frac{1}{4}=1$.

Bar heights are probabilities and add to 1.

Worked example: a game costs $2. Roll a die and win $9 on a six or $3 on a four or five. Then $E=9\cdot\frac{1}{6}+3\cdot\frac{2}{6}=\frac{15}{6}=2.5$, so the expected net result is $2.50-2.00=\$0.50$ per play.

Counting, Permutations, Combinations and Probability: frequently asked questions

How do I know whether to use a permutation or a combination?

Ask whether swapping the chosen items gives a different result. Officers of a club (president, treasurer) form a permutation; a committee of the same size is a combination. Use $_{n}P_{r}$ for the first and $_{n}C_{r}$ for the second.

Why does the second probability change when drawing without replacement?

Removing an item changes both the count of favorable outcomes and the total. With 5 red and 3 blue marbles, the chance of a second red is 4/7 after a red was removed, not 5/8.

What is the general multiplication rule used for?

It finds the probability that two events both happen when they may depend on each other. Multiply the probability of the first event by the probability of the second given the first: $P(A\text{ and }B)=P(A)\cdot P(B\mid A)$.

Does an expected value have to be a possible outcome?

No. Expected value is a long-run average, so it can be a value that never occurs. A die has expected value 3.5 even though no roll gives 3.5.

How do I use combinations to find a probability?

Count the favorable selections with combinations, count all selections with combinations, and divide. For two hearts from a deck, $\frac{_{13}C_{2}}{_{52}C_{2}}=\frac{78}{1326}=\frac{1}{17}$.

How does the counting principle work?

Suppose a lunch offers 3 salads, 4 entrees and 2 drinks. Each choice is independent of the others, so you multiply: $3\cdot 4\cdot 2=24$ meals. A phone PIN of 4 digits has $10^{4}=10{,}000$ possibilities, because repeated digits are allowed. Therefore, always ask whether a choice may repeat before you multiply.

When should you use a permutation or a combination?

Ask one question: does order matter? Awarding first and second place among 6 runners counts $_{6}P_{2}=30$ outcomes, since the two medals differ. Choosing 3 of 8 students for a committee gives $_{8}C_{3}=56$ groups, because the members hold equal roles. Repeated letters need a further division, as in LEVEL, where $\frac{5!}{2!\,2!}=30$.

How do you find a probability by counting?

Count the favorable outcomes, then count all outcomes, and divide. For two dice, 4 of the 36 cells sum to 9, so the chance is $\frac{1}{9}$. Without replacement, the second draw changes the total. With 3 red and 2 blue marbles, two reds have probability $\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{10}$.

What is expected value?

Multiply each outcome by its probability, then add the products. A $3 game that pays $12 on a six has an expected win of $12\cdot\frac{1}{6}=2$, so the net result is $-\$1$ per play.

How can you keep practicing?

Make up one counting question from your own day, such as the ways to order a pizza with toppings. Decide whether order matters, and then solve it twice, once by listing and once by formula. Matching both answers builds trust in the formulas.