Free ACT Math Statistics and Probability

ACT Math · Statistics and Probability · free lesson

Probability

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Outcomes and events

A probability is a number from 0 to 1 that says how likely an event is: 0 means it cannot happen, 1 means it is certain. When all the outcomes are equally likely, it is a count divided by a count.

Definition.

$P(\text{event}) = \dfrac{\text{number of outcomes in the event}}{\text{number of possible outcomes}}$ , when every outcome is equally likely.

A whole number from 1 to 20 is chosen at random. What is the probability that it is prime? The primes are circled below; the engine counts them, $k$ , and divides:

\begin{aligned} k &= \operatorname{piSieve}(20) \\ &= 8 \\ p &= \frac{k}{20} \\ &= \frac{8}{20} \\ &= \frac{2}{5} \end{aligned}

The probability is $\frac{2}{5}$ . The number 1 is not prime: counting it gives $\frac{9}{20}$ , a usual wrong option. Counting the outcomes is the whole difficulty when there are many; topic 44 (Counting) has the tools for that.

Not, or

An event either happens or it does not, so $P(\text{not } A) = 1 - P(A)$ . This is the complement rule. The probability that the number is not prime is $q$ :

\begin{aligned} q &= 1 - p \\ &= 1 - \frac{2}{5} \\ &= \frac{3}{5} \end{aligned}

For " $A$ or $B$ " (one or the other or both), add the two probabilities and take off the overlap, which was counted twice:

Rule.

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ . When $A$ and $B$ cannot happen together, they are mutually exclusive: the overlap is 0 and the probabilities simply add.

From 1 to 20, the number is prime with probability $\frac{2}{5}$ and even with probability $\frac{10}{20}$ ; it is both only for 2. So "prime or even" has probability $r$ :

\begin{aligned} r &= \frac{k}{20} + \frac{10}{20} - \frac{1}{20} \\ &= \frac{8}{20} + \frac{10}{20} - \frac{1}{20} \\ &= \frac{17}{20} \end{aligned}

Two-way tables and "given that"

The ACT often gives counts in a two-way table. Here 200 students are sorted by grade and by whether they play a sport.

	Plays a sport	Does not	Total
Grade 11	$48$	$52$	$100$
Grade 12	$36$	$64$	$100$
Total	$84$	$116$	$200$

A student is chosen at random from all 200. The probability of a grade 12 student who plays a sport uses one cell over the grand total, $u$ . But if you are told the student is in grade 12, only the grade 12 row counts: this is a conditional probability, written $P(\text{sport} \mid \text{grade 12})$ , $v$ . Turned round, of the students who play a sport, the share in grade 12 is $w$ :

\begin{aligned} u &= \frac{c}{a + b + c + d} \\ &= \frac{36}{48 + 52 + 36 + 64} \\ &= \frac{9}{50} \\ v &= \frac{c}{c + d} \\ &= \frac{36}{36 + 64} \\ &= \frac{9}{25} \\ w &= \frac{c}{a + c} \\ &= \frac{36}{48 + 36} \\ &= \frac{3}{7} \end{aligned}

Warning.

The three answers differ only in the bottom number: the grand total, the row total or the column total. The words after "given that", "of the" or "if the student is" choose it.

"And": a tree diagram

Events are independent when one happening does not change the probability of the other, as with two coin tosses. Then $P(A \text{ and } B) = P(A) \cdot P(B)$ . When the first changes the second, multiply by the conditional probability instead.

A bag holds 3 red and 2 blue marbles. Two are drawn, one after the other, without putting the first back. Watch the tree grow: the first draw, then the second, with the probabilities on the branches. After a red, only 2 of the 4 marbles left are red.

Multiply along a path for "and"; add the paths for "or". Both red is $s$ ; one of each colour, in either order, is $m$ :

\begin{aligned} s &= \frac{3}{5} \times \frac{2}{4} \\ &= \frac{3}{10} \\ m &= \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} \\ &= \frac{3}{5} \end{aligned}

With the first marble put back, the draws would be independent and both red would be $\frac{3}{5} \cdot \frac{3}{5}$ , which is $\frac{9}{25}$ , not $\frac{3}{10}$ .

Probability from areas

If a point is chosen at random in a region, the probability that it lands in a smaller part is the part's area over the whole area. A dart lands at random on a square board 10 inches on a side with a circle of radius $\rho$ inches at its center. Drag the slider: the probability is $\frac{\pi\rho^2}{100}$ .

For a radius of 3 inches the probability $G$ is $\frac{9\pi}{100}$ ; the engine gives it to three places:

\begin{aligned} G &= \operatorname{decimal}(\frac{\pi \times 3^{2}}{10^{2}}, 3) \\ &= 0.283 \end{aligned}

Note.

On the ACT this topic asks: a simple probability from a list or a range (watch the number 1 with primes); the complement; a probability from a two-way table, often "given that"; "and" and "or", and whether events are mutually exclusive or independent; experimental against theoretical probability (a spinner landing on red 30 times in 100 spins gives an experimental probability of 0.30, which may differ from the designed one); a sample space you must build, such as the 36 rolls of two dice; and probability from areas. Expected value, the average outcome of a game, is topic 46.

Worked examples

Each example is one of the kinds of question the ACT asks about probability. The numbers in the working are found by the engine.

Example.

1. Building the sample space. Two fair six-sided dice are rolled. What is the probability that the sum of the two numbers is 8?

The question gives no list of outcomes, so build one. The first die shows 1 to 6 and so does the second: every pair is equally likely, and the grid has $6 \times 6$ cells. The cells with sum 8 are shaded.

For a first die $a$ , the second must show $8 - a$ , which is a face of the die only when $a$ is 2 or more. The engine counts the first faces that work, $k$ , and divides by the number of cells, $N$ :

\begin{aligned} k &= \operatorname{sum}(a \ge 2) \\ &= 5 \\ N &= 6 \times 6 \\ &= 36 \\ p &= \frac{k}{N} \\ &= \frac{5}{36} \end{aligned}

The probability is $\frac{5}{36}$ . Listing sums instead of pairs (2 to 12, eleven sums) and answering $\frac{1}{11}$ is the trap: the sums are not equally likely.

Example.

2. Given that, from a two-way table. A town surveyed 250 adults about how they get to work.

	Drives	Takes the bus	Total
Under 40	$54$	$66$	$120$
40 or over	$81$	$49$	$130$
Total	$135$	$115$	$250$

If a person who takes the bus is chosen at random, what is the probability that the person is under 40?

"Who takes the bus" is the given: only the bus column counts, so the bottom number is that column's total. The top is the cell in that column for under 40. The probability is $p$ :

\begin{aligned} p &= \frac{b}{b + d} \\ &= \frac{66}{66 + 49} \\ &= \frac{66}{115} \end{aligned}

The answer is $\frac{66}{115}$ . The wrong totals give the usual wrong options: over the under-40 row, $\frac{11}{20}$ ; over everyone, $\frac{33}{125}$ .

Example.

3. "And" and "or" for independent events. A spinner has 5 equal sections, 2 of them red. It is spun once and a fair coin is tossed. What is the probability of red and heads? Of red or heads?

The spinner does not affect the coin, so the events are independent and "and" multiplies. They are not mutually exclusive (both can happen), so "or" subtracts the overlap:

\begin{aligned} r &= \frac{2}{5} \\ h &= \frac{1}{2} \\ u &= rh \\ &= \frac{2}{5} \times \frac{1}{2} \\ &= \frac{1}{5} \\ v &= r + h - u \\ &= \frac{2}{5} + \frac{1}{2} - \frac{1}{5} \\ &= \frac{7}{10} \end{aligned}

Red and heads has probability $\frac{1}{5}$ ; red or heads, $\frac{7}{10}$ . Adding without taking off the overlap gives $\frac{9}{10}$ , which is too big: it counts red-and-heads twice. A sum above 1 would show the slip at once.

Example.

4. At least one: use the complement. A fair die is rolled 3 times. What is the probability of at least one 6?

"At least one" covers many cases (one 6, two, three). Its complement is a single case: no 6 at all. Each roll misses 6 with probability $\frac{5}{6}$ , and the rolls are independent, so no 6 in three rolls is $z$ , and at least one 6 is $p$ :

\begin{aligned} z &= (\frac{5}{6})^{3} \\ &= \frac{125}{216} \\ p &= 1 - z \\ &= 1 - \frac{125}{216} \\ &= \frac{91}{216} \end{aligned}

The probability is $\frac{91}{216}$ , about $0.421$ . Adding $\frac{1}{6}$ three times gives $\frac{1}{2}$ , which is wrong: it counts the rolls with two or three sixes more than once.

Example.

5. Experimental against theoretical. A spinner is designed with 4 equal sections: red, blue, green and yellow. Ana spins it 200 times and records the results.

Color	Red	Blue	Green	Yellow
Spins	$46$	$62$	$47$	$45$

What is the experimental probability of blue, and how does it compare with the designed (theoretical) probability?

The experimental probability is the share of the $200$ spins that landed on blue, $e$ . The theoretical one is 1 section in 4, $t$ , and the count the design leads you to expect is $E$ :

\begin{aligned} e &= \frac{b}{n} \\ &= \frac{62}{200} \\ &= \frac{31}{100} \\ t &= \frac{1}{4} \\ E &= nt \\ &= 200 \times \frac{1}{4} \\ &= 50 \end{aligned}

Blue came up $62$ times against $50$ expected, an experimental probability of $0.31$ against $0.25$ . In the bars below the dashed line is the expected count. A gap this size can happen by chance; if blue kept coming up far more often over thousands of spins, the spinner would not be fair.

Example.

6. Probability from areas. A point is chosen at random inside a rectangle 12 units wide and 8 units tall. A triangle inside it has its base along the bottom edge, the full 12 units, and its top vertex on the top edge. What is the probability that the point lies inside the triangle? And if instead a circle of radius 2 is drawn inside the rectangle, what is the probability of landing in the circle?

The probability is the shaded area over the whole area. The rectangle's area is $R$ and the triangle's is $T$ (base 12, height 8), giving $p$ :

\begin{aligned} R &= 12 \times 8 \\ &= 96 \\ T &= \frac{1}{2} \times 12 \times 8 \\ &= 48 \\ p &= \frac{T}{R} \\ &= \frac{48}{96} \\ &= \frac{1}{2} \end{aligned}

Wherever the top vertex sits, the triangle is half the rectangle: the probability is $\frac{1}{2}$ . The circle has area $4\pi$ (that is, $\pi \cdot 2^2$ ), so its probability is $\frac{4\pi}{96}$ , or $q$ :

\begin{aligned} q &= \operatorname{decimal}(\frac{4\pi}{96}, 3) \\ &= 0.131 \end{aligned}

About 0.131. Using the circumference or the diameter in place of the area is the usual slip.

Practice

Answer ACT-style questions on probability. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise probability with new numbers