Elementary Probability (Senior 6 Basic Mathematics): free lessons | Ganzaa

Elementary Probability

Lessons

Senior 6 · Basic Mathematics

Elementary Probability

5 lessons

Lesson 1 of 5

Experiments, sample spaces, events and the probability of an event

25 min

Measuring chance

Before a farmer plants a whole field, she can test a sample of her seed: plant 400 seeds in a tray and count how many come up. If 352 germinate, she estimates the chance that any one seed germinates as $\frac{352}{400} = 0.88$ , and expects about $0.88 \times 2500 = 2200$ plants from a bag of 2,500 seeds (illustrative figures). In Unit 2 you learned to count outcomes without listing them. Probability uses those counts to measure how likely an outcome is, on a scale from 0 (impossible) to 1 (certain).

Definition

A random experiment is an action whose result cannot be predicted with certainty, although all possible results are known: rolling a die, tossing a coin, choosing a student by lottery.
An outcome is one possible result. The sample space $S$ is the set of all possible outcomes; $n (S)$ is the number of outcomes in it. For one die, $S = \{1, 2, 3, 4, 5, 6\}$ and $n (S) = 6$ .
An event is any subset of the sample space. "An even number" is the event $A = \{2, 4, 6\}$ . A simple event has one outcome; a compound event has more than one.
Outcomes are equally likely when none is favoured over another, as with a fair die or a well-shuffled pack of cards.
The complement of $A$ , written $A^{'}$ , is the event " $A$ does not happen": all outcomes in $S$ that are not in $A$ .

The probability of an event

If all outcomes in $S$ are equally likely, the classical probability of an event $A$ is

P(A) = \frac{n(A)}{n(S)} = \frac{\text{number of outcomes in } A}{\text{number of outcomes in the sample space}}

Why it works. If $n (S)$ outcomes are equally likely, each has probability $\frac{1}{n(S)}$ , and an event made of $n (A)$ of them collects $n (A)$ such shares.

Because $0 \le n(A) \le n(S)$ , every probability satisfies $0 \le P(A) \le 1$ , with $P(\varnothing) = 0$ (impossible) and $P (S) = 1$ (certain). Since $A$ and $A^{'}$ share no outcome and together make up $S$ , $n (A) + n (A^{'}) = n (S)$ , and dividing by $n (S)$ gives the complement rule:

P (A^{'}) = 1 - P (A)

Experimental probability

When outcomes are not equally likely (a seed germinating, a bus arriving late), we estimate probability from data. If an event happens $f$ times in $N$ trials, its relative frequency is $\frac{f}{N}$ . As $N$ grows, the relative frequency settles close to the true probability (the law of large numbers), which is why the farmer tests 400 seeds rather than 4.

Two dice: a sample space you can see

Each of the 36 cells is equally likely when both dice are fair. To find a probability, count the cells in the event and divide by 36.

Worked example

Worked example 1. Two fair dice are rolled. Find the probability that the total is (a) 7; (b) at least 10; (c) a prime number.

(a) The shaded diagonal has 6 cells: $P(7) = \frac{6}{36} = \frac{1}{6}$ .

(b) Totals 10, 11 and 12 occur in 3, 2 and 1 cells: $P(\text{at least } 10) = \frac{3 + 2 + 1}{36} = \frac{6}{36} = \frac{1}{6}$ .

P(\text{prime}) = \frac{1 + 2 + 4 + 6 + 2}{36} = \frac{15}{36} = \frac{5}{12}

Worked example

Worked example 2: using the complement. Three fair coins are tossed. Find the probability of (a) exactly two heads; (b) at least one head.

By the product rule, $n (S) = 2^{3} = 8$ : HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

(a) Exactly two heads: HHT, HTH, THH, so $P = \frac{3}{8}$ .

(b) "At least one head" is the complement of "no heads", which is the single outcome TTT:

P(\text{at least one head}) = 1 - P(\text{TTT}) = 1 - \frac{1}{8} = \frac{7}{8}

Counting the seven outcomes directly gives the same answer, but the complement is far quicker when the event has many outcomes.

Worked example

Worked example 3: a pack of cards. A standard pack has 52 cards: four suits (hearts and diamonds are red; clubs and spades are black) of 13 cards each, ace, 2 to 10, jack, queen, king. One card is drawn at random. Find $P(\text{king})$ , $P(\text{heart})$ and $P(\text{red picture card})$ .

P(\text{king}) = \frac{4}{52} = \frac{1}{13}, \qquad P(\text{heart}) = \frac{13}{52} = \frac{1}{4}, \qquad P(\text{red picture card}) = \frac{6}{52} = \frac{3}{26}

There are 6 red picture cards: the jack, queen and king of hearts and of diamonds.

Warning

Equally likely, or not? $P(A) = \frac{n(A)}{n(S)}$ only works when the outcomes are equally likely. The totals 2 to 12 of two dice are 11 outcomes, but they are not equally likely, so $P(7) \ne \frac{1}{11}$ . Always count in a sample space of equally likely outcomes, such as the 36 cells.

Tip

When a question says "at least one", "not" or "fewer than", think of the complement first: $P (A) = 1 - P (A^{'})$ .

Khan Academy: "Probability explained". Watch how the sample space is listed before any probability is calculated, then check one of the worked examples above in the same way.

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Practice

A letter is chosen at random from the word KIGALI. Find the probability that it is (a) a vowel; (b) the letter L.
Two fair dice are rolled. Find $P(\text{total} = 8)$ and $P(\text{a double})$ .
One card is drawn from a standard pack. Find the probability that it is (a) an ace; (b) not a club; (c) a black card higher than 8 (count 9 and 10 only).
A bag holds 7 red, 5 green and 3 yellow beads. One is drawn at random. Find $P(\text{green})$ and $P(\text{not red})$ .
Two fair dice are rolled. Find the probability that the product of the scores is even. (Hint: use the complement.)
A drawing pin is dropped 250 times and lands point up 95 times. Estimate the probability that it lands point up, and the number of times it would be expected to land point up in 1,000 drops.
Explain why the probability of an event can never be greater than 1.

Show answersHide answers

(a) The vowels are I, A, I: $\frac{3}{6} = \frac{1}{2}$ . (b) $\frac{1}{6}$ .
$\frac{5}{36}$ (2 + 6, 3 + 5, 4 + 4, 5 + 3, 6 + 2); $\frac{6}{36} = \frac{1}{6}$ .
(a) $\frac{4}{52} = \frac{1}{13}$ (b) $1 - \frac{13}{52} = \frac{3}{4}$ (c) the 9 and 10 of clubs and of spades: $\frac{4}{52} = \frac{1}{13}$ .
$\frac{5}{15} = \frac{1}{3}$ ; $\frac{8}{15}$ .
The product is odd only when both scores are odd: $3 \times 3 = 9$ cells. $P(\text{even}) = 1 - \frac{9}{36} = \frac{27}{36} = \frac{3}{4}$ .
$\frac{95}{250} = 0.38$ ; about $0.38 \times 1000 = 380$ times.
An event cannot contain more outcomes than the sample space, so $n(A) \le n(S)$ and $\frac{n(A)}{n(S)} \le 1$ .

Important

For equally likely outcomes, $P(A) = \frac{n(A)}{n(S)}$ ; always $0 \le P(A) \le 1$ and $P (A^{'}) = 1 - P (A)$ . Data give relative frequencies $\frac{f}{N}$ , which approach the probability as $N$ grows.

Sources and credits

Text, diagram, worked examples and exercises: Written by Ganzaa. The seed-germination figures are illustrative.

Video: Probability explained | Independent and dependent events, Khan Academy, embedded from YouTube.

1Measuring chance

2The probability of an event

3Experimental probability

4Two dice: a sample space you can see

Measuring chance

The probability of an event

Experimental probability

Two dice: a sample space you can see