The Mathematics of Maybe: Introduction to Probability

Every probability sits somewhere between 0 and 1. A ranking is a judgement about where on that line the event belongs, so the reason matters as much as the label.

Ranking, lowest first: (ii) and (iii) near 0  →  (iv) close to 1  →  (i) exactly 1.

This is experimental probability: the sample of 30 is the evidence, and the relative frequency of each colour is our estimate of its probability.

(i) Green sweets in the sample: 8 out of 30.

So the probability is \(\frac{4}{15}\), about 0.267 or 26.7%.

(ii) The sample says \(\frac{7}{30}\) of the sweets are yellow. Apply that fraction to the whole bag of 600.

About 140 yellow sweets.

(i) Arts Club: 11 students out of the 40 surveyed.

The probability is \(\frac{11}{40}\) = 0.275, that is 27.5%.

(ii) Scale the sample proportion up to the population of 800.

About 180 students in the school are likely to prefer the Sports Club.

This is an activity, so your own numbers will be different. Here is one real run of 20 tosses, written up the way yours should be.

(i) Heads came up 11 times.

(ii) Tails came up 9 times, and \(11 + 9 = 20\) as it must.

(iii) Experimental probability uses what actually happened:

(iv) For the next toss the answer is the theoretical probability, because the coin is fair and each toss is independent:

The sample space has three outcomes: \(S = \{\text{bottom},\ \text{top},\ \text{side}\}\), so \(n(S) = 3\). Here is one run of 100 tosses; your own counts will differ.

The three probabilities add to \(0.22 + 0.08 + 0.70 = 1\), which is the check that no toss was missed.

The die is fair, so all six faces are equally likely and this is a theoretical probability.

Favourable outcomes (even numbers): \(2, 4, 6\) — that is 3 of them.

The probability is \(\frac{1}{2}\), that is 0.5 or 50%.

(i) Experimental probability comes from what actually happened: a ‘3’ on 3 of the 12 rolls.

(ii) Theoretical probability assumes a fair die with six equally likely faces.

(iii) They differ because 12 rolls is a very small number of trials. With so few rolls, chance alone can easily push the count away from the expected \(\frac{1}{6}\times 12 = 2\) — getting 3 instead of 2 is nothing unusual. Nothing is wrong with the die.

The die can show any one of its six faces, and no face can be listed twice.

The sample size is 6.

(i) Each of the 6 die faces can pair with each of the 2 coin faces, so \(6\times 2 = 12\) outcomes.

(ii) The integers strictly between \(-5\) and \(+5\):

(iii) Only the colour of the ball is recorded, so there are just two outcomes:

These two outcomes are not equally likely. Out of 12 balls,

and \(\frac{5}{12} + \frac{7}{12} = 1\).

(i) Every snack can go with every drink, so there are \(3\times 2 = 6\) combinations.

(ii) The event ‘Samosa is the snack’ keeps only the rows beginning with S:

If every combination is equally likely, then

(i) The first stage is Basket A. It holds three fruits, not two — the apple and the two separate oranges, written O1 and O2 — so the tree starts with three branches, each of probability \(\frac{1}{3}\). The second stage is Basket B, with two branches of probability \(\frac{1}{2}\).

(ii) Reading the six paths from left to right:

All six are equally likely, each with probability \(\frac{1}{6}\).

(iii) Only one path gives an apple with a banana.

The same answer follows by multiplying along that single path:

The box holds \(3 + 4 + 2 = 9\) pens. Because the first pen is put back, the second draw faces exactly the same box, so the two stages are independent and carry the same three probabilities:

(i) Three colours at each stage give \(3\times 3 = 9\) ordered outcomes:

Multiplying along each path gives the probability of that outcome (all nine have denominator 81):

The nine entries add to \(\frac{81}{81} = 1\).

(ii) ‘Same colour’ means the diagonal of that table: RR, BB or GG.

So there is roughly a 36% chance that the two pens match.

(i) 0. An impossible event has no favourable outcome at all, so the numerator is 0.

(ii) sample space (written \(S\)). Its number of elements \(n(S)\) is the sample size.

(iii) 1. Every outcome is favourable, so the fraction is \(\frac{n(S)}{n(S)} = 1\).

(iv) \(\frac{1}{2}\), that is 0.5 or 50%, because a fair coin has two equally likely faces.

The count itself — 15 — is the frequency. The second blank asks for a fraction or a decimal, so it is the relative frequency.

First blank: relative frequency. Second blank: \(\frac{3}{10}\) = 0.3 (30%).

Equally likely: (ii) and (iii). Not equally likely: (i) and (iv). (v) is very nearly equally likely and is treated as such.

(i) Two coins:

‘At least one head’ means one head or two — every outcome except TT.

(ii) Ten cards:

Even numbers: \(2, 4, 6, 8, 10\) — 5 of them.

(iii) One die, \(S = \{1,2,3,4,5,6\}\). Greater than 4 means \(\{5, 6\}\).

(iv) The bag holds \(3 + 2 + 1 = 6\) balls. ‘Not red’ means the 2 blue and the 1 green, that is 3 balls.

Check with the complement: \(P(\text{red}) = \frac{3}{6} = \frac{1}{2}\), and the two add to 1.

(v) Three coins give \(2\times 2\times 2 = 8\) outcomes.

Exactly two heads (so the third coin must be a tail): HHT, HTH, THH.

One candy of each kind, picked at random, so the three outcomes are equally likely.

The probability is \(\frac{1}{3}\), about 33.3%.

Every shirt can be worn with every pair of pants, so the count is \(2\times 3 = 6\).

The six outfits are (Red, Jeans), (Red, Khakis), (Red, Shorts), (Blue, Jeans), (Blue, Khakis) and (Blue, Shorts).

First check the data: \(20 + 210 + 325 + 445 = 1000\), so every case is accounted for and the total is the denominator throughout.

(i) Less than 4000 km: 20 cases.

(ii) Between 4000 and 14000 km covers the two middle columns: \(210 + 325 = 535\) cases.

(iii) More than 14000 km: 445 cases.

Check: \(0.02 + 0.535 + 0.445 = 1\). The three events cover every tyre with no overlap, so they must add to 1.

PEACE has 5 letters, so there are 5 cards. Note that E appears twice — two separate cards, not one.

(i) Favourable cards: the P, both E cards and the C — that is \(1 + 2 + 1 = 4\) cards. Only the A is left out.

(ii) ‘Not an E’ leaves the P, the A and the C: 3 cards.

Check with the complement: \(P(E) = \frac{2}{5}\), and \(\frac{2}{5} + \frac{3}{5} = 1\).

The eight sectors are equal, so each number has probability \(\frac{1}{8}\).

(i) Only one sector is an 8.

(ii) Odd numbers: \(1, 3, 5, 7\) — 4 of the 8.

(iii) Greater than 2 means \(3, 4, 5, 6, 7, 8\) — 6 sectors. (2 itself is not included.)

(iv) Every number on the spinner is less than 9, so this event is certain.

(v) Multiples of 3 on the spinner: \(3\) and \(6\) — 2 sectors. (9 is not on the spinner.)

There are \(4 + 5 = 9\) balls at the first draw. The first ball is laid aside, so only 8 balls remain for the second draw — and how many of each colour depends on what came out first.

First draw: \(P(R) = \frac{4}{9}\) and \(P(B) = \frac{5}{9}\). Second draw, out of the 8 balls left:

(i) Red then blue is the second path.

(ii) Two blue balls: 5 blue out of 9, then 4 blue out of the 8 that remain.

Two dice give \(6\times 6 = 36\) equally likely outcomes, each written as an ordered pair (first die, second die).

The 36 outcomes, arranged by their sum:

An event with probability 0 — anything that cannot happen. For example:

The smallest possible sum is \(1 + 1 = 2\), so \(E_1\) has no favourable outcome.

‘The sum is 13’ works just as well, because the largest sum is \(6 + 6 = 12\).

An event with probability 1 — anything that must happen. For example:

‘Each die shows a whole number from 1 to 6’ is another.

(i) Two dice: \(n(S) = 36\). The possible sums run from 2 to 12; the primes among them are 2, 3, 5, 7 and 11, so a prime greater than 5 means a sum of 7 or 11.

(ii) The bag holds \(4 + 3 + 2 = 9\) balls. Two are drawn without replacement, and order does not matter, so the sample space is all unordered pairs:

It is quicker to count the pairs of the same colour and subtract.

(iii) Three coins give the 8 outcomes listed in Question 4(v). We need the first coin to be a head and the total number of heads to be exactly two.

HTT has a head first but only one head; THH has two heads but starts with a tail. Both conditions must hold.

(iv) Using each of 1, 2, 3, 4 exactly once gives

A number is even when its last digit is even, and here that means a 2 or a 4 — 2 of the 4 digits. Whichever is placed last, the other three digits can be arranged in \(3! = 6\) ways.

(v) Each question is guessed independently, with \(P(\text{correct}) = \frac{1}{4}\) and \(P(\text{wrong}) = \frac{3}{4}\). All three questions together give \(4\times 4\times 4 = 64\) equally likely answer sheets.

‘Exactly 2 correct’ means one question is wrong, and there are 3 choices for which one:

The same result as a product of probabilities:

(i) With replacement. The box is back to 4 balls for the second draw, so every one of the 4 first branches splits into 4.

The 16 ordered pairs, including the four repeats (1,1), (2,2), (3,3) and (4,4):

(ii) Without replacement. The ball already drawn cannot come out again, so each first branch splits into only 3.

It is the same table with the four diagonal entries removed.

(iii) The sample sizes are 16 and 12.

The difference is exactly the four repeated pairs, which only the with-replacement experiment allows.

The coin gives 2 outcomes and the cards give 6, and the two happen together, so the multiplication principle applies.

All 12 elements are equally likely, each with probability \(\frac{1}{12}\).

What is being recorded is the number of heads, not the pattern. With three coins that count can be 0, 1, 2 or 3 — and every one of those actually happens:

The correct sample space is (iv) \(\{0, 1, 2, 3\}\). It lists every possible count exactly once and lists nothing impossible.

(i) \(\{1, 2, 3\}\) — fails: it leaves out 0, but TTT is perfectly possible.

(ii) \(\{0, 1, 2\}\) — fails: it leaves out 3, but HHH is perfectly possible.

(iii) \(\{0, 1, 2, 3, 4\}\) — fails: 4 heads is impossible with only three coins.

(iv) \(\{0, 1, 2, 3\}\) — correct.

This is a geometric probability: the dye is equally likely to land anywhere on the rectangle, so the probability is the ratio of the two areas.

Reading the figure: the rectangle measures 3 m by 2 m, and the circle inside it has diameter 1 m, so its radius is \(\frac{1}{2}\) m.

Taking \(\pi = \frac{22}{7}\):

So the probability is \(\frac{\pi}{24} \approx 0.131\), that is about a 13% chance.