The Mathematics of Maybe: Introduction to Probability: questions with answers

3 of the 40 slips are blank, so $P(\text{blank}) = \frac{3}{40}$. Winning a prize is much more likely than not, but that does not make a blank impossible -- 3 of the slips in the bowl are blank, and an outcome that is in the bowl cannot have probability 0. Treating a likely event as certain is the same mistake as treating its opposite as impossible.

One ticket of the 60 wins, and every ticket is equally likely to be drawn, so Ravi's chance is $\frac{1}{60}$. It is a small chance and it is a real one: his ticket is in the draw, so the probability cannot be 0. Calling an unlikely event impossible would put the same number on Ravi's ticket as on a ticket that was never bought.

Heads occurred 27 times in 80 tosses, so the experimental probability of a head is $\frac{27}{80}$. The remaining 53 tosses were tails, and $\frac{53}{80}$ measures tails instead. The event asked about goes on top; its opposite belongs to the other question.

The numerator is the number of times the event itself occurred, so it is $\frac{30}{100}$. The other 70 tosses landed some other way, and $\frac{70}{100}$ is the experimental probability of NOT landing on the bottom -- a different event. The two add to 1, which is why the wrong one so often looks reasonable on its own.

Every snack goes with every drink and every one of those pairs goes with every sweet, so the three counts multiply: $3 \times 3 \times 2 = 18$. The tree branches 3 ways, then 3, then 2, and the paths are counted at the ends. Adding gives the number of ITEMS on sale at the fair.

Each of the 4 shirts can be worn with each of the 3 pairs of trousers, so the counts multiply: $4 \times 3 = 12$ outfits. On the tree there are 4 first branches and each splits into 3, giving 12 path ends. Adding the two counts gives the number of GARMENTS in the wardrobe, which is not the number of ways of dressing.

There are 19 beads in the jar and 7 of them are green, so $P = \frac{7}{19}$. The four colours are not four equally likely outcomes: the beads are what is drawn, and they come in unequal numbers, so $\frac{1}{4}$ would be the answer only if the jar held the same number of each colour.

There are 13 balls altogether and 4 of them are green, so $P = \frac{4}{13}$. The denominator is the whole sample space, green balls included: $\frac{4}{9}$ compares the green balls with the yellow ones, which is the odds in favour and not a probability -- it can even exceed 1 when the event is the commoner one.

The chapter answers this directly. It says we would choose a larger sample, for example by asking 100 students. It also says to make sure the sample is more representative, such as including students from different classes or grades. More evidence, and evidence drawn from across the school, is the only thing that improves an estimate. Asking the same $50$ students again adds no new evidence, and a smaller sample gives less.

The chapter says "approximately 600" and calls the procedure sampling, because a different class of $50$ would have given a different figure. So $600$ is an estimate, and the true number is expected to be near it rather than equal to it. Only $20$ students in the class named mango, so $600$ is not the class's own count. It rests on the measured relative frequency $\frac{20}{50} = 0.4$, and that is what separates it from a guess.

"The coin does not keep track of what happened before. Every time you flip, the chance of heads or tails stays exactly the same -- like starting afresh." So $P(\text{head}) = \frac{1}{2}$ on the seventh toss, exactly as on the first. Saying the chance of a head is now less than $\frac{1}{2}$, so that the counts even out, is the Gambler's Fallacy the chapter names. Saying it is more than $\frac{1}{2}$ because of the run is the same error wearing the other coat. The value $\frac{1}{64}$ answers a different question: the chance of six heads in a row before any tossing began.

This is the Law of Large Numbers: "as the number of trials increases, the experimental probability tends to get closer to the theoretical probability". It works by dilution, not by correction. An early run of six heads is still six heads after a thousand tosses, but six out of a thousand barely moves the proportion. The coin never produces extra tails to make up a shortfall, and its chance of a tail is a half on every toss from the first to the last.

Neither figure is evidence, so no amount of comparing them produces one. The chapter's answer is to collect data: count how often teams in the same position have won in the recorded matches, and use that relative frequency. Confidence is not a measurement, averaging two opinions makes a third opinion, and refusing to estimate at all gives up the whole point of the subject.

Knowing the list of outcomes is where the study of a random experiment begins, not where the randomness ends. The chapter's definition needs both halves: "you may know all the possible outcomes, but you cannot say which one will definitely occur". Only knowing which face comes up next would end the randomness. The list of faces, their number and the die's fairness are all known already, and the roll stays random.

The favourable outcomes have to be listed before they can be counted: the multiples of 4 up to 15 are 3 in number, so $P = \frac{3}{15}$. Reading the whole phrase as one outcome would give $\frac{1}{15}$ however many multiples there were, which cannot be right for two boxes of different sizes.