Taking chance into account

“Could this difference be down to luck?”

A new wheat variety yielded 20 kg/da more than the old one in a trial. Is the new variety really better, or was it just lucky that year, on those plots?

Every statistical test asks exactly this question and answers it with a probability: “If there were no difference at all between the varieties, how likely would I be to see a difference this large by chance alone?” If that probability is very small, you conclude the difference is real. That is why probability is the foundation of every test you will see later.

What does probability measure?

Probability expresses how likely an event is as a number between 0 and 1. 0 means “never”, 1 means “certain”. You can obtain this number in three ways:

Approach	How is it found?	Example
Classical probability	If all outcomes are equally likely: favourable outcomes ÷ all outcomes	Rolling a 6: $\frac{1}{6}$
Relative frequency	Look at how often it happened in the past	Frost in April in 5 of the last 20 years: $\frac{5}{20} = 0.25$
Subjective probability	An expert’s estimate based on experience	“Prices are likely to fall this year”

If the probability that an event happens is $p$, the probability that it does not is $q = 1 - p$. If 85% of seeds germinate, the probability that a random seed does not germinate is 0.15.

Two basic rules: “and” multiplies, “or” adds

Multiplication rule. The probability that independent events (ones that don’t affect each other) all happen together is the product of their probabilities. Both dice showing 6: $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$.

Addition rule. The probability that any one of several mutually exclusive events (ones that can’t happen at the same time) happens is the sum of their probabilities. A 1 or a 2 on a single die: $\frac{1}{6} + \frac{1}{6} = \frac{2}{6}$.

Example:Two dice, double 6Drawing with replacement1 or 2 on a die

A bag has 5 red, 6 yellow and 7 green marbles. Three draws are made, returning the marble each time: what is the probability of red, then yellow, then green?

How do the events combine?

All together (and): multiplyAny one (or): add

Event	Favourable outcomes	All outcomes	Probability	Delete
			0.2778	×
			0.3333	×
			0.3889	×

Add event

Result

Probability that all occur 0.036

1. If the events do not affect each other (independent), multiply the probabilities.

   $$P(A \text{ and } B) = P(A) \cdot P(B)$$
2. $$P = \dfrac{5}{18} \cdot \dfrac{6}{18} \cdot \dfrac{7}{18} = \dfrac{210}{5832} = \dfrac{35}{972} = 0.036008$$

Counting: how many different ways are there?

Classical probability needs “the number of all outcomes”. When there are few, you count them one by one. Counting every way to choose a chair and a deputy from a class of 20, though, is not practical. Permutations and combinations are the shortcut.

One question decides which to use: does the order of the chosen items matter?

• If order matters, permutation. Choosing a chair and a deputy: Ali as chair with Ayşe as deputy is different from Ayşe as chair with Ali as deputy. $_{20}P_2 = 20 \cdot 19 = 380$.
• If order doesn’t matter, combination. Choosing a 3-person jury: Ali, Ayşe and Mehmet form the same jury whatever order they are picked in. $_{10}C_3 = 120$.

A combination is always smaller than the matching permutation. Because different orderings of the same group count as one group, the permutation is divided by $r!$.

Example:JuryChair and deputyPlot orderFertilizer mixes

In how many ways can a 3-person jury be chosen from 10 professors? Jury members have no order.

Does the order of the chosen items matter?

Yes (permutation)No (combination)Arrange all (n!)

Total (n)Chosen (r)

Result

Combination 120

1. Order does not matter: the same people form one group whatever order they are chosen in. Divide the number of permutations by the number of ways r people can be ordered among themselves (r!).

   $${}_nC_r = \dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!} = \dfrac{{}_nP_r}{r!}$$
2. $${}_{10}C_{3} = \dfrac{10!}{3!\,(10-3)!} = \dfrac{10!}{3!\cdot 7!}$$
3. $$= \dfrac{720}{3!} = \dfrac{720}{6} = 120$$

From counting to probability

If the choice in the class of 20 is completely random, what is the probability that Ali becomes chair and Ayşe deputy? There is only 1 favourable outcome out of 380:

$$P = \frac{1}{_{20}P_2} = \frac{1}{380} \approx 0.0026$$

The really important use of combinations comes in the next chapter: in the binomial distribution, $\binom{n}{k}$ counts how many different ways “k successes in n trials” can arise.