Probability

5.3 Probability

Probability by counting

When all outcomes are equally likely:

$\mathrm{P}(A) = \frac{\text{number of outcomes in } A}{\text{total number of outcomes}}$

Of the 36 equally likely pairs, $(3, 6)$, $(4, 5)$, $(5, 4)$ and $(6, 3)$ give 9. So the probability is $\dfrac{4}{36} = \dfrac{1}{9}$.

A table of outcomes for two dice, or a list, avoids missing any.

Using permutations and combinations

Counting methods give the numbers for the fraction.

$\frac{{}^4\mathrm{C}_2 \times {}^6\mathrm{C}_1}{{}^{10}\mathrm{C}_3} = \frac{6 \times 6}{120} = 0.3$

The rules of probability

$A \cup B$ means "$A$ or $B$ or both". $A \cap B$ means "both $A$ and $B$". The overlap is subtracted, because it has been counted twice.

Exclusive and independent events

These two ideas sound alike but are different.

$0.5 \times 0.4 = 0.2$, so they are independent. $\mathrm{P}(A \cap B) \neq 0$, so they are not exclusive.

$\mathrm{P}(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7$.

Tree diagrams

Multiply along the branches, and add between the paths that fit. Without replacement, the numbers on later branches change.

$\mathrm{P}(\text{RR}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} \qquad \mathrm{P}(\text{BB}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$

$\mathrm{P}(\text{same}) = \frac{26}{56} = \frac{13}{28}$

Conditional probability

$\mathrm{P}(A \mid B)$ means the probability of $A$, given that $B$ has happened.

The word "given", or wording such as "if it is known that", signals this.

$\mathrm{P}(\text{RR} \mid \text{same}) = \frac{\frac{20}{56}}{\frac{26}{56}} = \frac{10}{13}$

Conditional probability lets you test for independence another way: $A$ and $B$ are independent if $\mathrm{P}(A \mid B) = \mathrm{P}(A)$.

Reversing a tree

Questions often give the later branches and ask about an earlier one.

$\mathrm{P}(\text{late}) = 0.3 \times 0.4 + 0.7 \times 0.1 = 0.12 + 0.07 = 0.19$.

$\mathrm{P}(\text{rain} \mid \text{late}) = \frac{0.12}{0.19} = 0.632$

Venn diagrams

Venn diagrams help when two or three events overlap. Fill in from the centre outwards, and check the regions total 1.

French or Spanish: $20 + 15 - 5 = 30$. So 10 study neither, and the probability is $\dfrac{10}{40} = 0.25$.