1.1 Probabilities

When head is denoted as $\text{H}$ and $\text{T}$, the sample space for three coin tosses is

$$ \phantom{.} \ \left\{ \ \begin{array}{ c c c c } (\text{T,T,T}) & (\text{T,T,H}) & (\text{T,H,T}) & (\text{T,H,H}) \\[1em] (\text{H,T,T}) & (\text{H,T,H}) & (\text{H,H,T}) & (\text{H,H,H}) \end{array} \ \right\} \ . $$ $$ \phantom{.} \ \left\{ \ \begin{array}{ c c c c } (\text{T,T,T}) & (\text{T,T,H}) \\[1em] (\text{T,H,T}) & (\text{T,H,H}) \\[1em] (\text{H,T,T}) & (\text{H,T,H}) \\[1em] (\text{H,H,T}) & (\text{H,H,H}) \end{array} \ \right\} \ . $$

Assuming only two genders, male (M) and female (F), then the sample space for counting in a group is

$$ \phantom{.} \ \{ \ (0 \cdot \text{F}), (1 \cdot \text{F}), (2 \cdot \text{F}), \cdot \cdots (n \cdot \text{F}) \ \} \ , $$

i.e. the sample space is

$$ \phantom{.} \ \{ \ n \in \mathbb{N} \ | \ 1 \leq n \leq k \ \} \ , $$

for some $k \in \mathbb{N}$.

There are four aces in a deck of cards. Then there’s a change of having zero, one, two, three, or four cards in a hand. That is the sample space of having four cards in a hand of $13$ cards is $\{0,1,2,3,4\}$.

A person can have a birthday in any of the $365$ days of the year. So the sample space for person’s birthday is $\{ 1, 2, 3, \ldots, 365 \}$. If it’s a leap year, the sample space is $\{ 1, 2, 3, \ldots, 365, 366 \}$ due to the leap day.

The sample space is

$$ \phantom{.} \ \left\{ \ \begin{array}{c c c c} ( & \text{on time} & , & \text{satisfactorily} &) \\ ( & \text{late } & , & \text{satisfactorily} &) \\ ( & \text{on time} & , & \text{unsatisfactorily} &) \\ ( & \text{late } & , & \text{unsatisfactorily} &) \end{array} \ \right\} \ . $$

The sample space is

$$ \phantom{.} \ \left\{ \ \begin{array}{c} \text{(red,shiny)} \\ \text{(red,dull)} \\ \text{(blue,shiny)} \\ \text{(blue,dull)} \end{array} \ \right\} \ . $$

(a)

If the odds ratio is $1$, then

$$ \phantom{,} \ \dfrac{p}{(1 - p)} = 1 \ , $$

$$ \phantom{,} \ p = (1 - p) \ , $$

from where it follows that

$$ \phantom{.} \ 2p = 1 \ , $$

$$ p = \dfrac{1}{2}. $$

Substituting $p$ back to the odds ratio means that

$$ \phantom{.} \ \dfrac{(1/2)}{(1 - (1/2))} = \dfrac{(1/2)}{(1/2)} = 1 \ . $$

(b)

Similarily, if the odds ratio is $2$, then

$$ \phantom{,} \ \dfrac{p}{(1 - p)} = 2 \ . $$

Then

$$ \phantom{.} \ p = 2(1 - p) = 2 - 2p \ , $$

$$ \phantom{.} \ 2p + p = 3p = 2 \ , $$

i.e.

$$ \phantom{.} \ p = \dfrac{2}{3} \ . $$

Substituting $p$ back, shows that

$$ \phantom{.} \ \dfrac{(2/3)}{1 - (2/3)} = \dfrac{(2/3)}{(1/3)} = \dfrac{2}{3}\left( \dfrac{3}{1} \right) = 2 \ . $$

(c)

Now,

$$ \phantom{.} \ \dfrac{p}{(1 - p)} = 0.25 = \dfrac{1}{4} \ , $$

$$ \phantom{,} \ \dfrac{p}{(1 - p)} = \dfrac{1}{4} \ , $$

$$ \phantom{.} \ p = \dfrac{1}{4}(1 - p) = \dfrac{1}{4} - \dfrac{1}{4}p \ , $$

and therefore

$$ \phantom{.} \ p - \dfrac{1}{4}p = \dfrac{3}{4}p = \dfrac{1}{4} \ . $$

Dividing both sides with the $p$’s coefficient shows that

$$ \phantom{.} \ p = \dfrac{(1/4)}{3/4} = \dfrac{1}{4} \left( \dfrac{4}{3} \right) = \dfrac{1}{4} \ . $$

Substitution leads to

$$ \phantom{.} \ \dfrac{(1/4)}{(1 - (1/4))} = \dfrac{(1/4)}{(3/4)} = \dfrac{1}{4} \left( \dfrac{4}{3} \right) = \dfrac{1}{4} \ . $$

$$ \begin{array}{r c l} P(V) & = & 1 - P(I) - P(II) - P(III) - P(IV) \\[1em] & = & 1 - 0.13 - 0.24 - 0.07 - 0.38 \\[1em] & = & 0.18 \ . \end{array} $$ $$ \begin{array}{r c l} P(V) & = & 1 - P(I) - P(II) - P(III) \\[1em] & & \quad - \ P(IV) \\[1em] & = & 1 - 0.13 - 0.24 - 0.07 - 0.38 \\[1em] & = & 0.18 \ . \end{array} $$

If $P(IV)$ and $P(V)$ are equally likely and mean

$$ \begin{array}{r c l} P(IV) + P(V) & = & 1 - P(I) - P(II) - P(III) \\[1em] & = & 1 - 0.08 - 0.2 - 0.33 \\[1em] & = & 0.38 \ , \end{array} $$ $$ \begin{array}{c l} & & P(IV) + P(V) \\[1em] & = & 1 - P(I) - P(II) - P(III) \\[1em] & = & 1 - 0.08 - 0.2 - 0.33 \\[1em] & = & 0.38 \ , \end{array} $$

then the probability for both $P(IV)$ and $P(V)$ is their sum divided by $2$, i.e. $0.19$.

Since the $P(\text{I}) = 2P(\text{II})$, $P(\text{II}) = 3P(\text{III})$, and $P(\text{III})$ is the unknown, i.e. $P(\text{III}) = x$, and

$$ P(\text{I}) + P(\text{II}) + P(\text{III}) = 1 \ , $$

then

$$ \begin{array}{c l} P(\text{I}) + P(\text{II}) + P(\text{III}) & = & 2P(\text{II}) + 3P(\text{III}) + P(\text{III}) \\[1em] & = & 2 \cdot 3P(\text{III}) + 3P(\text{III}) + P(\text{III}) \\[1em] & = & 6x + 3x + x \\[1em] & = & 10x \\[1em] & = & 1 \ , \end{array} $$ $$ \begin{array}{c l} & & P(\text{I}) + P(\text{II}) + P(\text{III}) \\[1em] & = & 2P(\text{II}) + 3P(\text{III}) + P(\text{III}) \\[1em] & = & 2 \cdot 3P(\text{III}) + 3P(\text{III}) + P(\text{III}) \\[1em] & = & 6x + 3x + x \\[1em] & = & 10x \\[1em] & = & 1 \ , \end{array} $$

from where it can be seen that $x = \frac{1}{10}$. Substituting $x$ back to $6x + 3x + x = 1$, shows that the probabilities of the outcomes are $P(\text{I}) = 0.6$, $P(\text{II}) = 0.3$, and $P(\text{III}) = 0.1$.

$$ P(\text{low}) = 0.28 $$

and

$$ \phantom{.} \ P(\text{average}) = 0.55 \ , $$

then

$$ \begin{array}{r c l} P(\text{high}) & = & p \\[1em] & = & 1 - P(\text{low}) - P(\text{average}) \\[1em] & = & 1 - 0.28 - 0.55 \\[1em] & = & 0.17 \ . \end{array} $$