Let’s say that \(Ω=\{a_1, a_2, …,a_n\}\)

We showed that the sum of all outcomes’ probabilities is 1 \[ℙ(\{a_1\}) + ℙ(\{a_2\}) + … + ℙ(\{a_n\})=1\] If we know these values, we can calculate everything.

The for all \(i\) \[p(a_i) = ℙ(\{a_i\})= ℙ(\text{outcome is exactly
}a_i)\] is called the **distribution** of the
probability.

This definition makes sense only if we agree on what are all the possible outcomes.

In other words, we must agree on what is \(Ω\)

Then the probability distribution is a function \[p: Ω → [0,1]\]

Notice that there may be more than one way to define \(Ω\)

Let’s assume that we know how many cards of each color are in the deck

There are \(n_c\) cards of color \(c\in\){“red”,“green”,“blue”, “yellow”}

There are \(N=∑ n_c\) cards in total

If we do not have any solid reason to expect any particular order of cards, then each individual card has the same probability \(1/N\)

The probability of “first color \(c\)” is \[ℙ(\text{color is }c)=\frac{n_c}{N}\]

We will continue drawing cards, so the proportions will change

Let’s say we got color \(c_1\) in the first draw

Now we have \(N-1\) cards in total, and there are \(n_{c_1}-1\) cards of color \(c_1\)

The probability of “second color \(c\)” is

\[ℙ(\text{second color is }c|\text{first color is }c_1)=\begin{cases} \frac{n_c}{N-1} &\text{if }c≠c_1\\ \frac{n_c-1}{N-1} &\text{if }c=c_1 \end{cases}\]

It gets complicated

The formula applies when our measurement changes the experiment

There are two exceptions where proportions do not change

- If every \(n_c\) is large, then
\(n_c-1≈ n_c\) and \(N-1 ≈ N\)
- This is the case when we interview a few people from a large population

- If we put the card back into the deck and shuffle it again
- This is called
**sampling with replacement**

- This is called

In practice, we are often sample from a very large population and we
model it as a *sampling with replacement*

If we replace the card into the deck after we see it, we will have

\[ℙ(\text{second color is }c|\text{first color is }c_1)=ℙ(\text{color is }c)=\frac{n_c}{N}\]

Notice that this means that the second result is
**independent** of the first result, and so on

Moreover, the **distribution** is identical in each
case

This is a very important case, and we give it a name

Independent, Identically Distributed (i.i.d.)

- Has 2 sides: \(\Omega=\{\text{'Head'}, \text{'Tail'}\}\)
- Distribution given by \(ℙ(\text{'Head'})\) and \(ℙ(\text{'Tail'})\) such that \[ℙ(\text{'Head'}) + ℙ(\text{'Tail'})= 1\]

All can be reduced to the value \[p=ℙ(\text{'Head'})\]

We say that the *probability distribution* of the coin depends
on the *parameter* \(p\)

Real coins are usually fair, so \(p=0.5\)

We are interested in cases where \(p\) may be anything

This is the case when we have cards of two colors, with different proportions, and we do sampling with replacement

Still, we will call “coin” any experiment with two possible outcomes

(In math this is called a *Bernoulli* distribution)

- There is only one way to get 0 heads: TT
- this happens with probability \(q^2\)

- There are two ways of getting 1 head: HT and TH
- this happens with probability \(pq\)

- There is only one way to get 2 heads: HH
- this happens with probability \(p^2\)

we get \[1⋅ q^2,\quad 2⋅ p q,\quad 1⋅ p^2\]

- There is only 1 way to get 0 heads: TTT
- this happens with probability \(q^3\)

- There are 3 ways of getting 1 head: HTT, THT, and TTH
- this happens with probability \(pq^2\)

- There are 3 ways of getting 2 head: THH, HTH, and HHT
- this happens with probability \(p^2q\)

- There is only one way to get 3 heads: HHH
- this happens with probability \(p^3\)

we get \[1⋅ q^3,\quad 3⋅ pq^2,\quad 3⋅ p^2q,\quad 1⋅ p^3\]

The rule of combinations are the same as in the binomial theorem

We get \[ℙ(k\text{ Heads in }N\text{ coins})= \binom{N}{k} p^k q^{(N-k)}\]

The numbers \(\binom{N}{k}\) are found in Pascal’s triangle

One way to remember it is to use the formula \[(p+q)^N =\sum_{k=0}^N \binom{N}{k} p^k q^{(N-k)}\]

This is why we call it **Binomial distribution**

We know that \[ℙ(K=k|N\text{ in total})=\binom{N}{k} p^k(1-p)^{n-k}\] We can calculate \(\binom{N}{k}\) using Pascal’s triangle, even in Excel

Pascal’s Triangle

\[ℙ(𝑆≤k)=\sum_{j=0}^k ℙ(𝑆=j)\]

We have a bacteria whose *GC content* is 38%

What is the GC content of a random DNA fragment of length 100bp?

Worldwide proportion is 1%

And *world* is a population

We are 4 people in this course, including me

What is the probability that there are \(k\) persons with epilepsy in our class?

To test a new fertilizer, three plants were planted in a new environment

After one month, one died and the others survived

- What is the best estimation for the survival probability?
- What is the 95% interval for the survival probability?