Complex events

We are interested in non-trivial events, that are usually combinations of smaller events

For example, we may ask “what is the probability that, in a group of 𝑛 people, at least two persons have the same birthday”

Fortunately, any complex event can be decomposed into simpler events, combined with and, or and not connectors

Exercise: decompose the birthday event into simpler ones

Probability of not 𝐴

If the event 𝐴 becomes more and more plausible, then the opposite event not 𝐴 becomes less and less plausible

It can be shown that we always have \[ℙ(\text{not } A) = 1-ℙ(A)\]

Probability of 𝐴 and 𝐵

\[ℙ(A\text{ and } B)=\frac{\text{Number of cases where }(A\text{ and } B)\text{ is true}}{\text{Total cases of combinations of }A\text{ and } B}\]

If \(n_A\) and \(n_B\) are the total number of cases for \(A\) and \(B\), then the total number of cases is \(n_A⋅n_B\)

In the same way, if \(m_A\) and \(m_B\) are the number of cases where \(A\) and \(B\) are true, respectively, then the number of cases where \((A\text{ and }B)\) is true is \(m_A⋅m_B\)

\[ℙ(A\text{ and } B)=\frac{m_A⋅m_B}{n_A⋅n_B}=\frac{m_A}{n_A}⋅\frac{m_B}{n_B}\]

Interpretation

We could say that \[\frac{m_A}{n_A}=ℙ(A)\qquad\frac{m_B}{n_B}=ℙ(B)\] but we have to be careful. The result of A may affect \(m_B\) and \(n_B\). We better write \[\frac{m_A}{n_A}=ℙ(A)\qquad\frac{m_B}{n_B}=ℙ(B|A)\]

Rewriting the Probability of 𝐴 and 𝐵

\[ℙ(A\text{ and } B)=\frac{m_A}{n_A}⋅\frac{m_B}{n_B}=ℙ(A)⋅ℙ(B|A)\] To simplify, instead of \(ℙ(A\text{ and } B)\) we write \(ℙ(A, B)\)

Thus, we write \[ℙ(A,B)=ℙ(A)⋅ℙ(B|A)\] “Prob that (𝐴 and 𝐵) happens is Prob that 𝐴 happens times Prob that 𝐵 happens given that A happens”

Joint Probability

We know that \((A\text{ and } B)\) is always the same as \((B\text{ and } A)\)

There are two ways to calculate the probability of of 𝐴 and 𝐵 happening simultaneously

• Start with the prob. of \(A\) and then of \(B\) given that \(A\) is true \[ℙ(A,B)=ℙ(A)⋅ℙ(B|A)\]
• Start with the prob. of \(B\) and then of \(A\) given that \(B\) is true \[ℙ(A,B)=ℙ(B)⋅ℙ(A|B)\]

Exercises

• Prob of getting heads twice when throwing coins
• Prob of getting 6 and 6 on two dice
• Prob of getting heads and a 6
• Prob of getting a green card

Probability of 𝐴 or 𝐵

We know how to calculate \(ℙ(A\text{ and } B)\) and \(ℙ(\text{not } A)\)

We also know the De Morgan’s law, to swap ANDs with ORs
\[\text{not } (A \text{ or } B) = (\text{not } A) \text{ and } (\text{not } B)\]

Therefore we can write

\[ \begin{aligned} ℙ(A \text{ or } B) & = 1 - ℙ(\text{not }(A \text{ or } B))\\ & = 1-ℙ( (\text{not } A) \text{ and } (\text{not } B)) \end{aligned} \]

Using the multiplication rule

\[ℙ(A \text{ or } B) = 1-ℙ( (\text{not } A) \text{ and } (\text{not } B)) \\ = 1-ℙ(\text{not } A)⋅P(\text{not } B|\text{not } A)\]

using negation rule \[ \begin{aligned} ℙ(A \text{ or } B) & = 1-ℙ(\text{not } A)⋅(1- ℙ(B|\text{not } A)) \\ & = 1-ℙ(\text{not } A) + ℙ(\text{not } A)⋅P(B|\text{not } A) \end{aligned} \]

Using the multiplication rule again

\[ \begin{aligned} ℙ(A \text{ or } B) & = 1 -ℙ(\text{not } A) + ℙ(\text{not } A,B) \\ ℙ(A \text{ or } B) & = 1 -(1-ℙ(A)) + ℙ(\text{not } A|B)ℙ(B) \\ ℙ(A \text{ or } B) & = ℙ(A) + (1-ℙ(A|B))ℙ(B) \\ ℙ(A \text{ or } B) & = ℙ(A) + ℙ(B)-ℙ(A|B)ℙ(B) \\ ℙ(A \text{ or } B) & = ℙ(A) + ℙ(B)-ℙ(A,B) \end{aligned} \] You need to remember only the last line

The previous lines justify why the last one is always true

It gets complicated

If there are three compatible events, things get messy

It gets worse with more events

If A and B are incompatible

if A and B cannot happen at the same time, then \((A \text{ and } B)\) is impossible, therefore \(ℙ(A,B)=0\)

In that case (and only in that case) \[ℙ(A \text{ or } B) = ℙ(A) + ℙ(B)\]

Splitting a set into pieces

In particular we have \[ℙ(A) = ℙ(A\text{ and } (B \text{ or } \text{not } B)) = ℙ(A,B) + ℙ(A, \text{not } B)\] because

• \((A \text{ and } B)\) is incompatible with \((A \text{ and } \text{not } B)\),
• \((A \text{ and } (B \text{ or } \text{not } B))\) is equal to \(A\)

Splitting \(Ω\)

If we partition Ω into 𝑛 subsets \(A_i\), such that they cover all Ω \[\Omega=A_1 ∪ A_2 ∪ … ∪ A_n\] and each pair of events are mutually incompatible \[A_i ∩ A_j=\phi\] then we have \[ℙ(\Omega)=ℙ(A_1) + ℙ(A_2) + … + ℙ(A_n)=1\]

There is an easier way

Using De Morgan’s rule

\[\begin{aligned} & ℙ(A \text{ or } B \text{ or } C) \\ & 1 - ℙ((\text{not } A) \text{ and } (\text{not } B) \text{ and } (\text{not } C))\\ & 1 - ℙ(\text{not } A)⋅ℙ(\text{not } B | \text{not } A)⋅ℙ(\text{not } C | \text{not } A, \text{not } B)\\ & 1 - (1-ℙ(A))⋅(1-ℙ(B | \text{not } A))⋅(1-ℙ(C | \text{not } A, \text{not } B)) \end{aligned} \]

This is often easier to calculate

Example: Multiple Birthdays

Let’s say we have three people, with birthday \(x_1, x_2\) and \(x_3.\)

The probability that there are at least two people with the same birthday is \[ℙ(x_2=x_1 \text{ or } x_3=x_2 \text{ or } x_3=x_1)\] which can be rewritten as \[1-ℙ(x_2≠x_1 \text{ and } x_3≠x_2 \text{ and } x_3≠x_1)\]

Now we onlu have and combinations

We want to calculate \[1-ℙ(x_2≠x_1 \text{ and } x_3≠x_2 \text{ and } x_3≠x_1)\] We can separate like this (only the first and) \[1-ℙ(x_2≠x_1)⋅ℙ(x_3≠x_2 \text{ and } x_3≠x_1|x_2≠x_1)\] Assuming 365 possible birthdays, we have \[1-\frac{364}{365}⋅\frac{363}{365}\]

Exercise

• What is the probability that, in a group of N people, at least two of them share the same birthday?

• How many people do we need to have at least 50% probability of least two of them sharing the same birthday?