Havard Stat 101 Lecture 4 - Conditional Probability
Independence
Definition: Independence
Events \(A\), \(B\) are independent if \(P(A\cap B) = P(A)P(B)\) .
Note
It is completely different from disjointness.
\(A,B,C\) are independent if
\[
\begin{aligned}
&P(AB) = P(A)P(B), \\
&P(BC) = P(B)P(C), \\
&P(AC) = P(A)P(C), \\
&P(ABC) = P(A)P(B)P(C).
\end{aligned}
\]
Similarly for events \(A_1,A_2,\cdots,A_n\) , means all multiply rules hold.
Newton-Pepys Problem (1693)
18:22
We have a fair dice, which is most likely below?
- A. At least one \(6\) with \(6\) dices;
- B. At least two \(6\) with \(12\) dices;
- C. At least three \(6\) with \(18\) dices;
The answer is A. Here is the calculation:
\[
P(A) = 1-\left(\frac{5}{6}\right)^6 \approx 0.665
\]
\[
P(B) = 1-\left(\frac{5}{6}\right)^{12} - 12\left(\frac{5}{6}\right)^{11} \frac{1}{6}\approx 0.619
\]
\[
P(C) = 1- \sum\limits_{k=0}^2 \binom{18}{k}\left( \frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{18-k} \approx 0.597
\]
Conditional Probability
Definition: Conditional Probability
\[P(A|B) = \dfrac{P(A\cap B)}{P(B)}\]