Conditional Probability

• Conditional probability of $A$ given $B$, denoted by $\mathbb{P}(A\vert B)$ $$\mathbb{P}(A \vert B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$$
  • If $A$ and $B$ are independent events $$\mathbb{P}(A \vert B) = \mathbb{P}(A)$$
• Example Considering a gamer who rolls the dice twice without showing us the result. Event $B$ is the sum being greater or equal to 10, can be denoted by $B = \{(i,j)\vert i+j \ge 10\}$; Let $A$ be the event that the sum is even. $$\mathbb{P}(A\vert B) = \frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)} = \frac{4/36}{6/36} = \frac{2}{3} \\ \mathbb{P}(A^{C} \vert B) = \frac{\mathbb{P}(A^{C} \cap B)}{\mathbb{P}(B)} = \frac{2/36}{6/36} = \frac{1}{3}$$

The Law of Total Probability

$$\mathbb{P}(A\cap B) = \mathbb{P}(B)\mathbb{P}(A\vert B) \\\\ $$

• This is useful when there exists some partition of $\Omega$, namely, $\{B_{1}, B_{2}, \cdots\}$. A partition of a set $U$ is a collection of non-empty sets that are mutally disjoint and whose union is $U$. Such a partition allows us to represent $A$ as a diajoint union of the sets $A\cap B$, thus we have the law of total probability $$\mathbb{P} = \sum_{k=0}^{\infty}\mathbb{P}(A\cap B_{k}) = \sum_{k=0}^{\infty}\mathbb{P}(A\vert B_{k})\mathbb{P}(B_{k})$$
• Example
  • Considering the world of semi-conductor manufacturing. Room cleaness in the manufacturing process is critical, and dust particles are kept to a minimum. Let $A$ be the event of a manufacturing failure, and assume that it dependents on the number of dust particles via $$\mathbb{P}(A\vert B_{k}) = 1 - \frac{1}{k+1}$$
  • Where $B_{k}$ is the event of having $k$ dust particles in the room, further assume that $$\mathbb{P}(B) = \frac{6}{\pi^{2}(k+1)^{2}} \quad \text{for } k = 0,1,\cdots$$
  • $\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^{s}}$ is Riemann Zeta Funciton $$\begin{array}{ll} \because & \mathbb{P}(A) = \sum_{k=0}^{\infty}\mathbb{P}(A\vert B)\mathbb{P}(B) \\ \therefore & \mathbb{P}(A) = \sum_{k=0}^{\infty}\bigg(1 - \frac{1}{k+1}\bigg)\frac{6}{\pi^{2}(k+1)^{2}} \\ \because & \sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6} = \zeta(2)\\ \therefore & \mathbb{P}(A) = 1 - \frac{6}{\pi^{2}}\zeta(3) \\ \because & \zeta(3) \approx 1.2021 \\ \therefore & \mathbb{P}(A) \approx 0.2692 \\ \end{array}$$
  • Approximate by Julia

using SpecialFunctions

n = 2000

probAgivenB(k) = 1 - 1/(k+1)
probB(k) = 6/(pi^2*(k+1)^2)

numerical = sum([probAgivenB(k)*probB(k) for k in 1:n])
analytic = 1-6/pi^2*zeta(3)

println("Analytic: ", analytic, "\tNumerical: ", numerical)