MATH 2421: Tutorial 3

Date: 2024-09-24 18:06:07

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Tutorial

Tutorial
- $n$ evens are independent if: for every sub-collection $A_{i_{1}}, \dots, A_{i_{r}}$
  - and meaningful sub-collection: must have $\geq 2$ elements
  - i.e. $2^{n} - n - 1$
    - all subset - no. of individual events - empty set

Exercises

Independent events
- each independently component: function at $p_{i}$ prob.
- that is the prob. that any of $n$ component works?
- as, for independent events, $P (A B) = P (A) P (B)$
- Solution: $P (⋃) = \sum p_{i} - \sum_{i < j} p_{i} p_{j} + \dots + (- 1)^{n + 1} \prod_{1 < i < n} p_{i}$
- 👨‍🏫 or, more simply:
  - let $s$ : system functions
    - ans $s_{i}$ : component $i$ functions
  - $P (s) = 1 - P (s^{c}) = 1 - ⋂_{1 < i < n} s_{i}^{c} = 1 - \prod_{1 < i < n} (1 - p_{i})$
Bayes' theorem
- 3 types of flashlight
  - $P (T > 100∣ I) = 0.7$
    - $P (I) = 0.2$
  - $P (T > 100∣ II) = 0.4$
    - $P (II) = 0.3$
  - $P (T > 100∣ III) = 0.3$
    - $P (III) = 0.5$
- a) $P (T > 100)$ ?
  - sum of products: $P (T > 100∣ I) P (I) + P (T > 100∣ II) P (II) + P (T > 100∣ III) P (III) = 0.14 + 0.12 + 0.15 = 0.41)$
- b) $P (j ∣ T > 100)$ for $j = I, II, III$ ?
  - $= \frac{P ( T > 100∣ j ) P ( j )}{P ( T > 100 )}$
  - $I : 0.14/0.41 = 0.3415$
  - $II : 0.12/0.41 = 0.2927$
  - $III : 0.15/0.41 = 0.3659$
Month Hall problem
- 1 car and 2 goats
- you can choose a door, $1$
- after your choice, the host removes one door with a goat behind, $2$
- then, does changing the door make you more likely to get the car?
- let $P (D_{i} = C)$ probability that car is behind it
- let $P (O_{i})$ probability that door $i$ was opened
  - $P (O_{2} ∣ D_{1} = C) = \frac{1}{2}$
  - $P (O_{2} ∣ D_{2} = C) = 0$
  - $P (O_{2} ∣ D_{3} = C) = 1$ // no other box to open $P (D_{1} = C ∣ O_{2}) = \frac{\frac{1}{2} \frac{1}{3}}{\frac{1}{2} \frac{1}{3} + 0 \cdot \frac{1}{3} + 1 \cdot \frac{1}{3}} = 1/3$ $P (D_{2} = C ∣ O_{2}) = \frac{0 \cdot \frac{1}{3}}{\frac{1}{2} \frac{1}{3} + 0 \cdot \frac{1}{3} + 1 \cdot \frac{1}{3}} = 0$ $P (D_{3} = C ∣ O_{2}) = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2} \frac{1}{3} + 0 \cdot \frac{1}{3} + 1 \cdot \frac{1}{3}} = 2/3$
- thus: better to change
Random Monty Hall problem
- What if the host opens a door that we did not pick, but which might contain car?
- 1/2 $P (D_{1} = C ∣ D_{3} = G) = \frac{P ( D _{3} = G ∣ D _{1} = C ) \cdot P ( D _{1} = C )}{P ( D _{3} = G ∣ D _{1} = C ) P ( D _{1} = C ) + P ( D _{3} = G ∣ D _{1} = G ) P ( D _{1} = G )} = \frac{1 \cdot 1/3}{1 \cdot 1/3 + 1/2 \cdot 2/3} = \frac{1}{2}$

MATH 2421: Probability