MATH 2421: Tutorial 2

Date: 2024-09-21 22:25:32

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Tutorial 2

Properties of Probability
- 60% of students: wear neither a ring / necklace
- 20% wears ring; 30% wear a necklace
- what is the probability of student wearing
- a ring / necklace?
  - $P (R^{c} N^{c}) = 0.6$
  - $P (R) = 0.2; P (N) = 0.3$
  - $P (R \cup N) = 1 - P (R^{c} N^{c}) = 0.4$
- a ring & necklace?
  - $P (RN) = P (R) + P (N) - P (R \cup N) = 0.1$
Sample Spaces Having Equally Likely Outcomes
- Five people, designated as A, B, C, D, E, are arranged in linear order. Assuming that each possible order is equally likely, what is the probability that
- there is exactly one person between A and B?
  - $\frac{2 \cdot 3 ! \cdot 3}{5 !} = \frac{3}{10}$
- there are exactly two people between A and B?
  - $\frac{2 \cdot 3 ! \cdot 2}{5 !} = \frac{2}{10} = \frac{1}{5}$
- there are three people between A and B?
  - $\frac{2 \cdot 3 ! \cdot 1}{5 !} = \frac{1}{10}$
Conditional Probability
- from years of production: $2/1000$ of robots are defective.
- if robot is not defective, it tests negative for 97% of time
- if robot is defective, it tests positive for 99.5% of time
- probability that a randomly selected robot: negative result?
  - $T$ : event of being tested positive
  - $D$ : event of actually being defective
  - $P (T^{c}) = P (T^{c} ∣ D) P (D) + P (T^{c} ∣ D^{c}) P (D^{c}) = 0.005 \cdot 0.002 + 0.97 \cdot 0.998 = 96.807%$
- probability that a randomly selected robot is defective, given positive result?
  - $P (D ∣ T) = \frac{P ( D \cap T )}{P ( T )} = \frac{0.002 \cdot 0.995}{0.03193} = \frac{0.00199}{0.03193} = 6.23%$
General Multiplication Rule
- deck of 52: divided randomly into 4 distinct piles of 13 cards each
- probability that each pile: has exactly 1 ace?
  - $\frac{( 1 4 ) ( 12 48 )}{( 13 52 )} \frac{( 1 3 ) ( 12 36 )}{( 13 39 )} \frac{( 1 2 ) ( 12 24 )}{( 13 26 )} \frac{( 1 1 ) ( 12 12 )}{( 13 13 )}$
Inclusion-Exclusion Principle
- Suppose that there are $N$ distinct types of coupons and each time one obtains a coupon it is, independent of prior selections, equally likely to be any one of the $N$ types. Denote $D_{n}$ as the number of distinct types of coupons that are contained in the first $n$ selections. Find $P (D_{n} = k)$ .
- probability that distinct tickets $\leq k$ : $\frac{( k N ) k ^{n}}{N ^{n}}$
  - no. of ways to choose $k$ types * no. of ways to choosing $n$ tickets within $k$ / total ways of choosing $n$ tickets within $N$
  - must be multiplied by $P (A)$ in order to be correct.
- let $A$ : no empty category exists within $k$
- $A^{c}$ : $\exists$ empty slot
- $B_{i}$ : $i$ -th slot is empty
- $P (A^{c}) = P (⋃_{i = 1}^{k} B_{i})$ $P (i = 1 ⋃ k B_{i}) = t hm 1 \leq i \leq k \sum P (B_{i}) - 1 \leq i < j \leq k \sum P (B_{i} B_{j}) + \dots + (- 1)^{k + 1} P (B_{1} B_{2} \dots B_{k}) = (1 k) (\frac{k - 1}{k})^{n} - (2 k) (\frac{k - 2}{k})^{n} \dots + (- 1)^{k + 1} (k k) (\frac{k - k}{k})^{n} = r = 1 \sum k (- 1)^{r - 1} (r k) (\frac{k - r}{k})^{n}$
- final answer: $(k N) \frac{k ^{n}}{N ^{n}} \times r = 1 \sum k (- 1)^{r - 1} (r k) (\frac{k - r}{k})^{n} = (k N) r = 1 \sum k (- 1)^{r - 1} (r k) (\frac{k - r}{N})^{n}$

MATH 2421: Probability