1. Define Discrete Probability. Give an example. Discrete probability deals with outcomes that can be counted and listed. Each outcome has a probability, and the sum of all probabilities is 1. Example: Tossing a fair coin has two discrete outcomes: Heads or Tails, each with probability $\frac{1}{2}$. 2. What does Finite Probability mean? Finite probability refers to probability models where the number of possible outcomes is limited and countable. 3. How does an experiment differ from a sample space and an event? Explain. - Experiment: A procedure with uncertain outcomes (e.g., rolling a die). - Sample space: The set of all possible outcomes of the experiment (e.g., $\{1,2,3,4,5,6\}$ for a die). - Event: A subset of the sample space; outcomes of interest (e.g., rolling an even number $\{2,4,6\}$). 4. An urn contains four blue balls and five red balls. What is the probability that a ball chosen at random from the urn is blue? Total balls = $4+5=9$ Probability(blue) = $\frac{4}{9}$ 5. What is the probability that when two dice are rolled, the sum of the numbers on the two dice is 7? Total outcomes when two dice are rolled = $6 \times 6 = 36$ Favorable outcomes for sum 7: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ = 6 Probability = $\frac{6}{36} = \frac{1}{6}$ 6. What is the probability that a person picks the correct six numbers out of 40? Total ways to choose 6 numbers from 40 = $\binom{40}{6} = \frac{40!}{6!\times 34!}$ Probability(correct pick) = $\frac{1}{\binom{40}{6}}$ 7. Find the probability that a hand of five cards in poker contains four cards of one kind. Total 5-card hands = $\binom{52}{5}$ Number of hands with four of a kind: - Choose rank for four cards: 13 ways - Choose 4 cards of that rank: only 1 way (all 4 suits) - Choose 1 card from remaining 48 cards: 48 ways Number of such hands = $13 \times 1 \times 48 = 624$ Probability = $\frac{624}{\binom{52}{5}}$ 8. Probability that numbers 11, 4, 17, 39, and 23 are drawn in that order from 50 balls: (a) Without replacement: Probability = $\frac{1}{50} \times \frac{1}{49} \times \frac{1}{48} \times \frac{1}{47} \times \frac{1}{46}$ (b) With replacement: Probability = $\left(\frac{1}{50}\right)^5$ 9. A sequence of 10 bits is randomly generated. Probability at least one bit is 0? Total sequences = $2^{10} = 1024$ Probability all bits are 1 = $\left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$ Probability(at least one 0) = $1 - \frac{1}{1024} = \frac{1023}{1024}$ 10. Probability a positive integer selected from 1 to 100 is divisible by 2 or 5: Count divisible by 2: $\lfloor\frac{100}{2}\rfloor=50$ Count divisible by 5: $\lfloor\frac{100}{5}\rfloor=20$ Count divisible by 10 (both 2 and 5): $\lfloor\frac{100}{10}\rfloor=10$ By inclusion-exclusion: Numbers divisible by 2 or 5 = $50 + 20 - 10 = 60$ Probability = $\frac{60}{100} = \frac{3}{5}$