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Probability and Mathematical Statistics

Random Events and Probability

This chapter systematically studies the basic concepts of random events, probability, probability calculation, conditional probability, the law of total probability, Bayes’ theorem, and event independence.

Random Events and Sample Space

  • Random event: uncertainty in experiment outcomes
  • Sample space: the set of all possible outcomes
  • Event relations and operations: inclusion, equality, union, intersection, difference, complementary event

Definition and Properties of Probability

  • Classical model: P(A)=mnP(A) = \frac{m}{n}
  • Geometric model: probability = favorable length / total length (area/volume)
  • Basic properties: 0P(A)10 \leq P(A) \leq 1, P(Ω)=1P(\Omega) = 1, P()=0P(\varnothing) = 0
  • Addition formula: P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B)

Conditional Probability, Law of Total Probability, Bayes’ Theorem

  • Conditional probability: P(AB)=P(AB)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}
  • Law of total probability: P(B)=P(Ai)P(BAi)P(B) = \sum P(A_i)P(B|A_i)
  • Bayes’ theorem: P(AiB)=P(Ai)P(BAi)P(Aj)P(BAj)P(A_i|B) = \frac{P(A_i)P(B|A_i)}{\sum P(A_j)P(B|A_j)}

Independence of Events

  • P(AB)=P(A)P(B)P(A \cap B) = P(A)P(B)
  • Independent repeated trials

Exercises

  1. Toss a fair coin twice. Find the probability of getting heads exactly once.
  2. Given P(A)=0.4,P(B)=0.5,P(AB)=0.2P(A) = 0.4, P(B) = 0.5, P(A \cap B) = 0.2, find P(AB)P(A \cup B).
  3. Given P(A)=0.3,P(BA)=0.5P(A) = 0.3, P(B|A) = 0.5, find P(AB)P(A \cap B).
  4. There are three boxes containing 2, 3, and 5 balls, respectively. One box is chosen at random and a ball is drawn. Given that the ball came from the third box, what is the probability that it is white? (The third box has 3 white and 2 black balls, the others are all white.)
  5. Given P(A)=0.6,P(B)=0.5,P(AB)=0.3P(A) = 0.6, P(B) = 0.5, P(A \cap B) = 0.3, determine whether A and B are independent.
Reference Answers

1. Toss a fair coin twice. Find the probability of getting heads exactly once.

Sample space has 4 outcomes, 2 have exactly one head, probability 24=0.5\frac{2}{4}=0.5


2. P(AB)P(A \cup B)

0.4+0.50.2=0.70.4+0.5-0.2=0.7


3. P(AB)P(A \cap B)

0.3×0.5=0.150.3 \times 0.5 = 0.15


4. Conditional probability

Probability of drawing a white ball from the third box: 35\frac{3}{5}, total probability 13×1+13×1+13×35=13+13+15=1115\frac{1}{3}\times1 + \frac{1}{3}\times1 + \frac{1}{3}\times\frac{3}{5} = \frac{1}{3}+\frac{1}{3}+\frac{1}{5}=\frac{11}{15}

Conditional probability =13×351115=13×35×1511=311=\frac{\frac{1}{3}\times\frac{3}{5}}{\frac{11}{15}}=\frac{1}{3}\times\frac{3}{5}\times\frac{15}{11}=\frac{3}{11}


5. Independence

0.6×0.5=0.30.6\times0.5=0.3, which equals P(AB)P(A\cap B), so they are independent.