Lecture 4,5 slide about probability and statistics
- 1. Introduction to Probability Chapter 2: Conditional Probability Dr. Nitin Gupta Department of Mathematics Indian Institute of Technology Kharagpur, Kharagpur - 721 302, INDIA. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 1 / 17
- 2. Outline 1 Conditional Probability 2 Independence 3 Theorem of total probability 4 Bayes’ theorem N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 2 / 17
- 3. References 1 Probability and statistics in engineering by Hines et al (2003) Wiley. 2 Mathematical Statistics by Richard J. Rossi (2018) Wiley. 3 Probability and Statistics with reliability, queuing and computer science applications by K. S. Trivedi (1982) Prentice Hall of India Pvt. Ltd. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 3 / 17
- 4. Conditional Probability Consider a family having two children, then Ω = {BB, GB, BG, GG}, n(Ω) = 4. Consider the event A: both the children are girls. Then P(A) = 1/4. If some information in the form of ”E: at least one of the children is a girl” is known. Then reduced sample space is E = {GB, BG, GG}. Then the probability of A given the condition E is P(A|E) = 1/3. Note that P(A|E) ≥ P(A). P(A|E) = n(A∩E) n(E) = n(A∩E)/n(Ω) n(E)/n(Ω) = P(A∩E) P(E) , provided P(E) > 0. Definition Let probability model be (Ω, f , P). Then the conditional probability of A ∈ f given B is defined as P(A|B) = P(A ∩ B) P(B) , P(B) > 0. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 4 / 17
- 5. Multiplication rule The probability that n events A1, A2, . . . , An ∈ f occur in a sequence is P(∩n i=1Ai ) = P(A1)P(A2|A1)P(A3|A1 ∩ A2) · · · P(An|A1 ∩ A2 · · · ∩ An−1), provided P(A1) > 0, P(A1 ∩ A2) > 0, . . . , P(A1 ∩ A2 · · · ∩ An−1) > 0. Example A bag contains 5 red, 5 white and 4 blue balls. If someone draws 3 balls one by one without replacement, then the probability that three balls will be drawn in the sequence red-white-blue is P(R1 ∩ W2 ∩ B3) = P(B3|W2 ∩ R1)P(W2|R1)P(R1) = P(R1)P(W2|R1)P(B3|W2 ∩ R1) = 5 14 × 5 13 × 4 12 . Here note that R1, W2, B3 are dependent events. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 5 / 17
- 6. Example Example A bag contains 5 red, 5 white and 4 blue balls. If someone draws 3 balls one by one with replacement, then the probability that three balls will be drawn in the sequence red-white-blue is P(R1 ∩ W2 ∩ B3) = P(R1)P(W2)P(B3) = 5 14 × 5 14 × 4 14 . Here note that R1, W2, B3 are independent events. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 6 / 17
- 7. Example Example In a war game, submarine S1 targets S2, and both S2 and S3 target S1. The probabilities of S1, S2 and S3 hitting their targets are 1/2, 2/3 and 1/3 respectively. They shoot simultaneously. We want to determine the conditional probability that S2 hits the target and S3 does not given that S1 is hit. Let event Ei denote that submarine Si hitting its targets. Here the required probability is P(E2 ∩ ¯ E3|S1 is hit) = P(E2)P( ¯ E3) P(E2 ∪ E3) = 2 3 × 2 3 2 3 + 1 3 − 2 3 × 1 3 = 4 7 N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 7 / 17
- 8. Independence Two events are independent if the occurrence of one does not effect the occurrence or nonoccurrence of the other. Definition Events A and B are independent if P(A|B) = P(A). Hence P(A ∩ B) = P(A)P(B) and also P(B|A) = P(B). If A and B are independent, then A and B̄ are independent. If A and B are independent, then Ā and B are independent. If A and B are independent, then Ā and B̄ are independent. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 8 / 17
- 9. Example Example Suppose that P(A) = 0.4, P(B) = 0.5, and A and B are independent events. Determine P(Ac ∪ Bc). Note that P(Ac ∪ Bc ) = P(Ac ) + P(Bc ) − P(Ac ∩ Bc ) = 1 − P(A) + 1 − P(B) − P(Ac )P(Bc ) = 1 − 0.4 + 1 − 0.5 − (1 − 0.4)(1 − 0.5) = 0.8, it follows since A and B are independent, then Ā = Ac and B̄ = Bc are also independent. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 9 / 17
- 10. Definition The n events A1, . . . , An are mutually independent if and only if the probability of intersection of any 2, 3, . . . , n of these sets is product of their respective probabilities, i.e., for r = 2, 3, . . . , n, P(Ai1 ∩ Ai2 ∩ · · · ∩ Air ) = P(Ai1 )P(Ai2 ) · · · P(Air ). N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 10 / 17
- 11. Example Consider the following electronic system (see diagram), which shows the probabilities of the system components operating properly (i.e, the reliability of the components). Assume that each component operates independently. Find the system reliability, i.e., the probability that the entire system operates? 0.9 0.8 0.7 Solution: Since components are mutually independent. Hence System reliability = (1 − (1 − 0.9)(1 − 0.8)) × 0.7 = 0.686 N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 11 / 17
- 12. Example Consider the experiment of rolling two fair dice repeatedly and independently until a total of 5 or a total of 7 appears. We want to determine the probability that a total of 5 is rolled before a total of 7 is rolled. Solution: Let event Ai denote that 5 is rolled before a 7 in the ith trail and experiment terminates. P(Ai ) = P i−1 j=1 (5 ∪ 7)c ∩ 5 =indep 26 36 i−1 × 4 36 Now P(5 before 7) = P ∞ [ i=1 Ai ! =disjoint ∞ X i=1 P(Ai ) = ∞ X i=1 26 36 i−1 × 4 36 = 4/36 1 − 26 36 = 2 5 . N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 12 / 17
- 13. Theorem of total probability Events E1, . . . , En are mutually exclusive and exhustive, and event A is caused by happening of E1, . . . , En, then P(A) = n X i=1 P(A|Ei )P(Ei ), here P(Ei ) 0, i = 1, 2, . . . , n. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 13 / 17
- 14. Bayes’ theorem Events E1, . . . , En are mutually exclusive and exhustive, and event A is caused by happening of E1, . . . , En, then for i = 1, 2, . . . , n P(Ei |A) = P(A|Ei )P(Ei ) Pn j=1 P(A|Ej )P(Ej ) , here P(A) 0 and P(Ei ) 0, i = 1, 2, . . . , n. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 14 / 17
- 15. Example In a town there are 200 car drivers, 500 two-wheeler drivers and 20 bus drivers. There is a probability 0.01, 0.03 and 0.15 respectively for an accident involving car, two-wheeler and bus. One of the drivers meets with an accident, what is the probability that he/she was driving a car? Solution: Let event A,B,C denote the events that the chosen driver drives a car, a two-wheeler, a bus, respectively. Let event E denote an accident. Here P(A) = 200 200+500+20 = 20 72, P(B) = 50 72, P(C) = 2 72. Also P(E|A) = 0.01, P(E|B) = 0.03 and P(E|C) = 0.15. Now the required probability, using Bayes’ theorem, is P(A|E) = P(A)P(E|A) P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C) = 20 72 × 0.01 20 72 × 0.01 + 50 72 × 0.03 + 2 72 × 0.15 = 0.1 N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 15 / 17
- 16. Example Consider the experiment of rolling two fair dice repeatedly and independently until a total of 5 or a total of 7 appears. We want to determine the probability that a total of 5 is rolled before a total of 7 is rolled. Solution: Let event A denote that 5 is rolled before a 7. Then A = S∞ i=1(A ∩ Bi ), where Bi is the event that the game terminates in ith roll. Now the required probability, using theorem of total probability, is P(A) = ∞ X i=1 P(A ∩ Bi ) = ∞ X i=1 26 36 i−1 × 4 36 = 4/36 1 − 26 36 = 2 5 . N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 16 / 17
- 17. Summary Since there may be some information available about the outcome of the trail in a given experiment, hence we introduced the concept of the conditional probability. Also if this information is irrelevant to the event under considration from there comes the definiton of the independence of the events. These definitions can be used to find the reliabilities of the series-parallel or parallel-series structures. Hence examples are provided for the same. In the last theorem of total probability and Bayes’ theorem were presented. N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 17 / 17