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1. Mathematics

2. Session Probability - 3

3. Session Objectives • Random Variable • Probability Distribution • Applications of Probability • Class Exercise

4. Random Variable Let us consider a random experiment of tossing three coins (or a coin is tossing three times). Then sample space is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} We are interested in the number of heads in each outcome. Let X denote the number of heads in each outcome, then X takes the values

5. Random Variable (Cont.) X (HHH) = 3, X(HHT) = 2, X(HTH) = 2, X(THH) = 2 , X(HTT) = 1, X (THT) = 1, X (TTH) = 1 and X(TTT) = 0 i.e. X = 0, 1, 2 or 3 X is a variable obtained from the random experiment and called random variable.

6. P(X = 0) = Probability of getting no head = P(TTT) = P(X = 1) = Probability of getting one head = P(HTT or THT or TTH ) = P(X = 2) = Probability of getting two heads = P(HHT or THH or HTH ) = P(X = 3) = Probability of getting three heads = P(HHH) = Probability Distribution Now, the probability of each outcome in the example is

7. Cont. Probability distribution of X is: P(X) is called the probability distribution of the random variable X.

8. Example –1 Two cards are drawn successfully with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of aces. Solution:The total number of cards in a pack of cards is 52. Let X denote the number of aces, then X can take the values 0, 1 or 2. When X = 0 Both the cards are non-aces. P(X = 0) = P(First drawn card is non-ace and second drawn card is also non-ace)

9. Solution (Cont.) When X = 1 One card is an ace and other is a non-ace P(X = 1) = P(First drawn card is ace and second drawn card is non-ace or first drawn card is non-ace and second drawn card is ace)

10. Solution (cont.) = P(First drawn card is ace and second drawn card is non-ace) + P(first drawn card is non-ace and second drawn card is ace) When X = 2 Both the cards are aces Hence, the probability distribution of X is

11. Example –2 Four bad oranges are mixed accidently with 16 good oranges. Find the probability distribution of the number of bad oranges in a draw of two oranges. Solution: Total number of oranges = 4 (bad oranges) +16 (good oranges) = 20 Let X denote the number of bad oranges, then X can take the values 0, 1 or 2.

12. Solution (Cont.) P(X = 0) = Probability of getting no bad orange = Probability of getting 2 good oranges P(X = 1) = Probability of getting one bad orange

13. Solution (Cont.) P(X = 2) = Probability of getting two bad oranges The probability distribution of X is given by

14. Probability that the number appearing on the top is less than 3 in a single through of a die = Probability of success Example –3 A fair die is tossed twice. If the number appearing on the top is less than 3. It is a success. Find the probability distribution of number of successes. (CBSE 2004) Solution: Let X denote the number of successes (getting a number less 3) when a fair die is tossed twice. Then X takes the values 0, 1 or 2.

15. Probability of not getting a success Solution (Cont.)

16. Solution (Cont.) The probability distribution of X is given by

17. Example –4 From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs. (CBSE 2004) Solution: Total number of bulbs = 30 = 6 defective bulbs + 24 non-defective bulbs Let X denote the number of defective bulbs. Then X can take the values 0, 1, 2, 3 or 4.

18. Solution (Cont.)

19. Solution (Cont.) The probability distribution of X is given by

20. Example –5 An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement. Solution: Total number of balls = 4 red + 3 blue = 7 balls Let X denote the number of blue balls. Then X can take the values 0, 1, 2 or 3.

21. Solution (cont.)

22. Solution (cont.) The probability distribution of X is given by

23. Applications of Probability Probability theory has diverse use in science and technology these days. In Biology, genetics involves a lot of calculations using probability. It is also used in various fields of engineering.

24. Example –6

25. Solution The circuit would work if all the switches S1, S2 and S3 work together. Consider event S1 : switch S1 is working event S2 : switch S2 is working event S3 : switch S3 is working

26. Solution (Cont.) P(circuit in working condition) = P(S1, S2 and S3 working together) [As working of one switch does not effect the working of the other i.e. independent events]

27. Example –7 The probability that a person visiting a dentist will have his teeth cleaned is 0.44, the probability that he will have a cavity filled is 0.24. The probability that he will have his teeth cleaned or a cavity filled is 0.6. What is the probability that a person visiting his dentist will have his teeth cleaned and a cavity filled? Solution: Let event A = the person will have his teeth cleaned. event B = he will have the cavity filled.

28. Solution (Cont.) By addition theorem

29. Example-8 • A town has two fire extinguishing engines functioning independently. • The probability of availability of each engine, when needed is 0.95. • What is the probability that • (i) neither of these is available when needed. • (ii) exactly one engine is available when needed. Solution:Let event A = availability of one engine. event B = availability of other engine.

30. Solution (Cont.) P(A) = P(B) = 0.95 [Given] (i) P(neither of these is available when needed) [As availability of one engine is independent of the availability of the other] = (0.05) × (0.05) = 0.0025

31. Solution (Cont.) (ii) P(exactly one engine is available when needed)

32. A company has estimated that the probabilities of success for three products introduced in the market are respectively. Assuming independence, find (i) the probability that the three products are successful. (ii) the probability that none of the products is successful. Example –9 Solution: Let event A = First product is successful event B = Second product is successful event C = Third product is successful

33. Solution (Cont.) (i) P(three products are successful)

34. Solution (Cont.) (ii) P(none of the products are successful)

35. Example –10 • The odds against a husband who is 45 years old, living till he is 70 • are 7 : 5 and the odds against his wife who is now 36, living till she • is 61 are 5 : 3. Find the probability that • the couple will be alive 25 years hence, • none of them will be alive 25 years hence, • (iii) at least one of them will be alive 25 years hence.

36. Solution Let event A = the husband will be alive 25 years hence event B = the wife will be alive 25 years hence (i) P(couple will be alive 25 years hence)

37. Solution (cont.) (ii) P(none of them will be alive 25 years hence)

38. Solution (cont.) (iii) P(at least one of them will be alive 25 years hence)

39. Thank you