Discrete Probability Distributions Discrete Random Variables & Probability Distributions

1. PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS Discrete Probability DistributionsDiscrete Random Variables & Probability Distributions

2. Random Variable Definition - A random variable is a mathematical function that associates a number with every possible outcome in the sample space S. Notation - Capital letters, usually X or Y, are used to denote random variables. Corresponding lower case letters, x or y, are used to denote particular values of the random variables X or Y. Definition - A discrete random variable X is a random variable that can take on or assume a finite number of possible values, say x1, x2, …, xk

3. Probability Mass Function Associated with a discrete random variable X having possible values x1, x2, …, xn is a function, p, called the probability mass function. The probability mass function of X associates with each possible value of X the probability of its occurrence. This set of ordered pairs, each of the form, (value of x, probability of that value occurring) or ( x, p(x) ) is the probability mass function of X. i.e., p(x)=P(X=x).

4. Probability Mass Function (Continued) The function is the probability mass function of the discrete random variable X if, for each possible outcome x , 1) 2)

5. Probability Distribution Function The (cumulative) probability distribution function, , of a discrete random variable X with probability mass function is given by

6. Example – Toss a coin 3 times If an experiment is “Toss a coin 3 times in sequence” and the random variable X is defined to be the number of heads that result, determine and plot the probability mass function and probability distribution function for X if The coin is fair The coin is biased with P(H)=0.75

7. Example – Toss a coin 3 times 1) If the coin is fair 1 2 3 Outcome Likelihood x 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1 HHH HHT HTH HTT THH THT TTH TTT 3 2 2 1 2 1 1 0 H H T H H T T 0 H H T T H T T

8. Example Solution - Toss a coin 3 times Continued From the outcome tree, using rules of probability we determine the value of the probability mass function of X as follows: x p(x)=P(X=x) 0 1/8 1 3/8 2 3/8 3 1/8 1

9. Example Solution - Toss a coin 3 times - Continued The Probability Distribution Function of X, F(x), is determined by asuming the appropriate values of p(x), since Therefore, x F(x) 0 1/8 1 4/8 2 7/8 3 1

10. Example Solution Continued p(x) 3/8 1/8 Probability Mass Function x 0 1 2 3 F(x) 1 0.5 Probability Distribution Function 0 x 0 1 2 3

11. Example Solution - Probability Mass Function 2) If the coin is biased with P(H)=0.75 likelihood x 2 3 1 outcomes 27/64 9/64 9/64 3/64 9/64 3/64 3/64 1/64 1 HHH HHT HTH HTT THH THT TTH TTT 3 2 2 1 2 1 1 0 H H T H H T T 0 H H T T H T T

12. Example Solution - Continued From the outcome tree, and applying rules of probability we determine the value of the probability mass function of X as follows: x p(x)=P(X=x) 0 1/64 1 9/64 2 27/64 3 27/64 1

13. Example Solution - Continued Then the Probability Distribution Function of X, F(X), is determined by asuming the appropriate values of p(x), since Therefore, x F(x) 0 1/64 1 10/64 2 37/64 3 1

14. Example Solution - Calculation of Mean P(x) 3/8 1/8 x 0 1 2 3 Mean µ P(x) 27/64 x 0 1 2 3 Mean µ

15. Example Solution - Continued p(x) 27/64 9/64 Probability Mass Function x 0 1 2 3 F(x) 1 32/64 Probability Distribution Function 0 x 0 1 2 3

16. Mean or Expected Value of a Discrete Random Variable X Mean or Expected Value of X Note The interpretation of μ: the average value of X in the long term.

17. Example - Toss a coin 3 times If an experiment is “Toss a coin 3 times in sequence” and the random variable X is defined to be the number of heads that result, what is the mean or expected value of X if The coin is fair b. The coin is biased with P(H)=0.75

18. Example Solution - Calculation of Mean Fair coin Using the Definition = (0)(1/8) + (1)(3/8) + (2)(3/8) + (3)(1/8) = 12/8 = 1.5 Using a Tabular format x p(x)=P(X=x) x P(x) 0 1/8 0/8 1 3/8 3/8 2 3/8 6/8 3 1/8 3/8 1 µ=12/8=1.5 P(x) 3/8 1/8 x 0 1 2 3 Mean µ

19. Example Solution - Calculation of Mean Biased coin p=3/4 Using the definition = (0)(1/36) + (1)(9/64) + (2)(27/64) + (3)(27/64) = 2.25 Using a Tabular format x p(x)=P(X=x) x P(x) 0 1/64 0/64 1 9/64 9/64 2 27/64 54/64 3 27/64 81/64 1 µ=144/64=2.25 P(x) 27/64 x 0 1 2 3 Mean µ

20. Standard Deviation of a Discrete Random Variable X the Standard Deviation of the discrete random variable X Where var(X) is the variance of X and

21. Example- Toss a coin 3 times If an experiment is “Toss a coin 3 times in sequence” and the random variable X is defined to be the number of heads that result, what is the standard deviation of X if The coin is fair The coin is biased with P(H)=0.75

22. Example Solution – Toss a coin 3 times Fair coin x p(x)=P(X=x) x2 x2P(x) 0 1/8 0 0/8 1 3/8 1 3/8 2 3/8 4 12/8 3 1/8 9 9/8 24/8

23. Example Solution –Calculation of Standard Deviation Biased coin, p=3/4 x p(x)=P(X=x) x2 x2P(x) 0 1/64 0 0/64 1 9/64 1 9/64 2 27/64 4 108/64 3 27/64 9 243/64 360/64

24. Example – Family Planning In planning a family of 4 children, find the probability distribution of: X = the number of boys Y = the number of changes in sex sequence Find (i) the probability mass and distribution functions (and plot), (ii) the mean, and (iii) the standard deviation.

25. Example Solution– Family planning - tree Outcomes Prob. B BBBB BBBG BBGB BBGG BGBB BGBG BGGB BGGG GBBB GBBG GBGB GBGG GGBB GGBG GGGB GGGG 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 1/16 B G B B G G B B B G G 1/2 B G G B 0 B G B B G G 1/2 G B B G B G G G

26. Example Solution X Y B 4 3 3 2 3 2 2 1 3 2 2 1 2 1 1 0 0 1 2 1 2 3 2 1 1 2 3 2 1 2 1 0 B G B B G G B B B G G 1/2 B G G B 0 B G B B G G 1/2 G B B G B G G G

27. Example – Family Planning – Solution Continued From the outcome tree and use of rules of probability the probability mass function of X is: Then, 1

28. Example – Family Planning – Solution Continued P(x) F(x) 1 6/16 4/16 y x 0 1 2 3 4 0 1 2 3

29. Example – Family Planning – Solution Continued From the outcome tree and use of rules of probability the probability mass function of X is:

30. Example – Family Planning – Solution Continued P(y) F(y) 1 6/16 2/16 y y 0 1 2 3 0 1 2 3

31. Discrete Uniform Distribution Definition - If the random variable X assumes the values x1, x2, ... xk with equal probabilities, then X has a discrete uniform distribution with probability mass function

32. Discrete Uniform Distribution (Continued) If X has the discrete uniform distribution U(k), then the mean and standard deviation are and

33. Rules If a and b are constants and if  = E(X) is the mean and  is the standard deviation of the discrete random variable X, respectively, and if Y=aX+b then And

34. Example The discrete random variable X has probability mass function , p(x), as follows : x p(x)=P(X=x) 0 1/8 1 3/8 2 3/8 3 1/8 Define the discrete random variable Y as Y=2X+5 Find: a) pY(y) b) FY(y) c) µY d) σY

35. Example Solution The probability mass function of Y is obtained by evaluating Y=2x+5 for each value pY(y) of X as follows: x Y pY(y) FY(y) 0 5 1/8 1/8 1 7 3/8 4/8 2 9 3/8 7/8 3 11 1/8 1 1

36. Example Solution Plot of pX(x) and pY(y) are: Mean of Y µY= E(2X+5) = 2 µX +5 = 2(1.5) +5 =8 Standard Deviation of Y pX(x) pY(y) 3/8 3/8 x 1/8 1/8 y 0 1 2 3 0 1 2 3 4 5 6 7 8 9 10 11

37. Rules - Continued If Y = g(X) is a function of a discrete random variable X, then

38. Chebyshev’s Theorem The probability that any random variable X will assume a value within k standard deviations of the mean is at least , i.e., Remark: Chebyshev’s Theorem gives a conservative estimate of the probability that a random variable assumes a value within k standard deviations of its mean for any real number k.

39. PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS Joint Probability Distributions

40. Joint Probability Mass Function The function p(x, y) is a joint probability mass function of the discrete random variables X and Y if 1. p(x, y)  0 for all (x, y) 2. 3. P(X = x, Y = y) = p(x, y) For any region A in the xy plane, P[(X, Y)  A] =

41. Example Two refills for a ballpoint pen are selected at random from a box that contains 3 blue refills, 2 red refills and 3 green refills. If X is the number of red refills and Y is the number of green refills selected, find a. The joint probability mass function p(x, y), and b. P[(X, Y)  A], where A is the region {(x, y)|x + y  1}

42. Example - Solution The possible pairs of values (x, y) are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2), and (2, 0), where p(0, 1), for example, represents the probability that a red and a green refill are selected. The total number of equally likely ways of selecting any 2 refills from the 8 is: The number of ways of selecting 1 red from 2 red refills and 1 green from 3 green refills is

43. Example – Solution Continued Hence, p(0, 1) = 6/28 = 3/14. Similar calculations yield the probabilities for the other cases, which are presented in the following table. Note that the probabilities sum to 1.

44. Example – Solution Continued The joint probability mass function of X and Y may be expressed mathematically as: for x = 0, 1, 2; y = 0, 1, 2; 0  x + y  2. b. P[(X, Y)  A] = P(X + Y  1) = p(0, 0) + p(0, 1) + p(1, 0)

45. Marginal Distributions If a function p(x, y) is a joint probability mass function of the discrete random variables X and Y, the marginal probability mass functions of X alone and of Y are , respectively, and

46. Example Show that the column and row totals of the following table give the marginal distribution of X alone and of Y alone.

47. Example – Solution Continued For the random variable X, we see that

48. Example Solution g(x) 15/28 10/28 3/28 x 0 1 2 G(x) 1 25/28 10/28 x 0 1 2 3

49. Example – Solution Continued For the random variable Y, we see that

50. Example Solution h(y) 15/28 12/28 1/28 x 0 1 2 H(y) 1 27/28 15/28 x 0 1 2