1. Section 2.4 For any random variable X, the cumulative distribution function (c.d.f.) of X is defined to be F(x) = P(X  x).

2. 1.The random variable X in Class Exercise #5 of Section 2.3 was found to have p.m.f. Find the distribution function F(x). f(x) = if x = –2 if x = –1 if x = 0 if x = 1 1/10 1/5 if x = 22/5 F(x) = if x < –2 if –2  x < –1 if –1  x < 0 if 0  x < 1 0 1/10 1/5 2/5 if 1  x < 2 3/5 if 2  x 1

3. Section 2.4 A Bernoulli experiment is one which must result in one of two mutually exclusive and exhaustive outcomes often labeled in general as “success” and “failure”. The probability of “success” is often denoted as p, and the probability of “failure” is often denoted as q, that is, q = 1 − p. If X is a random variable defined to be one (1) when a success occurs and zero (0) when a failure occurs, then X is said to have a Bernoulli distribution with success probability p. The p.m.f. of X is f(x) = E(X) =Var(X) = The m.g.f. of X is p x (1 – p) 1–x if x = 0, 1. pp(1 – p) For any random variable X, the cumulative distribution function (c.d.f.) of X is defined to be F(x) = P(X  x).

4. 2. (a) (b) An envelope contains 7 one-dollar bills and 3 five-dollar bills. If 4 bills are randomly selected without replacement, find the probability that exactly 2 are five-dollar bills. If 4 bills are randomly selected with replacement, find the probability that exactly 2 are five-dollar bills. 10 4 7 2 3 2 (21)(3) ——— = 210 = 3 — or 10 (7)(6)(3)(2) ————— = (10)(9)(8)(7) 4! —— 2! 3 — 10 7 2 3 2 —— = 10 4 4! —— 2! 1323 —— = 5000 0.2646 or 7 — 10 3 — 10 22 4! —— 2! = 0.2646

5. A sequence of Bernoulli trials is a sequence of independent Bernoulli experiments where the probability of the outcome labeled “success”, often denoted as p, remains the same on each trial; the probability of “failure” is often denoted as q, that is, q = 1 – p. Suppose X 1, X 2,..., X n make up a sequence of Bernoulli trials. If X = X 1 + X 2 +... + X n, then X is a random variable equal to the number of successes in the sequence of Bernoulli trials, and X is said to have a binomial distribution with success probability p, denoted b(n, p). The p.m.f. of X is f(x) = E(X) = Var(X) = The m.g.f. of X is n!—— x! (n – x)! p x (1 – p) n–x if x = 0, 1, …, n

6. 3. (a) An urn contains 4 clear marbles and 10 colored marbles. If 3 marbles are randomly selected with replacement, find each of the following probabilities: P(exactly 2 marbles are clear) = P(exactly 2 marbles are colored) = P(at least one marble is colored) = 2 3! 2 5 —— — — 2! 1! 7 7 2 3! 5 2 —— — — 2! 1! 7 7 2 — 7 3 1 –

7. (b)If 7 marbles are randomly selected with replacement, find each of the following probabilities: P(exactly 2 marbles are clear) = P(exactly 2 marbles are colored) = P(at least one marble is colored) = P(between 2 and 4 marbles inclusive are clear) = 2 7! 2 5 —— — — 2! 5! 7 7 5 2 7! 5 2 —— — — 2! 5! 7 7 5 2 — 7 7 1 – 2 7! 2 5 —— — — + 2! 5! 7 7 53 7! 2 5 —— — — + 3! 4! 7 7 44 7! 2 5 —— — — 4! 3! 7 7 3

8. (c)Consider the random variable Q = “the number of clear marbles when 3 marbles are selected at random with replacement” with p.m.f. f(q). Find f(q), E(Q), and Var(Q). f(q) = E(Q) = Var(Q) = 3q3q 2 — 7 q 5 — 7 3 – q if q = 0, 1, 2, 3

9. (d)Consider the random variable V = “the number of clear marbles when 7 marbles are selected at random with replacement” with p.m.f. g(v). Find g(v), E(V), and Var(V). g(v) = E(V) = Var(V) = 7v7v 2 — 7 v 5 — 7 7 – v if v = 0, 1, 2, …, 7

10. (e)Consider the random variable W = “the number of colored marbles when 7 marbles are selected at random with replacement” with p.m.f. h(w). Find h(w), E(W), and Var(W). (Note that V + W = 7.) h(w) = E(W) = Var(W) = 7w7w 5 — 7 w 2 — 7 7 – w if w = 0, 1, 2, …, 7

11. 4. (a) (b) The probability that a certain type of fire-cracker will explode is 0.8 or 80%. By defining the random variables X = “number of explosions with n fire crackers”, and Y = “number of duds with n fire crackers”, use the table of binomial probabilities to find the probability that out of 6 fire crackers, exactly 2 explode, X has a b(6, 0.8) distribution; Y has a b(6, 0.2) distribution. P(X = 2) =P(Y = 4) = P(Y  4) – P(Y  3) = 0.9984 – 0.9830 =0.0154 out of 6 fire crackers, exactly 2 do not explode, X has a b(6, 0.8) distribution; Y has a b(6, 0.2) distribution. P(Y = 2) = P(Y  2) – P(Y  1) = 0.9011 – 0.6553 =0.2458

12. (c) (d) (e) out of 15 fire crackers, exactly 5 do not explode, X has a b(15, 0.8) distribution; Y has a b(15, 0.2) distribution. P(Y = 5) = P(Y  5) – P(Y  4) = 0.9389 – 0.8358 =0.1031 out of 15 fire crackers, exactly 13 explode, X has a b(15, 0.8) distribution; Y has a b(15, 0.2) distribution. P(X = 13) =P(Y = 2) = P(Y  2) – P(Y  1) = 0.3980 – 0.1671 =0.2309 out of 10 fire crackers, at least 3 do not explode, X has a b(10, 0.8) distribution; Y has a b(10, 0.2) distribution. P(Y  3) =1 – P(Y  2) = 1 – 0.6778 =0.3222

13. (f) (g) (h) out of 15 fire crackers, at least 12 explode, X has a b(15, 0.8) distribution; Y has a b(15, 0.2) distribution. P(X  12) =P(Y  3) = 0.6482 out of 10 fire crackers, between 5 and 8 inclusive explode. Suppose n fire crackers are ignited. Find the smallest value of n such that the probability of at least one dud is greater than 0.9. X has a b(10, 0.8) distribution; Y has a b(10, 0.2) distribution. P(5  X  8) =P(2  Y  5) =P(Y  5) – P(Y  1) = 0.9936 – 0.3758 =0.6178 X has a b(n, 0.8) distribution; Y has a b(n, 0.2) distribution. P(Y  1) = 1 – P(Y = 0) =1 – (0.8) n 1 – (0.8) n > 0.9  0.1 > (0.8) n  ln(0.1) > ln[(0.8) n ] 

14. ln(0.1) > n ln(0.8)  ln(0.1) ——— < n  ln(0.8) n  11