Homework 3: Probability And Statistics

11/19/23, 11:36 PM HW03-Nyi Nyi Moe Htet Student: Nyi Nyi Moe Htet Date: 11/19/23 Instructor: Xiaolong Han Course: CSUN_MATH340_HAN (1) Assignment: HW03 1. Let A ⊂ B. Express the following probabilities as simply as possible. P(A∣B). Choose the correct answer below. A. 1 B. 0 D. P(B) G. P A c c P(B) P(AB) I. c L. P(AB) P(B) P(B) P(A) . Choose the correct answer below. P AB c B. F. P(B) C. P B P(B) P(A) G. P AB J. M. P(AB) c D. 1 P(A) P(A) I. P(A) P(A) K. P B P(B) E. F. c c P(AB) A. P(AB) H. P A B P(A) J. P A∣ B E. C. P B c P(AB) P(A) H. P(A) K. P A c L. c N. 0 O. P AB c P. P B P B c c P(A) P(A) P(B) P(B∣A). Choose the correct answer below. c c B. P(A) A. P A B P(B) P(A) P(B) P(AB) P(B) H. P A c I. K. P B c L. P(AB) 1 P(A) P(B) J. c . Choose the correct answer below. c A. F. P(A) G. P B∣ A E. 0 P(AB) D. C. P A B B. P(B) E. 1 F. P(B) P(A) P(A) C. c P(A) D. P A B P(B) G. 0 P(B) H. I. P(AB) P(B) M. P(AB) J. P A c K. P(B) N. P(B) L. P A c c P(A) c O. https://tdx.acs.pearsonprd.tech/api/v1/print/highered P(AB) P A P A B P A P. P B c c 1/5 11/19/23, 11:36 PM HW03-Nyi Nyi Moe Htet 2. Complete parts (a) and​(b). ​(a) Prove that if E and F are mutually​exclusive, then the following equation is true. P(E∣E ∪ F) = P(E) P(E) + P(F) Based on the definition of conditional probability, P(E∣E ∪ F) = Based directly on an axiom of probability, P(E{E ∪ F}) P(E ∪ F) . if E and F are mutually​exclusive, P(E ∪ F) = P(E) + P(F). ​Logically, P(E{E ∪ F}) = P(E). ​Therefore, the given equation is true. ​(b) Prove that if Ei ​, i ≥ 1 are mutually​exclusive, then the following equation is true. ∞ P Ej ∣ ∪ Ei = i=1 P Ej ∞ ∑ P Ei i=1 ∞ Based on the definition of conditional probability, P Ej ∣ ∪ ∞ Ei = P Ej i=1 an axiom of probability, P ∪ ∪ ∞ Ei i=1 divided by P ∪ Ei . i=1 ∞ ∞ By ∩ Ei i=1 = ∑ P Ei . ​Logically, since Ei ​, i ≥ 1 are mutually exclusive and Ej = Ei i=1 ∞ for exactly one​i, P Ej ∩ ∪ Ei i=1 = P Ej . ​Therefore, the given equation is true. https://tdx.acs.pearsonprd.tech/api/v1/print/highered 2/5 11/19/23, 11:36 PM HW03-Nyi Nyi Moe Htet 3. Consider two independent tosses of a fair coin. Let A be the event that the first toss results in​heads, let B be the event that the second toss results in​heads, and let C be the event that in both tosses the coin lands on the same side. Show that the events​A, B, and C are pairwise independent—that ​is, A and B are​independent, A and C are​independent, and B and C are independent—but not independent. Which of the following must be true of any events​A, B, and C if and only if they are pairwise​independent, but is not sufficient to conclude that the three events are​independent? A. P(AB) = P(A∣B)​, P(BC) = P(B∣C)​, and P(AC) = P(A∣C) B. P(AB) = P(BC)​, P(BC) = P(AC)​, and P(AC) = P(AB) C. P(A∣B) = P(B∣C)​, P(B∣C) = P(C∣A)​, and P(C∣A) = P(A∣B) D. P(AB) = P(A)P(B)​, P(BC) = P(B)P(C)​, and P(AC) = P(A)P(C) E. P(A∣B) = P(B)​, P(B∣C) = P(C)​, and P(C∣A) = P(A) In the order they appear in the previous​step, the six expressions across the three equations in the previous step are equal to the​following: 1 4 ​, 1 4 ​, 1 4 1 ​, 4 ​, 1 ​, and 4 1 4 . ​Therefore, the events​A, B, and C are pairwise independent. ​(Simplify your​answers.) Which of the following must be true of any events​A, B, and C if and only if they are​independent? A. P(ABC) = P(A)P(B∣A)P(C∣B) B. P(A)P(B)P(C) = P(AB)P(BC)P(AC) C. P(ABC) = P(A)P(B)P(C) D. P(A)P(B∣A)P(C∣AB) = P(ABC) The left side of the equation in the previous step is equal to 1 4 and the right side of the equation in the previous step is equal 1 to . Since these are not​equal, events​A, B, and C are not independent. 8 ​(Simplify your​answers.) 4. In each of n independent tosses of a​coin, the coin lands on heads with probability p. How large need n be so that the probability of 1 obtaining at least one head is at least ​? 2 Write an expression for the probability that there is at least one head in n tosses of the coin. 1 − (1 − p) n To find how large n must be so that the probability of obtaining at least one head is at least 1 2 ​, solve an appropriate inequality using the expression for the probability that there is at least one head in n tosses of the coin. n≥ − log 2 log (1 − p) https://tdx.acs.pearsonprd.tech/api/v1/print/highered 3/5 11/19/23, 11:36 PM HW03-Nyi Nyi Moe Htet 5. The probability of getting a head on a single toss of a coin is p. Suppose that A starts and continues to flip the coin until a tail shows​up, at which point B starts flipping. Then B continues to flip until a tail comes​up, at which point A takes​over, and so on. Let Pn,m denote the probability that A accumulates a total of n heads before B accumulates m. Show that the following equation is true. Pn,m = pPn − 1,m + (1 − p) 1 − Pm,n Let H1 denote the event that​A's first flip is heads and let E denote the event that A accumulates a total of n heads before B accumulates m. The most useful first step is to apply the law of total probability to note that Pn,m is equal to P H1 P E∣H1 c plus c P H 1 P E∣ H 1 . The probability that​A's first flip is heads is p . If​A's first flip is​heads, then the second overall flip is made by A. In order for the player making the second overall flip to accumulate their designated total of heads​first, the player making the second overall flip must accumulate n−1 more heads after​A's first flip before the other player accumulates m more heads after​A's first flip. ​(Simplify your​answers.) The probability that​A's first flip is not heads is 1−p . If​A's first flip is​heads, then the second overall flip is made by B. In order for the player making the second overall flip to accumulate their designated total of heads​first, the player making the second overall flip must accumulate m more heads after​A's first flip before the other player accumulates n more heads after​A's first flip. ​(Simplify your​answers.) If x is the probability that B accumulates their designated total of heads​first, then 1−x their designated total of heads first. Combining all of this​information, the given equation is true. ​(Simplify your​answer.) https://tdx.acs.pearsonprd.tech/api/v1/print/highered is the probability that A accumulates 4/5 11/19/23, 11:36 PM HW03-Nyi Nyi Moe Htet 6. Independent trials that result in a success with probability p are successively performed until a total of r successes is obtained. Show that the probability that exactly n trials are required is given by the following expression. n−1 r−1 r p (1 − p) n−r Use this result to solve the problem of the​points, which seeks to identify the probability that r successes occur before m failures. Hint​: In order for it to take n trials to obtain r​successes, how many successes must occur in the first n − 1 ​trials? In order for it to take n trials to obtain r​successes, the number of successes that must occur in the first n − 1 trials is the number of failures that must occur in the first n − 1 trials is occurs in the first n − 1 trials. ​(Simplify your​answers.) n−r r−1 and . Let B be the event that this many successes and failures Consider any particular sequence of the first n − 1 outcomes containing the numbers of successes and failures identified in the previous step. Each one of these sequences​will, by the assumed independence of​trials, occur with probability p The number of such sequences is equal to the number of combinations of ​(Simplify your​answers.) r−1 r−1 (1 − p) items selected from n−r . n−1 items. The probability that it takes exactly n trials to obtain r successes is equal to the probability of each sequence times the number of combinations of such sequences. required to obtain a total of r successes is equal to pP(B). The probability that exactly n trials are ​So, the given expression is the probability that exactly n trials are required. Now use the expression to solve the problem of the​points, which seeks to identify the probability that r successes occur before m failures. In terms of r and​m, the minimum number of trials that could be run for r successes to occur before m failures is r ​, and the maximum number of trials that could be run for r successes to occur before m failures is m+r−1 . ​(Simplify your​answers.) ​So, the probability that r successes occur before m failures is m+r−1 n−1 r ∑ p (1 − p) r−1 n= r ​(Simplify your​answers.) n−r . 7. Prove directly that P(E∣F) = P(E∣FG)P(G∣F) + P E∣FGc P Gc ∣F . The probability P(E∣F) can be rewritten as P(E∣F) = P E∣FG c c P G ∣F = P EFG P(F) P(EF) P(F) . ​Similarly, P(E∣FG)P(G∣F) = P(EFG) P(F) c . The given statement is true since P(EF) = P(EFG) + P EFG c and . 8. This question has been removed from this assignment by your instructor; you have received full credit. 9. This question has been removed from this assignment by your instructor; you have received full credit. 10. This question has been removed from this assignment by your instructor; you have received full credit. https://tdx.acs.pearsonprd.tech/api/v1/print/highered 5/5