f00soln2 - Statistics 265 Elements of Probability Theory...

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Unformatted text preview: Statistics 265 Elements of Probability Theory Fall Term 2000 Assignment 2 Due Friday September 29, 2000 Solutions 2.52 In the de nition of the independence of two events, you were given three equalities to check: P (A j B ) = P (A) or P (B j A) = P (B ) or P (A \ B ) = P (A) P (B ): If any one of these equalities holds, A and B are independent. Show that if any of these equalities hold, the other two also hold. Solution: We assume rst that P (A) > 0 and that P (B ) > 0 so that the conditional probabilities are de ned. (i.) P (A j B ) = P (A) implies P (B j A) = P (B ) Suppose that P (A j B ) = P (A) then A\ B P P (B j A) = P (P (A)B ) = P (A jP () )P (B ) = P (A)(A)(B ) = P (B ): A P (ii.) P (B j A) = P (B ) implies P (A \ B ) = P (A) P (B ) Suppose that P (B j A) = P (B ) then P (A \ B ) = P (B j A) P (A) = P (A) P (B ) (iii.) P (A \ B ) = P (A) P (B ) implies P (A j B ) = P (A) Suppose that P (A \ B ) = P (A) P (B ) then A\ P P (A j B ) = P (P (B )B ) = P (A)(B )(B ) = P (A): P 2.54 Suppose that A B and that P (A) > 0 and P (B ) > 0: Are A and B independent? Prove your answer. Solution: Since A B then P (A \ B ) = P (A) and A\ P (B j A) = P (P (A)B ) = P (A) = 1: P (A) Therefore, A and B are not independent unless P (B ) = 1: 2.57 If A and B are independent events, show that A and B are also independent. Are A and B independent? Solution: We have P (A) = P (A \ B ) + P (A \ B ) = P (A) P (B ) + P (A \ B ) so that P (A \ B ) = P (A) ; P (A) P (B ) = P (A) (1 ; P (B )) = P (A) P (B ): Therefore, A and B are independent. Also, if we use the above result with A replaced by B and B replaced by A then we see that since B and A are independent, then B and A are also independent. 2.64 Show that Theorem 2.6, the additive law of probability, holds for conditional probabilities. That is, if A B and C are events such that P (C ) > 0 prove that P (A B j C ) = P (A j C ) + P (B j C ) ; P (A \ B j C ): Hint: make use of the distributive law (A B ) \ C = (A \ C ) (B \ C ): Solution: From the distributive law we have P ((A B ) \ C ) = P (A \ C ) + P (B \ C ) ; P (A \ B \ C ) and dividing both sides of this equation by P (C ) we get P ((A B ) \ C ) = P (A \ C ) + P (B \ C ) ; P (A \ B \ C ) P (C ) P (C ) P (C ) P (C ) that is, P (A B j C ) = P (A j C ) + P (B j C ) ; P (A \ B j C ): 2.68 If A and B are two events, prove that P (A \ B ) 1 ; P (A) ; P (B ): Note: This is a simpli ed version of the Bonferroni inequality. Solution: Since P (A \ B ) 0 we have P (A B ) = P (A) + P (B ) ; P (A \ B ) P (A) + P (B ) and therefore P (A \ B ) = 1 ; P (A \ B ) = 1 ; P (A B ) 1 ; P (A) ; P (B ): 2.72 Three radar sets, operating independently, are set to detect any aircraft ying through a certain area. Each set has a probability of 0:02 of failing to detect a plane in its area. a. If an aircraft enters the area, what is the probability that it goes undetected? b. If an aircraft enters the area, what is the probability that it is detected by three radar sets? Solution: De ne F as the event that a radar set fails to detect a plane. The events of interest are intersections of independent sets a. P (aircraft undetected) = P (all three sets fail to detect) = P (F ) P (F ) P (F ) = (0:02)3: b. P (all three sets detect aircraft) = P (F ) P (F ) P (F ) = (0:98)3: 2.90 Two methods, A and B are available for teaching a certain industrial skill. The failure rate is 20% for A and 10% for B: However, B is more expensive and hence is used only 30% of the time. (A is used the other 70%): A worker was taught the skill by one of the methods but failed to learn it correctly. What is the probability that she was taught by method A? Solution: De ne the event F as \failure to learn," the event A as \method A was used," and the event B as \ method B was used," then P (F j A) = 0:20 P (F j B ) = 0:10 P (A) = 0:70 P (B ) = 0:30 therefore Baye's rule gives F ( P (A j F ) = P (F j A) PP((Aj)A) PPFA)B ) P (B ) +(j (0:20)(0:70) = (0:20)(0:70) + (0:10)(0:30) = 0::14 = 14 : 0 17 17 2 2.94 Following is a description of the game of craps. A player rolls two dice and computes the total of the spots showing. If the player's rst toss is a 7 or an 11 the player wins the game. If the rst toss is a 2 3 or 12 the player loses the game. If the player rolls anything else (4 5 6 8 9 or 10) on the rst toss, that value becomes the player's point. If the player does not win or lose on the rst toss, he tosses the dice repeatedly until he obtains either his point or a 7: He wins if he tosses his point before tossing a 7 and loses if he tosses a 7 before his point. What is the probability that the player wins a game of craps? Solution: De ne the events A : player wins B : sum of k on rst toss (or any toss) C : obtain a sum of k before obtaining a 7 k k From the law of total probability, we have P (A) = Now 12 X P (A \ B ): i i=2 P (A \ B2 ) = P (A \ B3 ) = P (A \ B12 ) = 0 and 6 2 P (A \ B7 ) = P (B7 ) = 36 and P (A \ B11 ) = P (B11) = 36 : Also, for k = 4 5 6 8 9 10 we have P (A \ B ) = P (B \ C ) = P (B ) P (C ) k k k k k since the tosses are independent. Now we have to calculate the P (C )'s. We do this as follows. Let F be the event that we do not throw a k or a 7 for k = 4 5 6 8 9 10: Since the events B and F are independent, for these values of k the event C of obtaining a k before obtaining a 7 has probability k k k k k P (C ) = P (B ) + P (F ) P (B ) + P (F ) P (F ) P (B ) + P (F ) P (F ) P (F ) P (B ) + + P (F ) P (F ) P (B ) + k k k k k k k k = P (B ) 1 + P (F ) + P (F )2 + P (F )3 + PB) = 1 ; (P (F ) k k k k k k + P (F ) + k k k that is, PB) P (C ) = 1 ; (P (F ) k k for k = 4 5 6 8 9 10: k 3 k n k k k Now, it is easy to see that 3 P (B4) = P (B10) = 36 and P (C4) = P (C10) = 1 3 4 and P (C ) = P (C ) = 2 P (B5) = P (B9 ) = 36 5 9 5 5 and P (C ) = P (C ) = 5 P (B6) = P (B8 ) = 36 6 8 11 so that P (A \ B4 ) = P (C4 \ B4 ) = P (C4) P (B4 ) = 1 3 P (A \ B5 ) = P (C5 \ B5 ) = P (C5) P (B5 ) = 2 5 5 P (A \ B6 ) = P (C6 \ B6 ) = P (C6) P (B6 ) = 11 31 36 = 36 42 36 = 45 5 25 36 = 396 and 25 P (A \ B8 ) = P (A \ B6 ) = 396 2 P (A \ B9 ) = P (A \ B5 ) = 45 1 P (A \ B10 ) = P (A \ B4 ) = 36 Therefore, P (A) = 12 X P (A \ B ) k k =2 1 2 25 6 25 2 1 2 = 0 + 0 + 36 + 45 + 396 + 36 + 396 + 45 + 36 + 36 + 0 = 0:4929 not a safe bet. 4 3.4 Consider a system of water owing through valves from A to B: (See the accompanying diagram.) Valves 1 2 and 3 operate independently, and each correctly opens on signal with probability 0:80: Find the probability distribution for Y the number of open paths from A to B after the signal is given. (Note that Y can take on the values 0 1 and 2:) 1 A B 2 Solution: 3 De ne the following events: A : valve 1 fails B : valve 2 fails C : valve 3 fails Now, p(0) = P (Y = 0) = P (A \ (B C )) since valve 1 must fail and valves 2 or 3 (or both) must also fail. Now, since the valves operate independently, we have p(0) = 0:2 0:2 + 0:2 ; (0:2)2 = (0:20)(0:36) = 0:072: Also, ; ; p(1) = P (Y = 1) = P A \ (B \ C ) + P A \ (B C ) since either valve 1 fails and valves 2 and 3 do not fail, or valve 2 or valve 3 (or both) fail and valve 1 does not fail. Again, since the valves operate independently, we have p(1) = (0:2)(0:8)2 + (0:8) 0:2 + 0:2 ; (0:2)2 = (0:2)(0:64) + (0:8)(0:36) = 0:416: Finally, p(2) = P (Y = 2) = P (A \ B \ C ) = P (A) P (B ) P (C ) since none of the valves must fail. Therefore, p(2) = (0:8)3 = 0:512: 3.6 Five balls, numbered 1 2 3 4 and 5 are placed in an urn. Two balls are randomly selected from the ve, and their numbers noted. Find the probability distribution of the following: a. the largest of the two sampled numbers b. the sum of the two sampled numbers ;5 Solution: There are 2 = 10 sample points, all equally likely: (1 2) (1 3) (1 4) (1 5) (2 3) (2 4) a. Let Y = largest of the two sample numbers, then 1 2 p(2) = 10 p(3) = 10 3 p(4) = 10 b. Let Y = sum of the two sample numbers then 1 1 2 2 p(3) = 10 p(4) = 10 p(5) = 10 p(6) = 10 5 (2 5) (3 4) (3 5) (4 5) 4 p(5) = 10 2 p(7) = 10 1 p(8) = 10 1 p(9) = 10 3.21 Let Y be a discrete random variable with mean and variance 2 : If a and b are constants, use Theorems 3.3 through 3.6 to prove that a. E (aY + b) = aE (Y ) + b = a + b b. V (aY + b) = a2V (Y ) = a2 2 : Solution: a. Let g1(Y ) = aY and g2 (Y ) = b then from Theorem 3.5 we have E (aY + b) = E (g1 (Y ) + g2 (Y )) = E (g1 (Y )) + E (g2 (Y )) = E (aY ) + E (b) and from Theorems 3.4 and 3.3 we get E (aY + b) = aE (Y ) + E (b) = a + b: b. From De nition 3.5, we have ; V (aY + b) = E aY + b ; (a + b)]2 ; = E aY ; a + b ; b]2 ; = E aY ; a ]2 : ; From Theorem 3.4, this equals a2 E Y ; ]2 which equals a2 V (Y ) = a2 2 by De nition 3.5. 3.22 The manager of a stockroom in a factory has constructed the following probability distribution for the daily demand (number of times used) for a particular tool. y p(y) 0 0:10 1 0:50 2 0:40 It costs the factory $10 each time the tool is used. Find the mean and variance of the daily cost for use of the tool. Solution: The mean cost is E (10 Y ) = 10 E (Y ) and E (Y ) = 0 0:10 + 1 0:50 + 2 0:40 = 1:30 and the mean cost is $13.00. The variance of the cost is V (10 Y ) = 100 V (Y ): We use the shortcut formula V (Y ) = E (Y 2) ; nd the variance. We have E (Y 2 ) = 02 0:10 + 12 0:50 + 22 0:40 = 2:10 and V (Y ) = E (Y 2 ) ; so the variance of the cost is 110 0:41 = 41: 2 = 2:10 ; (1:30)2 = 0:41 6 2 to ...
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