i=1 A i i=1 P (A c B c ) = 1 P ((A c B c ) c ) = 1 P (A B) = 1 (P (A) + P (B) P (A B)) =

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1 Homework 1 Solutions ( n P i1 A i ) [ n 1 P i1 A i ( n ) 1 P A i 1 i1 n P (A i ) i1 ] by countable subadditivity property Let A be the event that we select a male factory worker, and let B be the event that we select a married worker. The given information is then P (A) 0.65, P (B) 0.70, P (A B) 0.5. (c) P (A c B) P (B) P (A B) P (A c B c ) 1 P ((A c B c ) c ) 1 P (A B) 1 (P (A) + P (B) P (A B)) 1 ( ) P (A B) P (A) + P (B) P (A B) These answers can also be derived from the box method. The given information can be displayed as Event A A c Total B B c Total 0.65 Then, we can complete the table of probabilities using the addition rule Event A A c Total B B c Total The basic approach is to consider permutations of hats. 1

2 Person Hat Therefore, P (No match) Therefore, P (B) 0.6, P (L) 0.3, P (T ) 0.1, P (D B) 0., P (D L) 0.3, P (D T ) 0.9. P (D B)P (B) + P (D L)P (L) + P (D T )P (T ) P (B D) P (B D) (c) Let D i be the event that i of them are detained. Then, P (D B)P (B) P (D D 5 ) P (D ) + P (D 5 ) ( ) 5 [P (D T )] [1 P (D T )] + [P (D T )] 5 5(0.9) (0.1) + (0.9) Let K be the event that he knows the answer, and R be the event that he gets the answer right. Then, P (K) 0.7, P (K ) 0.3, P (R K) 1, P (R K ) 0.. P (R) P (R K)P (K) + P (R K )P (K ) P (K R) P (K R) P (R) P (R K)P (K) P (R) (c) Let R be the event that he gets exactly two of the first three questions right. Then, ( ) 3 P (R ) [P (R)] [1 P (R)] ( ) 3 (0.76) (0.)

3 1.8.8 Let D be an event such that the widget is defective. Then, P (D A) 0.0, P (D B) 0.03, P (D C) 0.05, P (A) 0.5, P (B) 0.3, P (C) 0.. P (D A)P (A) + P (D B)P (B) + P (D C)P (C) P (A D) P (B D) P (C D) P (A D) P (D A)P (A) P (B D) P (D B)P (B) P (C D) P (D C)P (C) Michael can draw one good sock and one bad sock in two different ways: the good sock can come first, or it can come second. Respectively, the probabilities of these sequences of events are and Therefore, the probability that he draws one good sock and one bad sock in his first two draws is The probability that the first two socks are of the same type is , using the complement rule along with the answer from part. The probability that the first two socks are good is P (both socks good both socks of same type) P (both socks good both socks of same type) P (both socks of same type) P (both socks good) P (both socks of same type) 6/55 7/55 9. (c) If Michael is to draw the second good sock on the fifth draw, then in his first four draws, he must have drawn exactly one good sock and exactly three bad socks. The probability of this combination of four socks being drawn is Then, Michael needs to draw a second good sock from the seven socks remaining in the drawer. This happens with probability 3 7. Therefore, the probability that Michael draws the second good sock on the fifth draw is Let T D be the event that Tom beats Dick, T H be the event that Tom beats Harry, and DH be the event that Dick beats Harry. Then, we have the following two possible outcomes (E 1 and E ) so that each player wins (W) one and loses (L) one. E 1 Tom Dick Harry Tom - W L Dick L - W Harry W L - E Tom Dick Harry Tom - L W Dick W - L Harry L W - 3

4 Thus, P ({Each player wins one and loses one}) P (E 1 ) + P (E ) P (T D)P (T H )P (DH) + P (T D )P (T H)P (DH ) ( ) ( ) ( ) ( ) ( ) ( ) Let C be chew tobacco, and B be bad breath. Then, we have the following probability table. From the table, it follows that or so EVENT C C TOTAL B B TOTAL P (C B ) 0.05, P (C) P (C B) + P (C B ), P (C B ) P (C B) P (C B) P (B) The magician can either draw a bunny first and then a squirrel, or a squirrel first and then a bunny. Respectively, these sequences of events occur with probabilities 3 5 and 5 3. Thus, the probability that the magician draws one of each animal to begin his act is The probability that the first two animals are the same is , using the complement rule along with the answer from part. The probability that the first two draws are bunnies is P (both are bunnies both animals are the same) P (both are bunnies both animals are the same) P (both animals are the same) P (both are bunnies) P (both animals are the same) 3/10 /5 3. (c) If the magician is to draw the second bunny on the fourth draw, then in his first three draws, he must have drawn exactly one bunny and exactly two squirrels. The probability of this combination of three animals being drawn is Then, the magician needs to draw a second bunny from the two animals remaining in the hat. This happens with probability. Therefore, the probability that the magician draws the second bunny on the fourth draw is

5 1.8.7 Consider a tree where stage 1 represents placing the prize randomly behind doors 1,, and 3, and stage represents the contestant choosing a door at random. Each path of this tree has probability 1/9 of occurring. If we take the switching strategy, then any path where the stage 1 and stage door numbers do not match will result in winning the prize. That is, if we initially guess incorrectly, the other incorrect door is eliminated by Monty, leaving only the correct door for us to switch to. Six paths in the tree do not have matching door numbers, so the probability of winning with the switching strategy is Consider a tree where stage 1 represents the stop chosen by the first bum, stage represents the stop chosen by the second bum, and stage 3 represents the stop chosen by the third bum. There are 7 equally likely paths in this tree. If the bums must exit at different stops, the first bum has three unique stops to choose from, but the second bum may only choose from two unique stops, and the third bum must exit at the last unclaimed stop. Thus, there are only paths in the tree that result in the bums exiting at three different stops, and the probability that this occurs is