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Unformatted text preview: EE 131A Problem Set #3: Solutions Probabilities Instructor: Vwani Roychowdhury 1. The amount of time cars are parked in a parking lot follows an exponential probability law with average time of 1 hour. The charge for the parking is $1 for each half-hour or less. Find the probability that a car pays k dollars. In order to pay k dollars for parking, the cars parking time in hours (denoted as PT) should be k- 1 2 hours < PT k 2 hours. The parking time follows an exponential probability law with average time of 1 hour, which means that P [ PT t ] = e- t with the units of t being hours. Using the result of example 2.10 of the textbook (about the probability of a lifetime that follows an exponential law to be in a specific interval) we have: P [ k- 1 2 hours < PT k 2 hours ] = P [ PT > k- 1 2 ]- P [ PT > k 2 ] P [ k- 1 2 hours < PT k 2 hours ] = e- k- 1 2- e- k 2 = e- k 2 ( e . 5- 1) P [ k- 1 2 hours < PT k 2 hours ] . 6487 e- k 2 . 1 2. Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second card are colored black, and one side of the third card is colored red and the other side black. The 3 cards are mixed up in a hat, and 1 card is randomly selected and put down on the ground. If the upper side of the chosen card is colored red, what is the probability that the other side is colored black? Let RR , BB , and RB denote, respectively, the events that the chosen card is the all red, all black, or the red-black card. Letting R be the event that the upturned side of the chosen card is red, we have that the desired probability is obtained by P [ RB | R ] = P [ RB R ] P [ R ] = P [ R | RB ] P [ RB ] P [ R | RR ] P [ RR ] + P [ R | RB ] P [ RB ] + P [ R | BB ] P [ BB ] = 1 2 1 3 1 1 3 + 1 2 1 3 + 0 1 3 = 1 3 Note that, given that we have a red upper side, the probability that we have red or black on the other side is 2/3 and 1/3 respectively and not 1/2 for each event, as some students might expect....
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- Spring '05