EE 422: Digital Communications Home work # 1 1) A player has a loaded die which

Question

EE 422: Digital Communications Home work # 1 1) A player has a loaded die which turns up the number 1I with probability 2/3 and the number 2 to 6 with probability 1/15 each. Unfortunately, he left his loaded die in a box with two honest dice and couldn't tell them apart. He picked one die (at random) from it once, and the number I appeared. Conditional on t l he picked up the honest die? The player rolled the die once more, and it came up l What this resu It, what is the probability that again is the probability after this second rolling that he has picked up the honest die? 2) You are one of 3 people. An award is to be given to the person who draws the shortest straw from a set of 3 straws placed in a box. What place in line should you take to maximize your chance of winning? ) 100 integrated circuits of type A, 200 of type B, and 300 of type C are mixed together ype A chips are known to be defective with probability 1/10, type B with probability of 1/20, and type C with probability of 1/30. a) If one chip is selected randomly, what is the probability of its being defective? Given that the selected chip in part (a) is defective, what is the probability that it is from type A? b) A student takes a train to get to a school's campus and then to class. The probability the student will arrive at class on time is 0.95 provided the train is on time. If the train is known to be on schedule 70% of the time, what is the probability the student will be on time to class? 4) 5) A random experiment consists of drawing a ball from an um that contains 4 red balls numbered 1, 2, 3, 4 and three black balls numbered 1, 2, 3. The following events are defined Ei- The number on the ball is even. E2 The color of the ball is red, and its number is greater than 1 = The number on the ball is less than 3. a) Determine P(E2), P(Es Ez), and PIE EsEs)? b) Are Es and Es independent?

Explanation / Answer

Ans:

3)P(A)=100/600=1/6

P(B)=200/600=2/6

P(C)=300/600=3/6

P(defective/A)=1/10=0.1

P(defective/B)=1/20=0.05

P(defective/C)=1/30

a)P(defective)=(1/6)*0.1+(2/6)*0.05+(3/6)*(1/30)=0.0499

b)P(A/defctive)=(1/6)*0.1/0.0499=0.0167/0.0499=0.3347

4)P(arrive on time/train on time)=0.95

P(train on time)=0.7

P(arrive on time and train arrive on time)=0.95*0.7=0.665

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