Alice has an unbiased 5-sided die and 5 different coins with her. The probabilities of obtaining a headon tosses of these coins are 16 , 26 , 36 , 46 and 56 respectively. She likes to observe patterns insubsequent tosses of these coins. Alice performs the folowing experiment. She rolls the 5-sided dieand if the ith side turns up, she chooses the ith coin and starts tossing this coin repeatedly.8 510 5(a) What is the expected number of tosses required for obtaining 6 consecutive heads given that theside 1 turned up during the roll of the die?(b) What is the expected number of tosses required for obtaining 6 consecutive heads whileperforming this random experiment?(c) What is the probability that in the first n tosses, she obtains n consecutive heads?(d) In the first n-tosses, she obtains n consecutive heads. Given this outcome, calculate the probabilitythat the ith side turned up during the roll of the die (the closed form expression for arbitrary i). Howdoes these probabilities behave as n → 0?

There are an unbiased -sided die and different coins with Alice.The probability that a head can be…

Answered: Alice has an unbiased 5-sided die and 5…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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You have entered an axe-throwing competition. To enter the competition, you must pay a flat fee of $20. Each contestant is allotted eight throws. You left your glasses at home and thus are facing greater odds than would normally be expected from a talented athlete such as yourself. Hitting the target in the center wins you $5. You have a 1/8 probability of hitting the center Hitting the target in the second ring wins you $3. you have a 2/8 probability of hitting the second ring Hitting the target in the third ring wins you $1. You have a 4/8 probability of hitting the third ring Missing the target wins you $0. you have a 1/8 probability of missing the target. Each throw is considered an event, so the following equation tells us the expected outcome per throw, which can be multiplied by 8 for the total expected outcome of the game. E=-2.5(8/8)+5(1/8)+3(2/8)+1(4/8)+0(1/8) E=-2.5+(5/8)+(3/4)+(1/2)+0 E=(-5/8) E=-0.625 8E=-5 2. The expected value of the game is -$5 3. This is not…
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