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Suppose that in Problem 9.2, Al is agile enough to escape from a single car, but if he encounters two or more cars while attempting to cross the road, then he is injured. What is the probability that he will be unhurt if it takes him s seconds to cross? Do this exercise for s = 5, 10, 20, 30.

Short Answer

Expert verified

The probability of succesfully crossing the bridge in s seconds is

Ps=e-s/20+se-s/2020

Also the probability for different values of s is

P5=1.25e-0.25P10=1.5e-0.5P20=2e-1P30=2.5e-1.5

Step by step solution

01

Given Information

We have given that cars arrive on highway at the rate of๐œ†=3permin.

Al takes s seconds to cross the highway.

02

Simplifying

Let Ps denote the probability that he will return unhurt after s seconds.

โ‡’Ps=P(X<2|๐œ†=s20)โ‡’Ps=P(X=0|๐œ†=s20)+P(X=1|๐œ†=s20)โ‡’Ps=e-s/20+se-s/2020

03

Calculations

Now putting the values of s=5,10,20,30in above equation we get,

P5=1.25e-0.25P10=1.5e-0.5P20=2e-1P30=2.5e-1.5

which is the required probability for the various values of s.

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Most popular questions from this chapter

Show that for any discrete random variable Xand functionf

H(f(X))โ‰คH(X)

In transmitting a bit from location A to location B, if we let X denote the value of the bit sent at location A and Y denote the value received at location B, then H(X) โˆ’ HY(X) is called the rate of transmission of information from A to B. The maximal rate of transmission, as a function of P{X = 1} = 1 โˆ’ P{X = 0}, is called the channel capacity. Show that for a binary symmetric channel with P{Y = 1|X = 1} = P{Y = 0|X = 0} = p, the channel capacity is attained by the rate of transmission of information when P{X = 1} = 1 2 and its value is 1 + p log p + (1 โˆ’ p)log(1 โˆ’ p).

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and let Y equal the value of the first die. Compute (a) H(Y), (b) HY(X), and (c) H(X, Y).

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