Question

A sensor controlled car-robot completes a path maze a couple of times, learning after each try.

It's success probability is:

P(x)=1- e^-x/2

For the x'th try (x = 1,2,3...) the cars actions are not dependent on success of previous tries.
A random var R is created - as the outcome of attempted number of tries.
If a successful event is called '1' ( meaning at least 1) and failed try is '0' (no success)
i). What is the expectation value for R of the first 2 consecutive attempts?
ii). What is the probability of success in first 2 consecutive attempts?

Answer 1

let X₁ be a random variable where

X₁=1 the first attempt results in a success

0 the attempt results in a failure

then

X₁=1 P(X)=1-e^(-x/2)

0 1-P(x)=e^(-x/2)

similarly

X₂ be the random variable corresponding to the second attempt

X2=1 P(X)=1-e^(-x/2)

0 1-P(x)=e^(-x/2)

Then R is a random variable for a attempted no. of tries

then

I) expected value of R in first two consecutive attempt

meaning R=X₁+X₂

E(R)=E(X₁)+E(X₂)

now if we carefully look at the distribution of X₁ and X₂ we will see that both of them follow bernoulli distribution with probability of success 1-e^(-x/2)

expectation of a bernoulli(p) distribution is p

then

E(R)=1 - e^(-x/2) + 1 - e^(-x/2)

=2 -2 e^(-x/2)

2) now since X₁ and X₂ follows bernoulli(p) then R follows a binomial distribution with parameters n=2 and p=1-e^(x/2)

Then we need to find

  

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