A mosquito is walking at random on the nonnegative number line. She starts at $1$. When she is at $0$, she always takes a step $1$ unit to the right, but, from any positive position on the line, she randomly moves left or right $1$ unit with equal probability. What is the expected number of times the mosquito will visit $0$ before the first time she visits $4$?

Answers

Answer 1

The expected number of times the mosquito will visit 0 before the first time she visits 4 is 3/4 by using concept of probability and geometric series.

Given that repeatedly throwing a fair tetrahedral die, trying to get a 4 (success) before getting a number from 1 to 3 (failure), which follows a geometric distribution.

To find the expected number of visits to 0 before reaching 4, analyze two cases:

success (reaching 4 before 0) and failure (reaching 0 before 4).

By the given data, the probabilities of two cases are:

P(success) p = 1/4,

P(failure)=3/4.

The expected count of failures, X, can be found in terms of success probability p as

[tex]E[X]=\sum_{x=0}^{+\infty}[x(1-p)^xp][/tex]

On substituting p = 1/4 gives:

[tex]E(X) = 0(1/4)+1(3/4)(1/4)+2(3/4)^2(1/4)+3(3/4)^3(1/4)+.........[/tex]

On taking (1/4) gives

[tex]E(X) = 1/4[3/4+(3/4)^2+(3/4)^3+......+\infty]----(1)[/tex]

Consider [tex]3/4+(3/4)^2+(3/4)^3+......+\infty[/tex]

This forms a geometric series tends to infinity and the sum is given by :

[tex]S_\infty = a/(1-r)[/tex]

Here: a= 3/4

r = 3/4

Plugging these values into formula gives:

[tex]3/4+(3/4)^2+(3/4)^3+......+\infty = S_\infty[/tex]

[tex]S_\infty= 3/4/(1-3/4)[/tex]

On subtracting the denominator gives:

= 3/4/(1/4)

On taking reciprocal gives:

= 3/4 * 4

On simplifying gives:

= 3

Substituting these value into given equation 1 gives:

E(X) = 1/4*[3]

On multiplying gives:

E(X) = 3/4.

Therefore, the  expected number of times the mosquito will visit 0 before the first time she visits 4 is 3/4 by using concept of probability and geometric series.

Learn more probability about here:

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