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Suppose that the probability of a head on any toss of a certain coin isp(0<p <1), and suppose that the coin is tossed repeatedly. LetXndenote the total number ofheads that have been obtained on the firstntosses, and letYn=nXndenote the total number of tails on the firstntosses. Suppose that the tosses are stopped as soon as a numbernis reached such that eitherXn=Yn+3 orYn=Xn+3. Determine the probability thatXn=Yn+3when the tosses are stopped.

Short Answer

Expert verified

If tosses are stopped the probability that\({X_n} = {Y_n} + 3\)is\(\frac{1}{2}\)

Step by step solution

01

Given information

Probability of a certain coin being head in a toss is\(p;\left( {0 < p < 1} \right)\).

Total no. of head obtained on the first n tosses is \({X_n}\).

02

State the events

Let, A win one dollar if head obtained in a toss otherwise B win one dollar if tail occurs.

Assuming that the initial fortune of two persons A and B are 3 dollars.

Then, if\({X_n} = {Y_n} + 3\)then A won the game and if\({Y_n} = {X_n} + 3\)then B won the game.

So, here \(i = 3\)and \(k = 6\)

03

Compute the probability

Referring to equation 2.4.6 for the following equation.

If\(p = \frac{1}{2}\), then the probability is given by,

\(\begin{aligned}{c}a = \frac{i}{k}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{aligned}\)

If\(p \ne \frac{1}{2}\)then the probability is given by,

\(\frac{1}{{{{\left( {\frac{{1 - p}}{p}} \right)}^3} + 1}}\)

Hence, if tosses are stopped the probability that \({X_n} = {Y_n} + 3\) is \(\frac{1}{2}\).

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Most popular questions from this chapter

Suppose that in Example 2.3.5, the observed individual has the recessive trait. Determine the posterior probability that the parents have the genotypes of the event \({{\bf{B}}_{\bf{4}}}\).

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