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Chapter 3: Problem 65

A coin that comes up heads with probability \(p\) is flipped \(n\) consecutive times. What is the probability that starting with the first flip there are always more heads than tails that have appeared?

Short Answer

Expert verified
The probability of having more heads than tails at any point in the sequence of n flips (even n only) is given by: \[P(\text{more heads than tails}) = \frac{C_{n/2} \cdot p^{n/2 + 1} \cdot (1-p)^{n/2 - 1}}{2^n}\] Where \(C_{n/2}\) is the Catalan number for \(\frac{n}{2}\) and p is the probability of getting a head. For any odd n, the probability will be 0.

Step by step solution

01

Determine all possible outcomes

The first step is to determine the total number of possible outcomes for flipping a coin n times. Since there are 2 possible outcomes (heads and tails) for each flip, there are a total of 2^n possible outcomes.
02

Valid sequences

Now, we need to determine the number of valid sequences where we have more heads than tails at any given point. A valid sequence can only start with a head. For a sequence of length n, we have the following number of valid sequences: - If n is odd, there cannot be a valid outcome since there will always be an even number of tails. - If n is even, we can calculate the valid sequences using the Catalan numbers which are given by the formula \(C_{k} = \frac{1}{k+1} \binom{2k}{k}\), where k represents the position in the sequence of the Catalan numbers. In our case, k = n/2.
03

Calculate probability

To find the probability, we can divide the number of valid sequences by the total number of outcomes. Since we're given that the probability of getting a head on any given flip is p, the probability of getting a tail is (1-p). For the number of valid sequences, we use the Catalan number with k = n/2, as derived in the previous step. So, the probability of having more heads than tails at any point in the sequence is: \[P(\text{more heads than tails}) = \frac{C_{n/2} \cdot p^{n/2 + 1} \cdot (1-p)^{n/2 - 1}}{2^n}\] Keep in mind that the formula is valid only for even n, for any odd n the probability will be 0.

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Most popular questions from this chapter

Let \(I_{1}, \ldots, I_{n}\) be independent random variables, each of which is equally likely to be either 0 or 1. A well-known nonparametric statistical test (called the signed-rank test) is concerned with determining \(P_{n}(k)\) defined by $$ P_{n}(k)=P\left\\{\sum_{j=1}^{n} j I_{j} \leq k\right\\} $$ Justify the following formula: $$ P_{n}(k)=\frac{1}{1} P_{n-1}(k)+\frac{1}{2} P_{n-1}(k-n) $$

An unbiased die is successively rolled. Let \(X\) and \(Y\) denote respectively the number of rolls necessary to obtain a six and a five. Find (a) \(E[X]\). (b) \(E[X \mid Y=1]\), (c) \(E[X \mid Y=5]\).

In the list problem, when the \(P\), is known, show that the best ordering (best in the sense of minimizing the expected position of the element requested) is to place the elements in decreasing order of their probabilities. That is, if \(P_{1}>P_{2}>\cdots>P_{n}\), show that \(1,2, \ldots, n\) is the best ordering.

Suppose \(p(x, y, z)\), the joint probably mass variables \(X, Y\), and \(Z\), is given by $$ \begin{array}{ll} p(1,1,1)=\frac{1}{8}, & p(2,1,1)=\frac{1}{4} \\ p(1,1,2)=\frac{1}{6}, & p(2,1,2)=\frac{1}{16}, \\ p(1,2,1)=\frac{1}{16}, & p(2,2,1)=0, \\ p(1,2,2)=0, & p(2,2,2)=\frac{1}{4} \end{array} $$ What is \(E(X \mid Y=2] ?\) What is \(E[X \mid Y=2, Z=1] ?\)

The number of customers entering a store on a given day is Poisson distributed with mean \(\lambda=10.\) The amount of money spent by a customer is uniformly distributed over \((0,100)\). Find the mean and variance of the amount of money that the store takes in on a given day.
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