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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

A coin, having probability \(p\) of landing heads, is continually flipped until at least one head and one tail has been flipped. (a) Find the expected number of flips needed. (b) Find the expected number of flips that lands on heads. (c) Find the expected number of flips that land on tails. (d) Repeat part (a) in the case where flipping is continued until there has been a total of at least two heads and one tail.

(a) From the results of Section \(3.6 .3\) we can conclude that there are \(\left(\begin{array}{c}n+m-1 \\ m-1\end{array}\right)\) nonnegative integer valued solutions of the equation \(x_{1}+\cdots+x_{n}=n .\) Prove this directly. (b) How many positive integer valued solutions of \(x_{1}+\cdots+x_{m}=n\) are there? Hint: Let \(y_{i}=x_{i}-1\). (c) For the Bose-Einstein distribution, compute the probability that exactly \(k\) of the \(X_{i}\) are equal to \(0 .\)

Let \(X_{1}\) and \(X_{2}\) be independent geometric random variabies having same parameter \(p\). Guess the value of $$ P\left[X_{1}=i \mid X_{1}+X_{2}=n\right] $$ Hint: Suppose a coin having probability \(p\) of coming up heads is continually flipped. If the second head occurs on flip number \(n\), what is the conditional probability that the first head was on flip number \(i\). \(i=1, \ldots, n-1 ?\) Verify your auess analytically.

There are three coins in a barrel. These coins, when flipped, will come up heads with respective probabilities of \(0.3,0.5,0.7.\) A coin is randomly selected from among these three and is then flipped ten times. Let \(N\) be the number of heads obtained on the ten Rips. Find (a) \(P[N=0\\}\). (b) \(P[N=n], n=0,1, \ldots, 10\) (c) Does \(N\) have a binomial distribution? (d) If you win \$1 cach time a head appears and you lose \$1 each time a tail appears, is this a fair game? Explain.

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) \(E[X \mid Y=5]\).
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