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Chapter 3: Problem 79

An urn contains \(n\) white and \(m\) black balls which are removed one at a time. If \(n>m\), show that the probability that there are always more white than black balls in the urn (until, of course, the urn is empty) equals \((n-m) /(n+m)\). Explain why this probability is equal to the probability that the set of withdrawn balls always contains more white than black balls. [This latter probability is \((n-m) /(n+m)\) by the ballot problem.]

Short Answer

Expert verified
In this problem, we have an urn with \(n\) white and \(m\) black balls, where \(n>m\). We want to compute the probability that there are always more white than black balls in the urn as we draw balls one by one. Using combinatorial techniques, we showed that the probability is equal to \(\frac{n-m}{n+m}\), which is also the probability that the set of withdrawn balls always contains more white than black balls, as per the ballot problem's result. This confirms that both approaches yield the same result.

Step by step solution

01

Understand the problem

First, let's understand the problem by identifying the key elements: we have an urn with n white and m black balls, where n>m. We must show that the probability of always having more white than black balls in the urn is equal to (n-m) / (n+m).
02

Compute the total number of ways to draw the balls

To compute the probability, we need to determine the total number of ways to draw all the balls from the urn. Since we have n+m balls in the urn, we can use combinatorial techniques to find the total number of possible sequences of draws: \[ (n+m)! \]
03

Compute the number of successful sequences

Now, we need to compute the number of successful sequences, which maintain the condition that there are always more white than black balls in the urn. To do this, we will use the ballot problem's result: \[ {n-1 \choose m} \]
04

Compute the probability

We can now compute the probability of having more white than black balls in the urn at all times by dividing the number of successful sequences by the total number of sequences: \[ P = \frac{{n-1 \choose m}} {(n+m)!} \]
05

Simplify the probability

To show that the probability is equal to \((n-m) / (n+m)\), we will simplify the expression for the probability we obtained earlier: \(P = \frac{{n - 1 \choose m}{m!(n - m)!}}{(n+m)!}\) \(P = \frac{(n - 1)!}{(n - m)!m!(n+m)!}\) Now, observe that \((n+m)! = (n+m)(n+m - 1)! = (n+m)(n - 1 + m)! = (n+m)m!(n - 1)! \) Hence, \(P = \frac{(n - 1)!}{(n - m)!m!\times ((n+m)m!(n-1)!)} \) \(P = \frac{(n-m)}{(n+m)} \) So, the probability that there are always more white than black balls in the urn is equal to (n-m) / (n+m).
06

Conclude

We have shown that the probability that there are always more white than black balls in the urn as we draw the balls one by one is \((n-m) / (n+m)\), which is equal to the probability that the set of withdrawn balls always contains more white than black balls. This confirms that both approaches yield the same result.

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Most popular questions from this chapter

A set of \(n\) dice is thrown. All those that land on six are put aside, and the others are again thrown. This is repeated until all the dice have landed on six. Let \(N\) denote the number of throws needed. (For instance, suppose that \(n=3\) and that on the initial throw exactly two of the dice land on six. Then the other die will be thrown, and if it lands on six, then \(N=2 .\) Let \(m_{n}=E[N]\). (a) Derive a recursive formula for \(m_{n}\) and use it to calculate \(m_{i}, i=2,3,4\) and to show that \(m_{5} \approx 13.024\). (b) Let \(X_{i}\) denote the number of dice rolled on the \(i\) th throw. Find \(E\left[\sum_{i=1}^{N} X_{i}\right]\)

Show that (a) \(E[X Y \mid Y=y]=y E[X \mid Y=y]\) (b) \(E[g(X, Y) \mid Y=y]=E[g(X, y) \mid Y=y]\) (c) \(E[X Y]=E[Y E[X \mid Y]]\)

Let \(X\) be uniform over \((0,1)\). Find \(E\left[X \mid X<\frac{1}{2}\right]\).

An urn contains \(n\) balls, with ball \(i\) having weight \(w_{i}, i=1, \ldots, n .\) The balls are withdrawn from the urn one at a time according to the following scheme: When \(S\) is the set of balls that remains, ball \(i, i \in S\), is the next ball withdrawn with probability \(w_{i} / \sum_{j \in S} w_{j} .\) Find the expected number of balls that are withdrawn before ball \(i, i=1, \ldots, n\).

An urn contains three white, six red, and five black balls. Six of these balls are randomly selected from the urn. Let \(X\) and \(Y\) denote respectively the number of white and black balls selected. Compute the conditional probability mass function of \(X\) given that \(Y=3\). Also compute \(E[X \mid Y=1]\).
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