Question

Consider an urn that contains slips of paper each with one of the numbers \(1,2, \ldots, 100\) on it. Suppose there are \(i\) slips with the number \(i\) on it for \(i=1,2, \ldots, 100\). For example, there are 25 slips of paper with the number \(25 .\) Assume that the slips are identical except for the numbers. Suppose one slip is drawn at random. Let \(X\) be the number on the slip. (a) Show that \(X\) has the \(\operatorname{pmf} p(x)=x / 5050, x=1,2,3, \ldots, 100\), zero elsewhere. (b) Compute \(P(X \leq 50)\). (c) Show that the cdf of \(X\) is \(F(x)=[x]([x]+1) / 10100\), for \(1 \leq x \leq 100\), where \([x]\) is the greatest integer in \(x\).

Short answer

(a) The pmf of \(X\) is \(p(x) = x/5050\), for \(x = 1,2,3,...,100\). (b) \(P(X \leq 50) = 0.255\). (c) The cdf of \(X\) for \(1 \leq x \leq 100\) is \(F(x)=[x]([x]+1)/10100\).

Step-by-step solution

Prove the definition of the pmf

For this urn, the total number of slips is the sum of the first 100 natural numbers, given by \((100)(100+1)/2 = 5050\). Since there are \(x\) slips with the number \(x\) on it, the probability that a slip picked at random has the number \(x\) on it is given by \(p(x) = x/5050\), for \(x = 1,2,3,...,100\).

Compute the Probability of \(X \leq 50\)

To compute the probability of \(X \leq 50\), one must sum up the probabilities \(p(x)\) for all \(x\) between 1 and 50. So, \(P(X \leq 50) = \sum_{x = 1}^{50} p(x) = \sum_{x = 1}^{50} x/5050 = (50)(50+1)/(2*5050) = 0.255\).

Show the cdf form of \(X\)

The cumulative distribution function \(F(x)\) of \(X\) is defined as \(P(X \leq x) = F(x)\). For \(1 \leq x \leq 100\), summing up the probabilities for all \(x\) gives \(F(x) = \sum_{k = 1}^{[x]} p(k) = \sum_{k = 1}^{[x]} k/5050 = ([x]([x]+1))/(2*5050) = [x][x+1]/10100\).