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Chapter 1: Q9E (page 50)

Suppose that n letters are placed at random in n envelopes, and let \({q_n}\) denote the probability that no letter is placed in the correct envelope. For which of the following four values of n is \({q_n}\)largest: n = 10, n = 21, n = 53, or n = 300?

Short Answer

Expert verified

Among the following four values n,\({q_n}\) is the largest for 10 envelopes or letters.

Step by step solution

01

Given information

n letters are placed at random in n envelopes.

\({q_n}\)is the probability that no letter is placed in the correct envelopes.

02

Express the required probability

Consider the event that no letter is placed in the correct envelope as E.

Thus, the probability that at least one letter is placed in the correct envelope is represented as,

\(\begin{aligned}{}{\bf{P}}\left( {\bf{E}} \right){\bf{ = 1 - P}}\left( {{{\bf{E}}^{\bf{c}}}} \right)\\{{\bf{q}}_{\bf{n}}}{\bf{ = 1 - P}}\left( {{{\bf{E}}^{\bf{c}}}} \right)\\{{\bf{q}}_{\bf{n}}}{\bf{ = 1 - }}\left( {{\bf{1 - }}\frac{{\bf{1}}}{{{\bf{2!}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{3!}}}}{\bf{ - }}\frac{{\bf{1}}}{{{\bf{4!}}}}{\bf{ + }}...{\bf{ + }}{{\left( {{\bf{ - 1}}} \right)}^{{\bf{n + 1}}}}\frac{{\bf{1}}}{{{\bf{n!}}}}} \right)\\{{\bf{q}}_{\bf{n}}}{\bf{ = 1 - }}\left( {\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\left( {{\bf{ - 1}}} \right)}^{{\bf{i + 1}}}}\frac{{\bf{1}}}{{{\bf{i!}}}}} } \right)\end{aligned}\)

03

Compute the probability for different values of n

\(\sum\limits_{{\bf{i = 1}}}^{10} {{{\left( {{\bf{ - 1}}} \right)}^{{\bf{i + 1}}}}\frac{{\bf{1}}}{{{\bf{i!}}}}} \)is an oscillating sequence as n gets larger.

It implies that:

  • For even values of n it increases up to the limiting value of 0.63212 (at n=7).
  • For odd values of n it decreases up to the limiting value of 0.63212 (at n=7).

For the expression,

\({q_n} = {\bf{1 - }}\left( {\sum\limits_{{\bf{i = 1}}}^{10} {{{\left( {{\bf{ - 1}}} \right)}^{{\bf{i + 1}}}}\frac{{\bf{1}}}{{{\bf{i!}}}}} } \right)\)\(\)

The largest value of the series would give the lowest value of\({q_n}\)and vice-versa.

Thus, the largest value of the probability is expected to be attained at smallest value of n which is 10.

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Most popular questions from this chapter

For the conditions of Exercise 4, what is the probability that neither student A nor student B will fail the examination?

Question:Let 胃 denote the proportion of registered voters in a large city who are in favor of a certain proposition. Suppose that the value of 胃 is unknown, and two statisticians A and B assign to 胃 the following different prior p.d.f.鈥檚\({{\bf{\xi }}_{\bf{A}}}\left( {\bf{\theta }} \right)\)and \({{\bf{\xi }}_{\bf{B}}}\left( {\bf{\theta }} \right)\) respectively:

\({{\bf{\xi }}_{\bf{A}}}\left( {\bf{\theta }} \right){\bf{ = 2\theta ,}}\,\,\,{\bf{0 < \theta < 1}}\)

\({{\bf{\xi }}_{\bf{B}}}\left( {\bf{\theta }} \right){\bf{ = 4}}{{\bf{\theta }}^{\bf{3}}}\,\,{\bf{,0 < \theta < 1}}\)

In a random sample of 1000 registered voters from the city, it is found that 710 are in favor of the proposition.

a. Find the posterior distribution that each statistician assigns to 胃.

b. Find the Bayes estimate for each statistician based on the squared error loss function.

c.Show that after the opinions of the 1000 registered voters in the random sample had been obtained, the Bayes estimates for the two statisticians could not possibly differ by more than 0.002, regardless of the number in the sample who were in favor of the proposition

Consider a state lottery game in which each winning combination and each ticket consists of one set of k numbers chosen from the numbers 1 to n without replacement. We shall compute the probability that the winning combination contains at least one pair of consecutive numbers.

a. Prove that if\({\bf{n < 2k - 1}}\), then every winning combination has at least one pair of consecutive numbers. For the rest of the problem, assume that\({\bf{n}} \le {\bf{2k - 1}}\).

b. Let\({{\bf{i}}_{\bf{1}}}{\bf{ < }}...{\bf{ < }}{{\bf{i}}_{\bf{k}}}\)be an arbitrary possible winning combination arranged in order from smallest to largest. For\({\bf{s = 1,}}...{\bf{,k}}\), let\({{\bf{j}}_{\bf{s}}}{\bf{ = }}{{\bf{i}}_{\bf{s}}}{\bf{ - }}\left( {{\bf{s - 1}}} \right)\). That is,

\(\begin{array}{c}{{\bf{j}}_{\bf{1}}}{\bf{ = }}{{\bf{i}}_{\bf{1}}}\\{{\bf{j}}_{\bf{2}}}{\bf{ = }}{{\bf{i}}_{\bf{2}}}{\bf{ - 1}}\\{\bf{.}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{j}}_{\bf{k}}}{\bf{ = }}{{\bf{i}}_{\bf{k}}}{\bf{ - }}\left( {{\bf{k - 1}}} \right)\end{array}\)

Prove that\(\left( {{{\bf{i}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{i}}_{\bf{k}}}} \right)\)contains at least one pair of consecutive numbers if and only if\(\left( {{{\bf{j}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{j}}_{\bf{k}}}} \right)\)contains repeated numbers.

c. Prove that\({\bf{1}} \le {{\bf{j}}_{\bf{1}}} \le ... \le {{\bf{j}}_{\bf{k}}} \le {\bf{n - k + 1}}\)and that the number of\(\left( {{{\bf{j}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{j}}_{\bf{k}}}} \right)\)sets with no repeats is\(\left( {\begin{array}{*{20}{c}}{{\bf{n - k + 1}}}\\{\bf{k}}\end{array}} \right)\)

d. Find the probability that there is no pair of consecutive numbers in the winning combination.

e. Find the probability of at least one pair of consecutive numbers in the winning combination

Prove that for every two events A and B, the probability that exactly one of the two events will occur is given by the expression

\(\Pr \left( A \right) + \Pr \left( A \right) - 2\Pr \left( {A \cap B} \right)\)

The solution to Exercise 1 of Sec. 3.9 is the pdf. of \({{\bf{X}}_{\bf{1}}}{\bf{ + }}{{\bf{X}}_{\bf{2}}}\)in Example 6.1.2. Find the pdf. of\({{\bf{\bar X}}_{\bf{2}}}{\bf{ = }}\frac{{{{\bf{X}}_{\bf{1}}}{\bf{ + }}{{\bf{X}}_{\bf{2}}}}}{{\bf{2}}}\). Compare the probabilities that \({X_2}\) and \({X_1}\)are close to 0.5. In particular, compute \({\bf{P}}\left( {\left| {{{\bf{X}}_{\bf{2}}}{\bf{ - 0}}{\bf{.5}}} \right|{\bf{ < 0}}{\bf{.1}}} \right)\)and\({\bf{P}}\left( {\left| {{{\bf{X}}_{\bf{1}}}{\bf{ - 0}}{\bf{.5}}} \right|{\bf{ < 0}}{\bf{.1}}} \right)\). What feature of the pdf \({X_2}\)makes it clear that the distribution is more concentrated near the mean?
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