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Chapter 7: Problem 10

What is the probability of these events when we randomly select a permutation of the 26 lowercase letters of the English alphabet? a) The first 13 letters of the permutation are in alphabetical order. b) \(a\) is the first letter of the permutation and \(z\) is the last letter. c) \(a\) and \(z\) are next to each other in the permutation. d) \(a\) and \(b\) are not next to each other in the permutation. e) \(a\) and \(z\) are separated by at least 23 letters in the permutation. f) \(z\) precedes both \(a\) and \(b\) in the permutation.

Short Answer

Expert verified
a) \( \frac{1}{13!} \) b) \( \frac{1}{650} \) c) \( \frac{1}{13} \) d) \( \frac{12}{13} \) e) \( \frac{1}{13} \) f) \( \frac{1}{3} \)

Step by step solution

01

- Total number of permutations

Find the total number of permutations of the 26 lowercase letters. Since there are 26 letters, the total number of permutations is given by \[ 26! \]
02

- Probability the first 13 letters are in alphabetical order

a) For the first 13 letters to be in alphabetical order, there is only 1 way to place these 13 letters among the possible permutations of the 13 letters in this position. Thus,\[ P = \frac{1}{13!} \]
03

- Probability that 'a' is the first and 'z' is the last letter

b) If 'a' must be the first letter and 'z' must be the last, the remaining 24 letters can be in any order.\[ P = \frac{24!}{26!} = \frac{1}{26 \times 25} = \frac{1}{650} \]
04

- Probability that 'a' and 'z' are next to each other

c) We can treat 'a' and 'z' as one single unit, which means we are permuting a set of 25 items (the 'a-z' block and 24 other letters). This block can either be 'az' or 'za'. Thus,\[ P = \frac{2 \times 25!}{26!} = \frac{2}{26} = \frac{1}{13} \]
05

- Probability that 'a' and 'b' are not next to each other

d) First, calculate the probability that 'a' and 'b' are next to each other, which is the same problem as in Step 4. Then, subtract this from 1:\[ 1 - \frac{1}{13} = \frac{12}{13} \]
06

- Probability that 'a' and 'z' are separated by at least 23 letters

e) If 'a' is at position i, 'z' must be at least position i+24 or i-24.- There are 3 valid ranges where 'a' can lie to comply; positions 1,2, and 3 expressly.So the answer is: \[ P = \frac{2}{26} = \frac{1}{13} \]
07

- Probability that 'z' precedes both 'a' and 'b'

f) Of the 3 positions 'a' and 'b' can lie on, 26! Of them manifest in 'z' logically occurring post being positioned, so:\[ P = \frac{1}{3} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
Probability is a way of expressing the likelihood of an event occurring. In this exercise, we calculate the probabilities for different arrangements of the 26 lowercase letters.
We use the total number of permutations of these letters as a starting point. In probability calculations, a crucial concept is the **total number of possibilities**. For 26 unique letters, this is represented by \(26!\). This notation means 26 multiplied by each consecutive integer down to 1: \[ 26! = 26 \times 25 \times 24 \times ... \times 1 \].
By understanding the total permutations, we can determine the probability of any specific arrangement or event.
Permutations
Permutations refer to the different ways in which a set of objects can be arranged. For our set of 26 letters, every unique sequence is a different permutation.
When calculating permutations, we often use factorial notation \(n!\).
For example, with 3 letters \('a', 'b', 'c'\), we have 3! permutations, which is \[ 3! = 3 \times 2 \times 1 = 6 \]. Permutations are important in probability calculations because they help determine the total number of possible arrangements.
Once we know the total permutations, we can then find the specific arrangements we are interested in, like having certain letters in specific positions.
Alphabetical Order
Alphabetical order means arranging letters as they appear in the alphabet. For example, the first 13 letters are 'a' to 'm'.
To find the probability of the first 13 letters being in alphabetical order, we need to consider that there is only **one** way they can be correctly ordered among themselves.
We use the following probability formula:
\[ P = \frac{1}{13!} \]
This formula tells us there is only one successful outcome where the first 13 letters are in alphabetical order, out of 13! possible combinations. This calculation shows the very low likelihood of this specific order happening randomly.
Event Probability
Event probability refers to the likelihood of a particular event occurring within the set of all possible outcomes. Let's analyze a few specific events:
To find the probability that 'a' is the first letter and 'z' is the last, we leave 24 letters to be arranged freely. Therefore, the probability is:
\[ P = \frac{24!}{26!} = \frac{1}{26 \times 25} = \frac{1}{650} \]
For 'a' and 'z' next to each other, treat them as one block (az or za), which gives 25 items to arrange. This probability is:
\[ P = \frac{2 \times 25!}{26!} = \frac{2}{26} = \frac{1}{13} \]
Each event's probability hinges on how we can manipulate the set of letters and the constraints provided.
Mathematical Problem Solving
Solving mathematical problems often involves breaking them down into smaller steps. For permutations in probability, follow these strategies:
  • **Identify the total number of permutations.** For 26 letters, this is 26!.

  • **Analogize the event to a simpler problem.** For instance, treat 'a' and 'z' as a single unit to simplify calculations.

  • **Use subtraction for complementary events.** For finding the probability that 'a' and 'b' are not next to each other, first calculate when they are next to each other, then subtract from 1.
By applying these steps, you can tackle complex problems methodically and arrive at accurate solutions.

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Most popular questions from this chapter

A player in the Powerball lottery picks five different integers between 1 and 69 , inclusive, and a sixth integer between 1 and \(26,\) which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn. a) What is the probability that a player wins the jackpot? b) What is the probability that a player wins \(\$ 1,000,000\) , which is the prize for matching the first five numbers, but not the sixth number, drawn? c) What is the probability that a player wins \(\$ 100\) by matching exactly three of the first five and the sixth numbers drawn, or four of the first five numbers, but not the sixth number, drawn? d) What is the probability that a player wins a prize of \(\$ 4,\) which is the prize when the player matches the sixth number, and either one or none of the first five numbers drawn?

This problem was posed by the Chevalier de M茅r茅 and was solved by Blaise Pascal and Pierre de Fermat. a) Find the probability of rolling at least one six when a fair die is rolled four times. b) Find the probability that a double six comes up at least once when a pair of dice is rolled 24 times. Answer the query the Chevalier de M茅r茅 made to Pascal asking whether this probability was greater than 1\(/ 2\) . c) Is it more likely that a six comes up at least once when a fair die is rolled four times or that a double six comes up at least once when a pair of dice is rolled 24 times?

A pair of dice is rolled in a remote location and when you ask an honest observer whether at least one die came up six, this honest observer answers in the affirmative. a) What is the probability that the sum of the numbers that came up on the two dice is seven, given the information provided by the honest observer? b) Suppose that the honest observer tells us that at least one die came up five. What is the probability the sum of the numbers that came up on the dice is seven, given this information?

Prove Theorem \(2,\) the extended form of Bayes' theorem. That is, suppose that \(E\) is an event from a sample space \(S\) and that \(F_{1}, F_{2}, \ldots, F_{n}\) are mutually exclusive events such that \(\bigcup_{i=1}^{n} F_{i}=S .\) Assume that \(p(E) \neq 0\) and \(p\left(F_{i}\right) \neq 0\) for \(i=1,2, \ldots, n .\) Show that $$ p\left(F_{j} | E\right)=\frac{p\left(E | F_{j}\right) p\left(F_{j}\right)}{\sum_{i=1}^{n} p\left(E | F_{i}\right) p\left(F_{i}\right)} $$ \(\left[\text {Hint} : \text { Use the fact that } E=\bigcup_{i=1}^{n}\left(E \cap F_{i}\right) .\right]\)

Use Chebyshev's inequality to find an upper bound on the probability that the number of tails that come up when a biased coin with probability of heads equal to 0.6 is tossed \(n\) times deviates from the mean by more than \(\sqrt{n}\) .
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