(a) Three typed letters and their envelopes are piled on a desk. If someone puts the letters into the envelopes at random (one letter in each), what is the probability that each letter gets into its own envelope? Call the envelopes \(A, B, C,\) and the corresponding letters \(a, b, c,\) and set up the sample space. Note that " \(a\) in \(C\) \(b\) in \(B, c\) in \(A "\) is one point in the sample space. (b) What is the probability that at least one letter gets into its own envelope? Hint: What is the probability that no letter gets into its own envelope? (c) Let \(A\) mean that \(a\) got into envelope \(A\), and so on. Find the probability \(P(A)\) that \(a\) got into \(A\). Find \(P(B)\) and \(P(C)\). Find the probability \(P(A+B)\) that either \(a\) or \(b\) or both got into their correct envelopes, and the probability \(P(A B)\) that both got into their correct envelopes. Verify equation (3.6)

Step by step solution

01

Determine the sample space

List all possible ways to distribute the letters to the envelopes. Since there are 3 letters and 3 envelopes, we have 3! = 6 possible arrangements: 1. a in A, b in B, c in C2. a in A, b in C, c in B3. a in B, b in A, c in C4. a in B, b in C, c in A5. a in C, b in A, c in B6. a in C, b in B, c in A

02

Calculate the probability of each letter getting in the correct envelope

There is only 1 arrangement where each letter gets into its correct envelope: a in A, b in B, c in C. The probability is the number of favorable outcomes (1) divided by the total possible outcomes (6). So the probability is: \[ P(\text{all correct}) = \frac{1}{6} \]

03

Calculate the probability that no letter gets into its correct envelope

Derangements are permutations where no element appears in its correct position. For 3 items, there are 2 derangements: 1. a in B, b in C, c in A2. a in C, b in A, c in BThe probability is the number of derangements (2) divided by the total possible outcomes (6). So the probability is: \[ P(\text{none correct}) = \frac{2}{6} = \frac{1}{3} \]

04

Find the probability that at least one letter gets into its own envelope

Use the complement rule. The probability that at least one letter is correctly placed is the complement of the probability that no letters are correctly placed. \[ P(\text{at least one correct}) = 1 - P(\text{none correct}) = 1 - \frac{1}{3} = \frac{2}{3} \]

05

Calculate individual event probabilities

To find the probability that a specific letter is placed correctly, note that each letter has 3 choices, and only 1 choice is correct. Thus, the probability: \[ P(A) = P(B) = P(C) = \frac{1}{3} \]

06

Calculate the probability that either a or b is correctly placed

Use the principle of inclusion-exclusion: \[ P(A + B) = P(A) + P(B) - P(AB) \] Since there are 3 envelopes, and with 1 correct letter the remaining 2 letters must be deranged. This occurs in \( \frac{2!}{3!} \). The result is: \[ P(A + B) = \frac{1}{3} + \frac{1}{3} - \frac{1}{6} = \frac{1}{2} \]

07

Calculate the probability that both a and b are correctly placed

Consider both events: \[ P(AB) = \frac{1}{6} \]

08

Verify Equation (3.6)

Equation (3.6) states that: \[ P(A + B) = P(A) + P(B) - P(AB) \] Substitute the obtained values: \[ P(A + B) = \frac{1}{3} + \frac{1}{3} - \frac{1}{6} = \frac{1}{2} \] This confirms that the calculated probabilities are consistent.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Derangements

A derangement is a permutation where none of the elements appear in their correct positions. In our letter and envelope problem, this means no letter is placed in its corresponding envelope. For example, with three letters (a, b, c) and three envelopes (A, B, C), there are 3! = 6 possible permutations. Out of these, only two arrangements qualify as derangements:

• a in B, b in C, c in A
• a in C, b in A, c in B

Thus, the probability of a derangement (no letter is correctly placed) for three letters is the number of derangements (2) divided by the total number of permutations (6). This gives us \( \frac{2}{6} = \frac{1}{3} \). Derangements play a crucial role in solving problems where we need to consider the number of ways to shuffle elements so that none of them appear in their original position.

Sample Space

The sample space in probability refers to the set of all possible outcomes of a given experiment. For our problem of randomly assigning three letters to three envelopes, we need to list all possible assignments. The sample space comprises all 3! = 6 possible permutations:

• a in A, b in B, c in C
• a in A, b in C, c in B
• a in B, b in A, c in C
• a in B, b in C, c in A
• a in C, b in A, c in B
• a in C, b in B, c in A

Each of these outcomes represents one way to distribute the letters among the envelopes. By understanding the sample space, we can effectively calculate the probabilities for different events by simply dividing the number of favorable outcomes by the total number of outcomes in the sample space.

Inclusion-Exclusion Principle

The inclusion-exclusion principle is a fundamental concept in combinatorics and probability. It helps to find the probability of the union of multiple events by summing the probabilities of individual events and subtracting the probabilities of their intersections. For our letter and envelope problem, let's find the probability that either 'a' or 'b' (or both) are correctly placed.

• Let P(A) be the probability that a is in A: \(\frac{1}{3}\).
• Let P(B) be the probability that b is in B: \(\frac{1}{3}\).
• Let P(AB) be the probability that both a and b are in A and B respectively: \(\frac{1}{6}\).

Using the inclusion-exclusion principle formula, we get: \(P(A + B) = P(A) + P(B) - P(AB)\), which simplifies to \(\frac{1}{3} + \frac{1}{3} - \frac{1}{6} = \frac{1}{2}\). This principle helps in calculating more complex probabilities by breaking them down into simpler, overlapping events.

Complement Rule

The complement rule in probability states that the probability of an event happening is equal to one minus the probability of it not happening. For our problem, to find the probability that at least one letter is correctly placed, we first need to calculate the probability that no letter is correctly placed (a derangement) and then use the complement rule.

• The probability of no letters being correctly placed is \(\frac{1}{3}\), as derived from the derangements.

Using the complement rule, the probability that at least one letter is correctly placed is: \(P(\text{at least one correct}) = 1 - P(\text{none correct})\), which results in \(1 - \frac{1}{3} = \frac{2}{3}\). This rule is very useful in situations where calculating the direct probability is more complex than calculating its complement.