Problem 38 - Algebra and Trigonometry with Analytic Geometry

Question

Letter and number experiment An experiment consists of selecting a letter from the alphabet and one of the digits 0 , \(1, \ldots, 9\). (a) Describe the sample space \(S\) of the experiment, and find \(n(S)\). (b) Suppose the letters of the alphabet are assigned numbers as follows: \(A=1, B=2, \ldots, Z=26\). Let \(E_{1}\) be the event in which the units digit of the number assigned to the letter of the alphabet is the same as the digit selected. Find \(n\left(E_{1}\right), n\left(E_{1}^{\prime}\right)\), and \(P\left(E_{1}\right)\). (c) Let \(E_{2}\) be the event that the letter is one of the five vowels and \(E_{3}\) the event that the digit is a prime number. Are \(E_{2}\) and \(E_{3}\) mutually exclusive? Are they independent? Find \(P\left(E_{2}\right), P\left(E_{3}\right), P\left(E_{2} \cap E_{3}\right)\), and \(P\left(E_{2} \cup E_{3}\right)\). (d) Let \(E_{4}\) be the event that the numerical value of the letter is even. Are \(E_{2}\) and \(E_{4}\) mutually exclusive? Are they independent? Find \(P\left(E_{2} \cap E_{4}\right)\) and \(P\left(E_{2} \cup E_{4}\right)\).

Step-by-Step Solution

Answer

(a) Sample space size is 260. (b) Event sizes: 10, 250; Probability: 1/26. (c) E2, E3 are not mutually exclusive, approximately independent; Probabilities: 5/26, 4/13, 1/13, 13/26. (d) E2 and E4 are not mutually exclusive or independent; Probabilities: 1/13, 8/13.

Step 1: Describe the Sample Space and Calculate n(S)

The sample space \( S \) consists of combinations of a letter from the alphabet (26 letters) and a digit from 0 to 9 (10 digits). Thus, the sample space is the set of all possible pairs like (A, 0), (A, 1),..., (B, 0), ..., until (Z, 9). The total number of outcomes \( n(S) \) is calculated as the product of the number of letters and digits: \( n(S) = 26 \times 10 = 260 \).

Step 2: Define E1 and Calculate n(E1), n(E1'), and P(E1)

Event \( E_1 \) occurs when the units digit of the assigned letter number is the same as the digit chosen. Numbers assigned to letters (A=1, ..., Z=26) have units digits from 0 to 9. For each number where the units digit is \( d \), there is 1 corresponding letter. So for each digit \( 0, 1, \, ..., 9 \), there is 1 matching letter, giving 10 outcomes for \( E_1 \). Thus, \( n(E_1) = 10 \) and \( n(E_1') = n(S) - n(E_1) = 260 - 10 = 250 \). The probability \( P(E_1) \) is \( \frac{n(E_1)}{n(S)} = \frac{10}{260} = \frac{1}{26} \).

Step 3: Define E2 and E3 and Check for Mutual Exclusivity and Independence

Define \( E_2 \) as the event of selecting a vowel (A, E, I, O, U), so \( n(E_2) = 5 \times 10 = 50 \). For \( E_3 \), the prime digits are 2, 3, 5, 7, giving \( n(E_3) = 26 \times 4 = 104 \). To check mutual exclusivity, note \( E_2 \) and \( E_3 \) are not mutually exclusive as they can have outcomes in common (e.g., choosing A and 2). To check independence, compute \( P(E_2 \cap E_3) = \frac{5 \times 4}{260} = \frac{1}{13} \), and compare \( P(E_2)P(E_3) = \frac{50}{260} \times \frac{104}{260} = \frac{5}{65} \times \frac{4}{10} \approx \frac{1}{13} \). Since \( P(E_2 \cap E_3) \approx P(E_2)P(E_3) \), they are approximately independent. Calculate \( P(E_2 \cup E_3) = P(E_2) + P(E_3) - P(E_2 \cap E_3) = \frac{5}{26} + \frac{4}{13} - \frac{1}{13} = \frac{5}{26} + \frac{3}{13} \approx \frac{13}{26} \).

Step 4: Define E4 and Analyze Relationship with E2

Event \( E_4 \) is the event that the numerical value of the letter is even. There are 13 even numbers from 1 to 26, so \( n(E_4) = 13 \times 10 = 130 \). \( E_2 \) and \( E_4 \) are not mutually exclusive as 'E' (which is a vowel) is even. To determine independence, check if \( P(E_2 \cap E_4) \equiv P(E_2)P(E_4) \). We know \( P(E_2) = \frac{50}{260} \), \( P(E_4) = \frac{130}{260} \), and \( n(E_2 \cap E_4) = 2 \times 10 = 20 \), so \( P(E_2 \cap E_4) = \frac{20}{260} = \frac{1}{13} \). Since \( \frac{1}{13} eq \frac{50}{260} \times \frac{130}{260} \), they are not independent. Finally, calculate \( P(E_2 \cup E_4) = P(E_2) + P(E_4) - P(E_2 \cap E_4) = \frac{50}{260} + \frac{130}{260} - \frac{20}{260} = \frac{160}{260} = \frac{8}{13} \).

Key Concepts

Sample Space

A sample space is essentially the set of all possible outcomes of a particular experiment. It helps us visualize and understand the breadth of possibilities. In the given exercise, we are experimenting with choosing letters from the alphabet and digits from 0 to 9.

Because there are 26 letters in the alphabet (A to Z) and 10 digits (0, 1, ..., 9), each letter can pair with any of the digits. This results in a total number of sample points in the sample space. Calculating the size of this sample space involves multiplying the number of letters by the number of digits:

• Total letters = 26
• Total digits = 10
• Sample space size, \(n(S) = 26 \times 10 = 260\)

This makes the sample space of the experiment contain 260 different outcomes such as (A,0), (A,1), ..., (Z, 9). It's important to grasp the idea of a sample space because it frames the "universe" of possibilities in which any event is considered.

Mutually Exclusive Events

Mutually exclusive events are events that cannot occur simultaneously. If one happens, the other cannot. An intuitive example would be tossing a coin where the outcomes "heads" and "tails" are mutually exclusive.

In the context of this exercise, we need to see if picking a vowel (\(E_2\)) and picking a digit that is a prime number (\(E_3\)) are mutually exclusive. Since an event is defined as a subset of the sample space, events \(E_2\) and \(E_3\) are not mutually exclusive because it is possible to both select a vowel and a prime number simultaneously, such as picking "A" and "2".

Similarly, examining \(E_2\) (selecting a vowel) and \(E_4\) (letter has an even number value), they are not mutually exclusive either since selecting "E" results in an even numeric value and a vowel simultaneously.

Mutual exclusivity is a simple yet powerful concept that helps in determining whether events can have any common outcomes.

Independent Events

Independent events are events where the outcome of one does not affect or alter the probability of the other occurring. Understanding this helps in accurately calculating the probabilities by using the multiplication rule.

To check independence between events \(E_2\) and \(E_3\), we need to verify if the probability of their intersection is equal to the product of their individual probabilities. If \(P(E_2 \cap E_3) = P(E_2) \times P(E_3)\), then they are independent. Here, we find that:

• \(P(E_2) = \frac{5}{26}\)
• \(P(E_3) = \frac{2}{5}\)
• \(P(E_2 \cap E_3) = \frac{1}{13}\)
• \(P(E_2) \times P(E_3) \approx \frac{1}{13}\)

This approximation shows they are approximately independent.

In another relationship, \(E_2\) and \(E_4\) are not independent as the probability of their intersection does not match the product of their individual probabilities. Independence is an essential assumption in calculating joint probabilities and needs careful verification.

Probability Calculation

Probability calculation allows us to quantify the likelihood of an event happening. To compute probabilities, you divide the number of successful outcomes by the total number of possible outcomes.

For event \(E_1\), where the letter's number ends in the digit selected, only 10 outcomes satisfy this over the full sample space of 260, so:

• \[P(E_1) = \frac{10}{260} = \frac{1}{26}\]

For calculating compound probabilities, the principle of inclusion-exclusion is useful as seen with \(E_2\) and \(E_3\):

• \[P(E_2 \cup E_3) = P(E_2) + P(E_3) - P(E_2 \cap E_3)\]
• This yields \(P(E_2 \cup E_3) = \frac{5}{26} + \frac{4}{13} - \frac{1}{13} = \frac{13}{26}\).

Such calculations help in understanding how likely different scenarios are, guiding decision-making based on the probabilities of events.