A number is selected at random from a list of the first 10 positive integers. What is the probability that...

1. Translate the problem requirements: We need to find the probability that a randomly selected number from \(\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\) is either odd OR divisible by 3 (or both).
2. Identify numbers satisfying each condition: List out which numbers are odd and which are divisible by 3 from our set of 10 numbers.
3. Apply the inclusion-exclusion principle: Count numbers that are odd OR divisible by 3, being careful not to double-count numbers that satisfy both conditions.
4. Calculate the final probability: Divide the count of favorable outcomes by the total number of possible outcomes (10).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in everyday language. We have a bag with 10 numbered balls: \(\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\). We're picking one ball at random and want to know: what's the chance that the number on the ball is either odd OR divisible by 3 (or both)?

The key word here is "OR" - this means we want numbers that satisfy at least one of these conditions:

• The number is odd, OR
• The number is divisible by 3, OR
• The number satisfies both conditions

Process Skill: TRANSLATE - Converting the probability language into concrete conditions we can check

2. Identify numbers satisfying each condition

Let's go through our list \(\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\) and identify which numbers meet each condition:

Odd numbers: These are numbers that don't divide evenly by 2

• 1 (odd) ✓
• 2 (even) ✗
• 3 (odd) ✓
• 4 (even) ✗
• 5 (odd) ✓
• 6 (even) ✗
• 7 (odd) ✓
• 8 (even) ✗
• 9 (odd) ✓
• 10 (even) ✗

So our odd numbers are: \(\{1, 3, 5, 7, 9\}\) - that's 5 numbers

Numbers divisible by 3: These divide evenly by 3 with no remainder

• \(1 ÷ 3 = 0\) remainder 1 ✗
• \(2 ÷ 3 = 0\) remainder 2 ✗
• \(3 ÷ 3 = 1\) remainder 0 ✓
• \(4 ÷ 3 = 1\) remainder 1 ✗
• \(5 ÷ 3 = 1\) remainder 2 ✗
• \(6 ÷ 3 = 2\) remainder 0 ✓
• \(7 ÷ 3 = 2\) remainder 1 ✗
• \(8 ÷ 3 = 2\) remainder 2 ✗
• \(9 ÷ 3 = 3\) remainder 0 ✓
• \(10 ÷ 3 = 3\) remainder 1 ✗

So our numbers divisible by 3 are: \(\{3, 6, 9\}\) - that's 3 numbers

3. Apply the inclusion-exclusion principle

Here's where we need to be careful! If we just add "5 odd numbers + 3 numbers divisible by 3", we get 8. But wait - some numbers appear in both lists!

Let's see which numbers satisfy BOTH conditions (odd AND divisible by 3):

• From our lists: \(\{1, 3, 5, 7, 9\} ∩ \{3, 6, 9\}\)
• The numbers that appear in both lists: \(\{3, 9\}\)
• That's 2 numbers that we've counted twice

Process Skill: CONSIDER ALL CASES - We must avoid double-counting numbers that satisfy both conditions

Using plain English logic:
Total favorable outcomes = (Numbers that are odd) + (Numbers divisible by 3) - (Numbers counted twice)
Total favorable outcomes = \(5 + 3 - 2 = 6\)

Let's verify by listing all favorable numbers directly:
\(\{1, 3, 5, 6, 7, 9\}\)

• 1: odd ✓
• 3: odd AND divisible by 3 ✓
• 5: odd ✓
• 6: divisible by 3 ✓
• 7: odd ✓
• 9: odd AND divisible by 3 ✓

Yes, that's exactly 6 numbers!

4. Calculate the final probability

Now we can find our probability using the basic definition:

Probability = (Number of favorable outcomes) ÷ (Total number of possible outcomes)

We found 6 favorable outcomes out of 10 total possible outcomes:
Probability = \(\frac{6}{10} = \frac{3}{5}\)

Looking at our answer choices, \(\frac{3}{5}\) corresponds to choice D.

Final Answer

The probability that a randomly selected number from \(\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\) is either odd or divisible by 3 is \(\frac{3}{5}\).

Answer: D

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "OR" as "AND"
Students often confuse the logical operators. When the question asks for numbers that are "odd OR divisible by 3," some students mistakenly think they need numbers that satisfy BOTH conditions simultaneously (odd AND divisible by 3). This leads them to only consider \(\{3, 9\}\) instead of all numbers satisfying at least one condition.

2. Not recognizing the need for inclusion-exclusion principle
Many students jump straight to adding the counts without realizing that some numbers appear in both categories. They might think: "5 odd numbers + 3 numbers divisible by 3 = 8 favorable outcomes" without considering that this double-counts certain numbers.

Errors while executing the approach

1. Incorrectly identifying odd numbers or multiples of 3
Students may make basic classification errors, such as thinking 2, 4, 6, 8, 10 are odd numbers, or missing that 6 and 9 are divisible by 3. These fundamental identification mistakes cascade into wrong final answers.

2. Arithmetic errors in inclusion-exclusion calculation
Even when students understand the concept, they might make calculation mistakes like: \(5 + 3 - 2 = 7\) instead of 6, or forget to subtract the overlap entirely, leading to \(5 + 3 = 8\) favorable outcomes.

3. Listing favorable outcomes incorrectly
When trying to verify by direct counting, students might create an incorrect list like \(\{1, 3, 5, 7, 9, 3, 6, 9\}\) (including duplicates) or miss some numbers entirely, leading to wrong counts.

Errors while selecting the answer

1. Not simplifying the fraction properly
Students might correctly calculate \(\frac{6}{10}\) but fail to reduce it to \(\frac{3}{5}\), then not recognize that \(\frac{3}{5}\) appears as answer choice D. They might instead look for \(\frac{6}{10}\) among the options and make an incorrect selection.

2. Confusing probability with count
Some students might arrive at the correct count of 6 favorable outcomes but then select an answer choice that represents 6 rather than the probability \(\frac{6}{10} = \frac{3}{5}\), especially if "6" appeared as an option in a different format.