A Box Contains 10 Tags Numbered 1 Through 10

Here's an exploration of probability scenarios involving a box containing ten tags numbered 1 through 10, offering detailed calculations and explanations.

Understanding the Basics of Probability

Probability, at its core, is the measure of the likelihood that an event will occur. It is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. The fundamental formula for calculating probability is:

Probability of an event = (Number of favorable outcomes) / (Total number of possible outcomes)

In our scenario, the "event" is drawing a tag from the box, and the "outcomes" are the numbers on the tags. Let's delve into various scenarios and calculate their probabilities.

Scenario 1: Drawing a Specific Number

Question: What is the probability of drawing the tag numbered 5?

Analysis:

• Favorable outcomes: There is only one tag with the number 5.
• Total possible outcomes: There are 10 tags in total.

Calculation:

Probability (drawing tag 5) = 1 / 10 = 0.1 or 10%

Scenario 2: Drawing an Even Number

Question: What is the probability of drawing a tag with an even number?

Analysis:

• Favorable outcomes: The even numbers between 1 and 10 are 2, 4, 6, 8, and 10. Thus, there are 5 favorable outcomes.
• Total possible outcomes: There are 10 tags in total.

Calculation:

Probability (drawing an even number) = 5 / 10 = 0.5 or 50%

Scenario 3: Drawing a Number Greater Than 7

Question: What is the probability of drawing a tag with a number greater than 7?

Analysis:

• Favorable outcomes: The numbers greater than 7 are 8, 9, and 10. Thus, there are 3 favorable outcomes.
• Total possible outcomes: There are 10 tags in total.

Calculation:

Probability (drawing a number greater than 7) = 3 / 10 = 0.3 or 30%

Scenario 4: Drawing a Prime Number

Question: What is the probability of drawing a tag with a prime number?

Analysis:

• Favorable outcomes: Prime numbers between 1 and 10 are 2, 3, 5, and 7. Thus, there are 4 favorable outcomes. (Remember, 1 is not considered a prime number).
• Total possible outcomes: There are 10 tags in total.

Calculation:

Probability (drawing a prime number) = 4 / 10 = 0.4 or 40%

Scenario 5: Drawing a Number Divisible by 3

Question: What is the probability of drawing a tag with a number divisible by 3?

Analysis:

• Favorable outcomes: Numbers divisible by 3 between 1 and 10 are 3, 6, and 9. Thus, there are 3 favorable outcomes.
• Total possible outcomes: There are 10 tags in total.

Calculation:

Probability (drawing a number divisible by 3) = 3 / 10 = 0.3 or 30%

Scenario 6: Drawing Two Tags Without Replacement (Conditional Probability)

This scenario introduces the concept of conditional probability, where the outcome of the first draw affects the probability of the second draw because we are not replacing the first tag.

Question: What is the probability of drawing the tag numbered 5 first, and then drawing the tag numbered 7, without replacing the first tag?

Analysis:

• Probability of drawing tag 5 first: As calculated before, this is 1/10.
• Probability of drawing tag 7 second, given that tag 5 was drawn first: Since tag 5 is removed, there are only 9 tags left. There is only one tag numbered 7. Therefore, the probability is 1/9.

Calculation:

Probability (drawing tag 5 then tag 7) = Probability (drawing tag 5) * Probability (drawing tag 7 | tag 5 already drawn) = (1/10) * (1/9) = 1/90 or approximately 1.11%

Scenario 7: Drawing Two Tags With Replacement

In this scenario, we replace the first tag before drawing the second. This means the outcome of the first draw does not affect the probability of the second draw.

Question: What is the probability of drawing the tag numbered 5 first, and then drawing the tag numbered 7, with replacement?

Analysis:

• Probability of drawing tag 5 first: This is 1/10.
• Probability of drawing tag 7 second: Since the tag is replaced, there are still 10 tags. The probability of drawing tag 7 is 1/10.

Calculation:

Probability (drawing tag 5 then tag 7 with replacement) = Probability (drawing tag 5) * Probability (drawing tag 7) = (1/10) * (1/10) = 1/100 or 1%

Scenario 8: Drawing Two Tags Without Replacement - General Case

Question: What is the probability of drawing an even number first, and then an odd number, without replacement?

Analysis:

• Probability of drawing an even number first: There are 5 even numbers (2, 4, 6, 8, 10). The probability is 5/10 = 1/2.
• Probability of drawing an odd number second, given that an even number was drawn first: After drawing an even number, there are 9 tags left. The number of odd numbers remains unchanged at 5. Therefore, the probability is 5/9.

Calculation:

Probability (drawing an even number then an odd number) = Probability (drawing an even number) * Probability (drawing an odd number | even number already drawn) = (1/2) * (5/9) = 5/18 or approximately 27.78%

Scenario 9: Drawing Two Tags With Replacement - General Case

Question: What is the probability of drawing an even number first, and then an odd number, with replacement?

Analysis:

• Probability of drawing an even number first: There are 5 even numbers (2, 4, 6, 8, 10). The probability is 5/10 = 1/2.
• Probability of drawing an odd number second: Since the tag is replaced, there are still 10 tags. There are 5 odd numbers (1, 3, 5, 7, 9). The probability is 5/10 = 1/2.

Calculation:

Probability (drawing an even number then an odd number with replacement) = Probability (drawing an even number) * Probability (drawing an odd number) = (1/2) * (1/2) = 1/4 or 25%

Scenario 10: Drawing Three Tags Without Replacement

This builds upon conditional probability, extending it to three events.

Question: What is the probability of drawing the tags numbered 1, 2, and 3, in that specific order, without replacement?

Analysis:

• Probability of drawing tag 1 first: 1/10
• Probability of drawing tag 2 second, given that tag 1 was drawn first: 1/9 (9 tags remaining)
• Probability of drawing tag 3 third, given that tags 1 and 2 were drawn: 1/8 (8 tags remaining)

Calculation:

Probability (drawing 1, 2, 3 in order) = (1/10) * (1/9) * (1/8) = 1/720 or approximately 0.14%

Scenario 11: Drawing Three Tags With Replacement

Question: What is the probability of drawing the tags numbered 1, 2, and 3, in that specific order, with replacement?

Analysis:

• Probability of drawing tag 1 first: 1/10
• Probability of drawing tag 2 second: 1/10 (tag is replaced)
• Probability of drawing tag 3 third: 1/10 (tag is replaced)

Calculation:

Probability (drawing 1, 2, 3 in order with replacement) = (1/10) * (1/10) * (1/10) = 1/1000 or 0.1%

Scenario 12: Drawing Three Tags Without Replacement - Any Order

Question: What is the probability of drawing the tags numbered 1, 2, and 3, in any order, without replacement?

Analysis:

This is a bit more complex. We need to consider all the possible orders in which we could draw 1, 2, and 3. These are:

• 1, 2, 3
• 1, 3, 2
• 2, 1, 3
• 2, 3, 1
• 3, 1, 2
• 3, 2, 1

There are 6 possible orders. We already know the probability of drawing them in one specific order (1, 2, 3) is 1/720. Since each order is equally likely, we multiply that probability by the number of possible orders.

Calculation:

Probability (drawing 1, 2, and 3 in any order) = 6 * (1/720) = 6/720 = 1/120 or approximately 0.83%

Alternatively, we can think about this in terms of combinations and permutations.

• Combinations: The number of ways to choose 3 tags out of 10 is 10C3 = (10!)/(3!7!) = (1098)/(321) = 120
• Favorable outcome: Only one of those combinations will yield the numbers 1, 2, and 3.
• Therefore, the probability is 1/120.

Scenario 13: Drawing Four Tags Without Replacement - Sum

Question: What is the probability of drawing four tags without replacement such that their sum is equal to 20?

Analysis:

This is a more challenging problem requiring us to identify all the combinations of four numbers between 1 and 10 that add up to 20. We need to be systematic to avoid missing any. Also, since we are drawing without replacement, a number can only appear once in each combination.

Here are the possible combinations:

• 1 + 2 + 7 + 10 = 20
• 1 + 3 + 6 + 10 = 20
• 1 + 3 + 7 + 9 = 20
• 1 + 4 + 5 + 10 = 20
• 1 + 4 + 6 + 9 = 20
• 1 + 4 + 7 + 8 = 20
• 1 + 5 + 6 + 8 = 20
• 2 + 3 + 5 + 10 = 20
• 2 + 3 + 6 + 9 = 20
• 2 + 3 + 7 + 8 = 20
• 2 + 4 + 5 + 9 = 20
• 2 + 4 + 6 + 8 = 20
• 2 + 5 + 6 + 7 = 20
• 3 + 4 + 5 + 8 = 20
• 3 + 4 + 6 + 7 = 20
• 3 + 5 + 6 + 6 = 20 (Invalid - 6 appears twice)
• 4 + 5 + 5 + 6 = 20 (Invalid - 5 appears twice)

We have identified 15 valid combinations.

Now we need to calculate the total number of ways to choose 4 tags from 10 without replacement. This is a combination problem: 10C4 = (10!)/(4!6!) = (10987)/(4321) = 210

Calculation:

Probability (sum of four tags equals 20) = (Number of favorable combinations) / (Total number of possible combinations) = 15 / 210 = 1/14 or approximately 7.14%

Scenario 14: Expected Value

Question: What is the expected value of a single draw from the box?

Analysis:

Expected value is the average outcome we'd expect if we repeated the experiment (drawing a tag) many times. To calculate it, we multiply each possible outcome by its probability and then sum the results. In this case, each tag has a probability of 1/10 of being drawn.

Calculation:

Expected Value = (1 * 1/10) + (2 * 1/10) + (3 * 1/10) + (4 * 1/10) + (5 * 1/10) + (6 * 1/10) + (7 * 1/10) + (8 * 1/10) + (9 * 1/10) + (10 * 1/10) = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) / 10 = 55 / 10 = 5.5

The expected value of a single draw is 5.5.

Scenario 15: Variance and Standard Deviation

Question: What are the variance and standard deviation of a single draw from the box?

Analysis:

Variance measures how spread out the data is from the expected value. Standard deviation is the square root of the variance and provides a more interpretable measure of spread.

To calculate the variance, we do the following:

1. Find the difference between each outcome and the expected value.
2. Square each of those differences.
3. Multiply each squared difference by its probability (1/10 in this case).
4. Sum the results.

Calculation:

Variance = [(1-5.5)^2 + (2-5.5)^2 + (3-5.5)^2 + (4-5.5)^2 + (5-5.5)^2 + (6-5.5)^2 + (7-5.5)^2 + (8-5.5)^2 + (9-5.5)^2 + (10-5.5)^2] / 10

= [20.25 + 12.25 + 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 + 12.25 + 20.25] / 10

= 82.5 / 10 = 8.25

Standard Deviation = √(Variance) = √8.25 ≈ 2.87

The variance is 8.25, and the standard deviation is approximately 2.87.

Conclusion

By exploring these diverse scenarios using a simple box of numbered tags, we've illustrated fundamental probability concepts, including simple probability, conditional probability, expected value, variance, and standard deviation. These principles are crucial for understanding risk, making informed decisions, and analyzing data in various fields, from finance and science to everyday life. Understanding these calculations can help anyone better assess the likelihood of various events and make more informed choices. Remember to always carefully define the "favorable outcomes" and "total possible outcomes" when approaching probability problems.