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Fifty raffle tickets are numbered 1 through 50 , and one of them is drawn at random. What is the probability that the number is a multiple of 5 or \(7 ?\) Consider the following "solution": Since 10 tickets bear numbers that are multiples of 5 and since 7 tickets bear numbers that are multiples of 7 , we conclude that the required probability is $$ \frac{10}{50}+\frac{7}{50}=\frac{17}{50} $$ What is wrong with this argument? What is the correct answer?

Short Answer

Expert verified
The error in the argument was that it failed to account for the overlap in multiples of the number 5 and 7, specifically the number 35, which caused double-counting. After identifying 10 multiples of 5 and 7 multiples of 7, it's crucial to subtract the one common multiple of both 5 and 7 – which is 35 – from these numbers. Therefore, the correct probability should be \( \frac{{10 + 7 - 1}}{50} = \frac{16}{50} \), which simplifies to \( \frac{8}{25} \).

Step by step solution

01

Identify multiple of 5 and 7

First, identify all the numbers between 1 and 50 that are multiples of 5 and count them. There should be a total of 10 tickets (5, 10, 15, ..., 50). Then, do the same for multiples of 7. There should be 7 such tickets (7, 14, 21, ..., 49).
02

Identify common multiples

Second, it is crucial to identify numbers that are multiples of both 5 and 7 because they have been counted twice in the first step. There are two such numbers on the tickets: 35 and 49. They are counted in the multiples of 5 and 7, so they must be subtracted once to balance the double-counting.
03

Calculate the correct probability

Now, the correct probability expression should be: \( \frac{{\text{{multiples of 5}} + \text{{multiples of 7}} - \text{{common multiples}}}}{{\text{{total numbers}}}} \). After inserting the values from Steps 1 and 2 it becomes: \( \frac{{10 + 7 - 1}}{50} = \frac{16}{50} \). Finally, simplify it to the lowest terms, which is \( \frac{8}{25} \). The main error in the original reasoning was failing to take into account the overlap between the multiples of 5 and 7, more specifically the number 35, which lead to overestimating the probability.

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Key Concepts

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

Multiples in Probability
When calculating probabilities involving multiples of numbers, it's like looking for specific patterns in a lottery. Imagine each ticket has a unique number, and we're interested in those with a number that fits into our specific pattern – like being a multiple of a particular number. In our raffle ticket example, the pattern we're looking at includes tickets numbered with a multiple of 5 or 7.

To find the probability, we count the total tickets fitting our criteria and divide by the grand total of tickets. However, it's vital to remember that some numbers could fit both patterns – meaning they're multiples of both 5 and 7. These are what we call common multiples, and they matter a lot because they can trick us into counting twice if we're not careful.
Common Multiples
Common multiples are numbers that are multiples of two or more other numbers. For instance, 35 and 70 are common multiples of 5 and 7 because they can be divided evenly by both numbers. In probability, it's essential to be on the lookout for these special numbers.

In the context of our ticket problem, identifying these common multiples is like finding where our patterns overlap. This is crucial because the odds of drawing a number that is both a multiple of 5 and 7 is part of our pattern – but it's easy to overcount if we treat them as separate events. To calculate the probability accurately, we subtract common multiples once to correct the overestimation.
Correcting Probability Overestimation
It's simple to misjudge the odds by not accounting for overlaps, also known as overestimation. Overestimation in our raffle ticket problem comes from counting the common multiples of 5 and 7 twice. This error inflates our expected probability, making it seem more likely than it really is to draw a multiple of 5 or 7.

To correct this, we recognize that the common multiples belong to both sets and hence, remove them from our count once. By doing so, we reach the true count and calculate the correct odds. Correcting for overestimation is a good habit in probability calculation and helps us arrive at more reliable results.
Probability Simplification
Simplification, like tidying up after a math party, makes the final answer clean and clear. To simplify, we take our adjusted count of qualifying tickets and put it over the total number of tickets, just as we would with any fraction.

In our case, after correcting the overestimation, we got a probability of \( \frac{16}{50} \) for drawing a number that is a multiple of 5 or 7. But it's not in its simplest form! Like reducing fractions in basic arithmetic, we simplify this fraction to get \( \frac{8}{25} \), which is easier to understand and interpret. Simplifying probabilities helps us better communicate the likelihood and make more intuitive comparisons between different probabilities.

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