Let S be the sample space of all five digit numbers. If p is the probability that a randomly selecte

Q: Wait, why do we use the formula $n = \frac{l - a}{d} + 1$? Can't we just divide the range by 7?

That is a common trap! Simply dividing the range (like $\frac{90000}{7}$) gives you an approximation, but it doesn't account for the specific boundaries of our five-digit range $[10000, 99999]$. The AP formula is the only way to be mathematically precise about the first and last terms within your specific interval.

Q: Why is the intersection $A \cap B$ equal to multiples of 35?

Great question! Since 7 and 5 are coprime (they share no common factors other than 1), any number that is a multiple of both must be a multiple of their product, which is $7 \times 5 = 35$. This is a fundamental property of divisibility.

Q: Why is the sample space 90,000? Shouldn't it be $99999 - 10000 = 89999$?

Ah, the classic 'fencepost error'! If you have a fence with posts at 1 and 2, you have 2 posts, but the difference is $2-1=1$. When counting integers in an inclusive range $[a, b]$, the count is always $b - a + 1$. So, $99999 - 10000 + 1 = 90000$. Always remember to add that 1!

Question

Let S be the sample space of all five digit numbers. If $p$ is the probability that a randomly selected number from S, is a multiple of 7 but not divisible by 5, then $9p$ is equal to

Options:

(A) 1.0146

(B) 1.2085

(C) 1.0285

(D) 1.1521

Sigma Percentile · Accepted Answer

### Analyzing the Universe of Numbers

Welcome, future engineer. Today, we aren't just solving a probability problem; we are embarking on a journey through the architecture of numbers.

When we look at the set of all five-digit numbers, we are looking at a vast, structured universe. The smallest five-digit number is $10000$, and the largest is $99999$.

Many students stumble here, falling into the 'fencepost error' trap. They subtract $10000$ from $99999$ and get $89999$. But remember, we must include both endpoints.

Thus, our sample space $S$ has a size of:
$$n(S) = 99999 - 10000 + 1 = 90000$$

### The Art of Exclusion: Set Theory

Now, let us define our conditions. We are interested in numbers divisible by $7$ but *not* by $5$.

Let $A$ be the set of numbers divisible by $7$, and $B$ be the set of numbers divisible by $5$. We are looking for the region in our Venn diagram that belongs to $A$ but lies outside $B$.

Mathematically, this is the set $A \cap B^c$. The most elegant way to calculate this is to take the total count of multiples of $7$ and subtract the count of numbers that are multiples of *both* $7$ and $5$:
$$n(A \cap B^c) = n(A) - n(A \cap B)$$

### The Arithmetic Progression Engine

To find $n(A)$, we need to count how many multiples of $7$ exist between $10000$ and $99999$. This is an arithmetic progression where the common difference $d = 7$.

First, we find the smallest multiple of $7$ in our range. Dividing $10000$ by $7$ gives approximately $1428.57$. The next integer is $1429$. Thus, our first term is:
$$a = 1429 	imes 7 = 10003$$

Next, we find the largest multiple. Dividing $99999$ by $7$ gives approximately $14285.57$. Taking the integer part, $14285$, we find our last term:
$$l = 14285 	imes 7 = 99995$$

Using the formula for the number of terms $n = \frac{l - a}{d} + 1$, we get:
$$n_1 = \frac{99995 - 10003}{7} + 1 = \frac{89992}{7} + 1 = 12856 + 1 = 12857$$

### The Intersection: The Multiples of 35

Now, we must subtract the 'intruders'—the numbers divisible by both $7$ and $5$. Since $7$ and $5$ are coprime, these are the multiples of $35$.

Smallest multiple: $10000 / 35 \approx 285.7$. The next integer is $286$. So, our first term is:
$$a' = 286 	imes 35 = 10010$$

Largest multiple: $99999 / 35 \approx 2857.1$. The integer part is $2857$. So, our last term is:
$$l' = 2857 	imes 35 = 99995$$

Calculating the number of terms $n_2$:
$$n_2 = \frac{99995 - 10010}{35} + 1 = \frac{89985}{35} + 1 = 2571 + 1 = 2572$$

### The Final Victory

We are almost at the finish line. The number of favorable outcomes is:
$$n(A \cap B^c) = n_1 - n_2 = 12857 - 2572 = 10285$$

The probability $p$ is the ratio of favorable outcomes to the total sample space:
$$p = \frac{10285}{90000}$$

Finally, the question asks for $9p$. Watch how the numbers dance for us:
$$9p = 9 	imes \frac{10285}{90000} = \frac{10285}{10000} = \mathbf{1.0285}$$

And there it is. By methodically breaking down the problem into sets and using the precision of arithmetic progressions, we have arrived at the solution.

Animated Solution for Mathematics - Probability: Let S be the sample space of all five digit numbers. If $p$ is the probability that a randomly selected number from S, is a multiple of 7 but not divisible by 5, then $9 p$ is equal to

Select Answer:

Visualized Solution

Defining the Sample Space $S$

Defining the Sets

The Logical Strategy

Understanding the Intersection

Finding Multiples of 7: First Term

Finding Multiples of 7: Last Term

Counting Multiples of 7 ( $n_{1}$ )

Finding Multiples of 35: First Term

Finding Multiples of 35: Last Term

Counting Multiples of 35 ( $n_{2}$ )

Final Count of Favorable Outcomes

Calculating Probability $p$

Finding the Final Value $9 p$

The Sigma Insight: Classical Definition of Probability

Analyzing the Universe of Numbers

The Art of Exclusion

The Arithmetic Progression Engine

The Intersection

The Final Victory

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