
Let 
X
X be a continuous random variable with PDF:
f
X
(
x
)
=
{
5
x
4
0
<
x
≤
1
0
otherwise
f 
X
​
 (x)={ 
5x 
4
 
0
​
  
0<x≤1
otherwise
​
 
Find 
P
(
X
≤
3
9
∣
X
>
1
9
)
.
P(X≤ 
9
3
​
 ∣X> 
9
1
​
 ).Enter your answer correct to three decimals accuracy.


We are tasked with finding \( P\left(X \leq \frac{3}{9} \mid X > \frac{1}{9}\right) \), which is a conditional probability. This can be calculated as:

\[
P\left(X \leq \frac{3}{9} \mid X > \frac{1}{9}\right) = \frac{P\left(\frac{1}{9} < X \leq \frac{3}{9}\right)}{P\left(X > \frac{1}{9}\right)}
\]

### Step 1: Calculate the cumulative distribution function (CDF)
We are given the probability density function (PDF):

\[
f_X(x) = 5x^4 \quad \text{for } 0 < x \leq 1
\]

To compute the required probabilities, we first find the CDF. The CDF \( F_X(x) \) is the integral of the PDF:

\[
F_X(x) = \int_0^x 5t^4 \, dt
\]

We calculate this integral:

\[
F_X(x) = 5 \int_0^x t^4 \, dt = 5 \left[\frac{t^5}{5}\right]_0^x = x^5
\]

Thus, the CDF is:

\[
F_X(x) = x^5
\]

### Step 2: Calculate the numerator
Now, we compute the probability \( P\left(\frac{1}{9} < X \leq \frac{3}{9}\right) \). This is:

\[
P\left(\frac{1}{9} < X \leq \frac{3}{9}\right) = F_X\left(\frac{3}{9}\right) - F_X\left(\frac{1}{9}\right)
\]

Substituting the values:

\[
F_X\left(\frac{3}{9}\right) = \left(\frac{3}{9}\right)^5 = \left(\frac{1}{3}\right)^5 = \frac{1}{243}
\]

\[
F_X\left(\frac{1}{9}\right) = \left(\frac{1}{9}\right)^5 = \frac{1}{59049}
\]

So, the numerator is:

\[
P\left(\frac{1}{9} < X \leq \frac{3}{9}\right) = \frac{1}{243} - \frac{1}{59049} = \frac{59049 - 243}{59049 \times 243} = \frac{58806}{59049 \times 243}
\]

### Step 3: Calculate the denominator
Next, we compute the probability \( P\left(X > \frac{1}{9}\right) \), which is:

\[
P\left(X > \frac{1}{9}\right) = 1 - F_X\left(\frac{1}{9}\right)
\]

Substitute the value of \( F_X\left(\frac{1}{9}\right) \):

\[
P\left(X > \frac{1}{9}\right) = 1 - \frac{1}{59049}
\]

Thus:

\[
P\left(X > \frac{1}{9}\right) = \frac{59048}{59049}
\]

### Step 4: Compute the conditional probability
Now we compute the conditional probability:

\[
P\left(X \leq \frac{3}{9} \mid X > \frac{1}{9}\right) = \frac{\frac{58806}{59049 \times 243}}{\frac{59048}{59049}} = \frac{58806}{59048 \times 243}
\]

### Step 5: Final calculation

Now we can calculate the final value. First, the value:

\[
P(X \leq \frac{3}{9} \mid X > \frac{1}{9}) = \frac{58806}{14374764} = 0.00409
\]

Rounding to three decimal places, the final answer is:

\[
P(X \leq \frac{3}{9} \mid X > \frac{1}{9}) \approx 0.004
\]

Let me know if you need further details or if anything is unclear.

---

Here are 5 related questions you might consider:

1. How does the conditional probability formula change if the intervals were different?
2. What would be the result if \( X \) had a different PDF, such as \( f_X(x) = 6x^5 \)?
3. How do you interpret the conditional probability in terms of real-world events?
4. What is the significance of finding the CDF when calculating probabilities?
5. How would you approach calculating \( P(X \leq a \mid X > b) \) for general \( a \) and \( b \)?

Tip: When dealing with conditional probabilities, always find the CDF to simplify complex calculations!

Let X be a continuous random variable with PDF: f_X(x) = { 5x^4 for 0 < x ≤ 1, 0 otherwise. Find P(X ≤ 3/9 | X > 1/9). Enter your answer correct to three decimals accuracy.

Learn how to calculate the conditional probability P(X ≤ 3/9 | X > 1/9) for a continuous random variable with PDF f_X(x) = 5x^4 for 0 < x ≤ 1. The solution covers key concepts in probability theory, including PDFs, CDFs, and the law of total probability, with a detailed step-by-step approach. Suitable for undergraduate students studying probability and statistics.

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Let X X be a continuous random variable with PDF: f X ( x )

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