If a random variable x is Gamma distribution with parameter λ = 3, compute the
value of
1. P(x ≤ 1) 2. P(1 ≤ x ≤ 2)

Given that \( x \) is a random variable that follows a Gamma distribution with parameter \(\lambda = 3\), we need to compute the following probabilities:

1. \( P(x \leq 1) \)
2. \( P(1 \leq x \leq 2) \)

Let's proceed step-by-step:

### Gamma Distribution

The Gamma distribution's probability density function (PDF) for a random variable \( x \) is given by:

\[
f(x; \alpha, \lambda) = \frac{\lambda^\alpha x^{\alpha - 1} e^{-\lambda x}}{\Gamma(\alpha)}
\]

where:
- \(\alpha\) is the shape parameter,
- \(\lambda\) is the rate parameter,
- \(\Gamma(\alpha)\) is the Gamma function, which generalizes the factorial function.

### Interpreting the Problem

The problem provides us with \(\lambda = 3\), but we do not have an explicit value for \(\alpha\) (shape parameter). If only \(\lambda\) is given and no \(\alpha\) is specified, a typical interpretation (or the most straightforward) would be to assume the standard exponential distribution, where \(\alpha = 1\). This is because the exponential distribution is a special case of the Gamma distribution with \(\alpha = 1\).

Hence, we assume:
- Shape parameter \(\alpha = 1\)
- Rate parameter \(\lambda = 3\)

### Exponential Distribution PDF

With \(\alpha = 1\), the Gamma distribution simplifies to an exponential distribution with PDF:

\[
f(x; \lambda) = \lambda e^{-\lambda x}
\]

For our problem, this becomes:

\[
f(x; 3) = 3 e^{-3x}
\]

### Computing the Probabilities

1. **Probability \( P(x \leq 1) \)**

\[
P(x \leq 1) = \int_0^1 3 e^{-3x} \, dx
\]

Using integration by substitution, let \( u = -3x \), hence \( du = -3dx \), which gives \( dx = -\frac{1}{3}du \). The integral becomes:

\[
P(x \leq 1) = \int_0^1 3 e^{-3x} \, dx = \left[-e^{-3x}\right]_0^1 = 1 - e^{-3}
\]

\[
P(x \leq 1) \approx 1 - e^{-3} \approx 1 - 0.0498 \approx 0.9502
\]

2. **Probability \( P(1 \leq x \leq 2) \)**

\[
P(1 \leq x \leq 2) = \int_1^2 3 e^{-3x} \, dx
\]

Following a similar method:

\[
P(1 \leq x \leq 2) = \left[-e^{-3x}\right]_1^2 = e^{-3} - e^{-6}
\]

\[
P(1 \leq x \leq 2) \approx 0.0498 - 0.0025 \approx 0.0473
\]

### Final Answers

1. \( P(x \leq 1) \approx 0.9502 \)
2. \( P(1 \leq x \leq 2) \approx 0.0473 \)

Would you like more details, or do you have any questions?

### Related Questions

1. How does the Gamma distribution differ from the exponential distribution?
2. What is the cumulative distribution function (CDF) of the exponential distribution?
3. How can we interpret the shape and rate parameters of the Gamma distribution?
4. How would the probability change if the shape parameter \(\alpha\) were different?
5. What are the mean and variance of a Gamma-distributed random variable?

### Tip

The exponential distribution is a special case of the Gamma distribution and is commonly used to model the time until the next event in a Poisson process.

Learn how to calculate probabilities for a Gamma distributed random variable with parameter λ = 3. Understand the relationship with the exponential distribution, integral calculus techniques, and probability theory principles.

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