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We are given the probability mass function (PMF) of a random variable \( X \), which is defined as:

\[
f(i) = \frac{c \lambda^i}{i!}, \quad i = 0, 1, 2, \dots
\]

where \( \lambda \) is a positive constant. We are asked to:

1. Find the value of \( c \).
2. Compute \( P(X = 0) \).
3. Compute \( P(X > 2) \).

### Part (a) Finding \( c \)

Since this is a probability mass function, the sum of the probabilities over all possible values of \( i \) must equal 1. That is:

\[
\sum_{i=0}^{\infty} f(i) = 1
\]

Substitute the expression for \( f(i) \):

\[
\sum_{i=0}^{\infty} \frac{c \lambda^i}{i!} = 1
\]

Recognize that the sum \( \sum_{i=0}^{\infty} \frac{\lambda^i}{i!} \) is the Taylor series expansion of \( e^\lambda \). Thus, the above equation becomes:

\[
c e^\lambda = 1
\]

Solving for \( c \):

\[
c = \frac{1}{e^\lambda}
\]

### Part (b) Finding \( P(X = 0) \)

Using the PMF, we can compute the probability \( P(X = 0) \):

\[
P(X = 0) = f(0) = \frac{c \lambda^0}{0!} = c
\]

From part (a), we know that \( c = \frac{1}{e^\lambda} \), so:

\[
P(X = 0) = \frac{1}{e^\lambda}
\]

### Part (c) Finding \( P(X > 2) \)

We want to calculate \( P(X > 2) \), which is the sum of the probabilities for \( i = 3, 4, 5, \dots \):

\[
P(X > 2) = 1 - P(X \leq 2)
\]

First, calculate \( P(X \leq 2) \), which is the sum of the probabilities for \( i = 0, 1, 2 \):

\[
P(X \leq 2) = f(0) + f(1) + f(2)
\]

We already know \( f(0) = \frac{1}{e^\lambda} \). Now calculate \( f(1) \) and \( f(2) \):

\[
f(1) = \frac{c \lambda^1}{1!} = \frac{1}{e^\lambda} \cdot \lambda
\]

\[
f(2) = \frac{c \lambda^2}{2!} = \frac{1}{e^\lambda} \cdot \frac{\lambda^2}{2}
\]

Thus, \( P(X \leq 2) \) is:

\[
P(X \leq 2) = \frac{1}{e^\lambda} + \frac{\lambda}{e^\lambda} + \frac{\lambda^2}{2e^\lambda}
\]

Factor out \( \frac{1}{e^\lambda} \):

\[
P(X \leq 2) = \frac{1}{e^\lambda} \left( 1 + \lambda + \frac{\lambda^2}{2} \right)
\]

Finally, subtract from 1 to get \( P(X > 2) \):

\[
P(X > 2) = 1 - \frac{1}{e^\lambda} \left( 1 + \lambda + \frac{\lambda^2}{2} \right)
\]

### Summary:
1. \( c = \frac{1}{e^\lambda} \)
2. \( P(X = 0) = \frac{1}{e^\lambda} \)
3. \( P(X > 2) = 1 - \frac{1}{e^\lambda} \left( 1 + \lambda + \frac{\lambda^2}{2} \right) \)

Do you have any questions or need further details?

---

**Relative questions:**
1. What is the significance of the parameter \( \lambda \) in this PMF?
2. How would \( P(X = k) \) be calculated for any arbitrary \( k \)?
3. What type of distribution does this PMF resemble?
4. How can we interpret \( P(X > 2) \) in terms of real-world applications?
5. How would the PMF change if we were dealing with a continuous random variable instead?

**Tip:** In many problems involving sums like this, recognizing series expansions (like \( e^\lambda \)) can simplify your work.

The probability mass function of a random variable X is given by f(i) = c * λ^i / i! for i = 0, 1, 2, ..., where λ is a positive constant. Find (a) value of c, (b) P(X = 0), and (c) P(X > 2).

Solve a probability mass function problem where f(i) = c * λ^i / i!. Calculate the value of c, the probability P(X=0), and P(X>2). This solution explores series expansions, the exponential distribution, and probability theory.

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