Answer :
Let's start by analyzing the given problem step by step.
### (a) Finding the parameter [tex]\( b \)[/tex]
The probability mass function (PMF) of a random variable [tex]\( X \)[/tex] is given by:
[tex]\[ p_x(x) = b \frac{\lambda^x}{x!}, \quad x = 0,1,2,\ldots \][/tex]
To determine the parameter [tex]\( b \)[/tex], we need to use the fact that the sum of all probabilities for the PMF must be equal to 1:
[tex]\[ \sum_{x=0}^{\infty} p_x(x) = 1 \][/tex]
Substituting the given PMF into the sum, we have:
[tex]\[ \sum_{x=0}^{\infty} b \frac{\lambda^x}{x!} = 1 \][/tex]
This is a series recognized as the Maclaurin series expansion for the exponential function [tex]\( e^\lambda \)[/tex]:
[tex]\[ \sum_{x=0}^{\infty} \frac{\lambda^x}{x!} = e^\lambda \][/tex]
Thus, we can rewrite the equation as:
[tex]\[ b \cdot e^\lambda = 1 \][/tex]
Solving for [tex]\( b \)[/tex], we get:
[tex]\[ b = \frac{1}{e^\lambda} \][/tex]
Given a specific value [tex]\( \lambda = 1 \)[/tex] (since [tex]\(\lambda > 0\)[/tex]), the value of [tex]\( b \)[/tex] calculates to:
[tex]\[ b = \frac{1}{e} \approx 0.36787944117144233 \][/tex]
### (b) Calculating [tex]\( P[X=1] \)[/tex]
To find [tex]\( P[X=1] \)[/tex], we substitute [tex]\( x = 1 \)[/tex] into the PMF:
[tex]\[ P[X=1] = p_x(1) = b \frac{\lambda^1}{1!} \][/tex]
Using the previously calculated value of [tex]\( b \)[/tex] and [tex]\( \lambda = 1 \)[/tex]:
[tex]\[ P[X=1] = \left(\frac{1}{e}\right) \cdot \frac{1^1}{1!} \][/tex]
Which simplifies to:
[tex]\[ P[X=1] = \frac{1}{e} \approx 0.36787944117144233 \][/tex]
### (c) Calculating [tex]\( P[X>3] \)[/tex]
To find [tex]\( P[X>3] \)[/tex], we first need to calculate [tex]\( P[X \leq 3] \)[/tex]. This can be done by summing the probabilities of [tex]\( P[X=0] \)[/tex], [tex]\( P[X=1] \)[/tex], [tex]\( P[X=2] \)[/tex], and [tex]\( P[X=3] \)[/tex]:
[tex]\[ P[X \leq 3] = P[X=0] + P[X=1] + P[X=2] + P[X=3] \][/tex]
Given the PMF, these probabilities are:
[tex]\[ P[X=0] = b \frac{\lambda^0}{0!} = \left(\frac{1}{e}\right) \cdot \frac{1}{1} = \frac{1}{e} \][/tex]
[tex]\[ P[X=1] = b \frac{\lambda^1}{1!} = \left(\frac{1}{e}\right) \cdot \frac{1}{1} = \frac{1}{e} \][/tex]
[tex]\[ P[X=2] = b \frac{\lambda^2}{2!} = \left(\frac{1}{e}\right) \cdot \frac{1^2}{2} = \frac{1}{2e} \][/tex]
[tex]\[ P[X=3] = b \frac{\lambda^3}{3!} = \left(\frac{1}{e}\right) \cdot \frac{1^3}{6} = \frac{1}{6e} \][/tex]
Adding these probabilities together:
[tex]\[ P[X \leq 3] = \frac{1}{e} + \frac{1}{e} + \frac{1}{2e} + \frac{1}{6e} \][/tex]
This simplifies to:
[tex]\[ P[X \leq 3] = \frac{1}{e} \left(1 + 1 + \frac{1}{2} + \frac{1}{6} \right) = \frac{1}{e} \left(\frac{6 + 6 + 3 + 1}{6} \right) = \frac{1}{e} \left(\frac{16}{6} \right) = \frac{8}{3e} \][/tex]
Using [tex]\( e \approx 2.71828 \)[/tex]:
[tex]\[ P[X \leq 3] \approx \frac{8}{3 \cdot 2.71828} \approx 0.9810118431238463 \][/tex]
Finally, [tex]\( P[X > 3] \)[/tex] is the complement of [tex]\( P[X \leq 3] \)[/tex]:
[tex]\[ P[X > 3] = 1 - P[X \leq 3] \approx 1 - 0.9810118431238463 = 0.01898815687615374 \][/tex]
Thus, the computed values are:
- (a) [tex]\( b \approx 0.36787944117144233 \)[/tex]
- (b) [tex]\( P[X=1] \approx 0.36787944117144233 \)[/tex]
- (c) [tex]\( P[X>3] \approx 0.01898815687615374 \)[/tex]
### (a) Finding the parameter [tex]\( b \)[/tex]
The probability mass function (PMF) of a random variable [tex]\( X \)[/tex] is given by:
[tex]\[ p_x(x) = b \frac{\lambda^x}{x!}, \quad x = 0,1,2,\ldots \][/tex]
To determine the parameter [tex]\( b \)[/tex], we need to use the fact that the sum of all probabilities for the PMF must be equal to 1:
[tex]\[ \sum_{x=0}^{\infty} p_x(x) = 1 \][/tex]
Substituting the given PMF into the sum, we have:
[tex]\[ \sum_{x=0}^{\infty} b \frac{\lambda^x}{x!} = 1 \][/tex]
This is a series recognized as the Maclaurin series expansion for the exponential function [tex]\( e^\lambda \)[/tex]:
[tex]\[ \sum_{x=0}^{\infty} \frac{\lambda^x}{x!} = e^\lambda \][/tex]
Thus, we can rewrite the equation as:
[tex]\[ b \cdot e^\lambda = 1 \][/tex]
Solving for [tex]\( b \)[/tex], we get:
[tex]\[ b = \frac{1}{e^\lambda} \][/tex]
Given a specific value [tex]\( \lambda = 1 \)[/tex] (since [tex]\(\lambda > 0\)[/tex]), the value of [tex]\( b \)[/tex] calculates to:
[tex]\[ b = \frac{1}{e} \approx 0.36787944117144233 \][/tex]
### (b) Calculating [tex]\( P[X=1] \)[/tex]
To find [tex]\( P[X=1] \)[/tex], we substitute [tex]\( x = 1 \)[/tex] into the PMF:
[tex]\[ P[X=1] = p_x(1) = b \frac{\lambda^1}{1!} \][/tex]
Using the previously calculated value of [tex]\( b \)[/tex] and [tex]\( \lambda = 1 \)[/tex]:
[tex]\[ P[X=1] = \left(\frac{1}{e}\right) \cdot \frac{1^1}{1!} \][/tex]
Which simplifies to:
[tex]\[ P[X=1] = \frac{1}{e} \approx 0.36787944117144233 \][/tex]
### (c) Calculating [tex]\( P[X>3] \)[/tex]
To find [tex]\( P[X>3] \)[/tex], we first need to calculate [tex]\( P[X \leq 3] \)[/tex]. This can be done by summing the probabilities of [tex]\( P[X=0] \)[/tex], [tex]\( P[X=1] \)[/tex], [tex]\( P[X=2] \)[/tex], and [tex]\( P[X=3] \)[/tex]:
[tex]\[ P[X \leq 3] = P[X=0] + P[X=1] + P[X=2] + P[X=3] \][/tex]
Given the PMF, these probabilities are:
[tex]\[ P[X=0] = b \frac{\lambda^0}{0!} = \left(\frac{1}{e}\right) \cdot \frac{1}{1} = \frac{1}{e} \][/tex]
[tex]\[ P[X=1] = b \frac{\lambda^1}{1!} = \left(\frac{1}{e}\right) \cdot \frac{1}{1} = \frac{1}{e} \][/tex]
[tex]\[ P[X=2] = b \frac{\lambda^2}{2!} = \left(\frac{1}{e}\right) \cdot \frac{1^2}{2} = \frac{1}{2e} \][/tex]
[tex]\[ P[X=3] = b \frac{\lambda^3}{3!} = \left(\frac{1}{e}\right) \cdot \frac{1^3}{6} = \frac{1}{6e} \][/tex]
Adding these probabilities together:
[tex]\[ P[X \leq 3] = \frac{1}{e} + \frac{1}{e} + \frac{1}{2e} + \frac{1}{6e} \][/tex]
This simplifies to:
[tex]\[ P[X \leq 3] = \frac{1}{e} \left(1 + 1 + \frac{1}{2} + \frac{1}{6} \right) = \frac{1}{e} \left(\frac{6 + 6 + 3 + 1}{6} \right) = \frac{1}{e} \left(\frac{16}{6} \right) = \frac{8}{3e} \][/tex]
Using [tex]\( e \approx 2.71828 \)[/tex]:
[tex]\[ P[X \leq 3] \approx \frac{8}{3 \cdot 2.71828} \approx 0.9810118431238463 \][/tex]
Finally, [tex]\( P[X > 3] \)[/tex] is the complement of [tex]\( P[X \leq 3] \)[/tex]:
[tex]\[ P[X > 3] = 1 - P[X \leq 3] \approx 1 - 0.9810118431238463 = 0.01898815687615374 \][/tex]
Thus, the computed values are:
- (a) [tex]\( b \approx 0.36787944117144233 \)[/tex]
- (b) [tex]\( P[X=1] \approx 0.36787944117144233 \)[/tex]
- (c) [tex]\( P[X>3] \approx 0.01898815687615374 \)[/tex]
