The problem involves a Poisson process describing customer arrivals with intensity $\lambda = 2$ per unit time.
Analyzing Variables X and Y
Let $X$ be the number of customers in the time interval $(1,2)$. The length of this interval is $t_X = 2 - 1 = 1$. Since the arrival process is Poisson with rate $\lambda=2$, $X$ follows a Poisson distribution with parameter $\mu_X = \lambda t_X = 2 \times 1 = 2$. So, $X \sim \text{Poisson}(2)$.
Let $Y$ be the number of customers in the time interval $(5,10)$. The length of this interval is $t_Y = 10 - 5 = 5$. $Y$ follows a Poisson distribution with parameter $\mu_Y = \lambda t_Y = 2 \times 5 = 10$. So, $Y \sim \text{Poisson}(10)$.
Evaluating Option B: Independence
The time intervals $(1,2)$ and $(5,10)$ are disjoint. A key property of Poisson processes is that the number of events occurring in disjoint time intervals are independent.
- Therefore, $X$ and $Y$ are independent. Option B is true.
Evaluating Option C: Sum Distribution
The sum of two independent Poisson random variables is also a Poisson random variable, with a parameter equal to the sum of their individual parameters.
- $X+Y$ follows a Poisson distribution with parameter $\mu_{X+Y} = \mu_X + \mu_Y = 2 + 10 = 12$.
- So, $X+Y \sim \text{Poisson}(12)$.
- The statement claims $X+Y$ is Poisson with parameter $6$. This contradicts our finding. Option C is false.
Evaluating Option D: Difference Distribution
The difference between two independent Poisson random variables ($X-Y$) does not follow a Poisson distribution. Poisson distributions are defined for non-negative integers.
- Option D is false.
Evaluating Option A: Conditional Probability
We need to evaluate $P(X = 0 \mid X + Y = 12)$.
Consider two independent Poisson random variables $N_1 \sim \text{Poisson}(\mu_1)$ and $N_2 \sim \text{Poisson}(\mu_2)$. If we condition on the total sum $N_1 + N_2 = n$, the conditional distribution of $N_1$ is Binomial:
$P(N_1 = k \mid N_1 + N_2 = n) = \binom{n}{k} p^k (1-p)^{n-k}$where $p = \frac{\mu_1}{\mu_1 + \mu_2}$. This represents a binomial distribution $B(n, p)$.
In our case, $N_1 = X \sim \text{Poisson}(2)$ ($\mu_1 = 2$) and $N_2 = Y \sim \text{Poisson}(10)$ ($\mu_2 = 10$). The sum is $n = 12$. The probability $p$ is:
$p = \frac{\mu_X}{\mu_X + \mu_Y} = \frac{2}{2 + 10} = \frac{2}{12} = \frac{1}{6}$We are interested in $P(X = 0 \mid X + Y = 12)$. This corresponds to the binomial probability with $k=0$, $n=12$, and $p=1/6$.
$P(X = 0 \mid X + Y = 12) = \binom{12}{0} \left(\frac{1}{6}\right)^0 \left(1 - \frac{1}{6}\right)^{12-0}$ $P(X = 0 \mid X + Y = 12) = 1 \times 1 \times \left(\frac{5}{6}\right)^{12} = \left(\frac{5}{6}\right)^{12}$- Therefore, Option A is true.
Conclusion
The true statements are A and B.
Important Questions from Discrete Probability
- Suppose two fair dice are thrown independently at random. Let $X$ and $Y$ be the numbers on the upper face of the first die and that of the second die, respectively. Then which of the following statements are true?
- A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability that the last ball drawn is black equals
Ten balls are put in 6 slots at random. Then the expected total number of balls in the two extreme slots is
- There are five empty boxes. Balls are placed independently one after another in randomly selected boxes. The probability that the fourth ball is the first to be placed in an occupied box equals
Suppose $A, B, C$ are events in a common probability space with
$P(A) = 0.2, \ P(B) = 0.2, \ P(C) = 0.3$, $P(A \cap B) = 0.1, \ P(A \cap C) = 0.1, \ P(B \cap C) = 0.1$.
Which of the following are possible values of $P(A \cup B \cup C)$?
