We begin by analyzing the given problem involving independent Poisson random variables \(X\) and \(Y\) with means 4 and 2 respectively. We need to evaluate which of the given statements are true.
1. **Conditional Distribution**: The statement claims that the conditional distribution of \(X\) given \(X + Y = 3\) is a Binomial distribution. For Poisson random variables \(X\) and \(Y\), their joint distribution given \(X + Y = n\) follows a binomial distribution.
The distribution is \(X | (X + Y = n) \sim \text{Binomial}(n, \frac{\lambda_1}{\lambda_1 + \lambda_2})\), where \(\lambda_1\) and \(\lambda_2\) are the means of \(X\) and \(Y\).
Here, \(\lambda_1 = 4\) and \(\lambda_2 = 2\), so \(X| (X + Y = 3) \sim \text{Binomial}(3, \frac{4}{6}) = \text{Binomial}(3, \frac{2}{3})\).
Therefore, the statement that the conditional distribution is \(\text{Binomial}(3, \frac{1}{3})\) is false.
2. **Probability Calculation**: \(P(X \le 1 | X + Y = 3)\) should be computed to verify if it equals \(\frac{7}{27}\).
The pmf for a Binomial random variable is \(P(Z = k) = \binom{n}{k} p^k (1-p)^{n-k}\).
Here, \(P(X = 0 | X + Y = 3) = \binom{3}{0}(\frac{2}{3})^0(\frac{1}{3})^3 = \frac{1}{27}\)
\(P(X = 1 | X + Y = 3) = \binom{3}{1}(\frac{2}{3})^1(\frac{1}{3})^2 = \frac{6}{27} = \frac{2}{9}\)
Thus, \(P(X \le 1 | X+Y=3) = \frac{1}{27} + \frac{6}{27} = \frac{7}{27}\).
This matches the given statement, so it is true.
3. **Expected Value Calculation**: \(E(X | X+Y = 3)\) is calculated using the mean of the binomial distribution, which is \(n \cdot p\).
Here, \(E(X | X + Y = 3) = 3 \cdot \frac{2}{3} = 2\).
This matches the given statement, so it is true.
4. **Characteristic Function**: The characteristic function for a Poisson distribution with mean \(\lambda\) at a point \(t\) is given by \(e^{\lambda(e^{it} - 1)}\).
The sum of two independent Poisson random variables is another Poisson random variable with mean equal to the sum of their means, i.e., mean 6 here.
Thus, the characteristic function of \(X+Y\) at \(t = \pi\) is \(e^{6(e^{i\pi} - 1)} = e^{6(-2)} = e^{-12}\), which aligns with the statement.
Based on the analysis, the true statements are:
- \(P(X \le 1 | X+Y=3) = \frac{7}{27}\)
- \(E(X | X+Y = 3) = 2\)
- The value of the characteristic function of \(X + Y\) at the point \(t = \pi\) is \(e^{-12}\)
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 $P(A \cup B \cup C)$?
