(a) Two cards are drawn at random and one by one with replacement from a well-shuffled pack of 52 playing cards. Find the probability distribution of the number of aces. Also, find its mean and variance. OR (b) It is given that 2% of the screws manufactured by a company are defective. Use Poisson distribution to find the probability that a packet of 100 screws contains (i) no defective screw, (ii) one defective screw.

Step 1: Identify parameters for Poisson distribution

We are given that the number of screws in a packet is $n = 100$ and the probability of a screw being defective is $p = 2\% = 0.02$. For a Poisson distribution, we need to calculate the mean $\lambda$.

Step 2: Calculate the mean (λ)

The mean $\lambda$ for a Poisson distribution, when approximating a binomial distribution, is given by the product of the number of trials $n$ and the probability of success $p$. In this case, $\lambda = 100 \times 0.02 = 2$.

Step 3: State the Poisson probability formula

The probability mass function for a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $k$ is the number of occurrences, $\lambda$ is the average rate of occurrence, and $e$ is Euler's number (approximately 2.71828).

Step 4: Calculate probability of no defective screws ($k=0$)

To find the probability of no defective screws, we substitute $k=0$ and $\lambda=2$ into the Poisson probability formula. Since $2^0 = 1$ and $0! = 1$, the probability simplifies to $e^{-2}$, which is approximately $0.1353$.

Step 5: Calculate probability of one defective screw ($k=1$)

To find the probability of one defective screw, we substitute $k=1$ and $\lambda=2$ into the Poisson probability formula. This gives us $2e^{-2}$, which is approximately $2 \times 0.1353 = 0.2706$.

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