Question 1174702
Let's break down each part of this problem.

**(a) Automatic Transmissions**

* **Binomial Distribution:** This situation follows a binomial distribution because:
    * There are a fixed number of trials (n = 5 buyers).
    * Each trial has only two outcomes (automatic or not).
    * The probability of success (automatic) is constant (p = 0.6).
    * The trials are independent.

* **Mean (μ):**
    * μ = np = 5 * 0.6 = 3

* **Standard Deviation (σ):**
    * σ = √(np(1-p)) = √(5 * 0.6 * 0.4) = √1.2 ≈ 1.095

* **Complete Distribution:**
    * P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
    * Where (n choose k) = n! / (k! * (n-k)!)

    * P(X = 0) = (5 choose 0) * (0.6)^0 * (0.4)^5 = 1 * 1 * 0.01024 = 0.01024
    * P(X = 1) = (5 choose 1) * (0.6)^1 * (0.4)^4 = 5 * 0.6 * 0.0256 = 0.0768
    * P(X = 2) = (5 choose 2) * (0.6)^2 * (0.4)^3 = 10 * 0.36 * 0.064 = 0.2304
    * P(X = 3) = (5 choose 3) * (0.6)^3 * (0.4)^2 = 10 * 0.216 * 0.16 = 0.3456
    * P(X = 4) = (5 choose 4) * (0.6)^4 * (0.4)^1 = 5 * 0.1296 * 0.4 = 0.2592
    * P(X = 5) = (5 choose 5) * (0.6)^5 * (0.4)^0 = 1 * 0.07776 * 1 = 0.07776

**(b) Poker Hand with 2 Kings**

* **Total Possible Hands:** (52 choose 5) = 2,598,960
* **Ways to Choose 2 Kings:** (4 choose 2) = 6
* **Ways to Choose 3 Other Cards (not Kings):** (48 choose 3) = 17,296
* **Probability:** (6 * 17,296) / 2,598,960 ≈ 0.0399

**(c) Poisson Random Variable**

* **Poisson Distribution:** X ~ Poisson(λ), where λ = 1.5.
* **Formula:** P(X = k) = (e^(-λ) * λ^k) / k!

* **P(X = 1):**
    * P(X = 1) = (e^(-1.5) * 1.5^1) / 1!
    * P(X = 1) ≈ (0.2231 * 1.5) / 1 ≈ 0.3347

* **P(X = 7):**
    * P(X = 7) = (e^(-1.5) * 1.5^7) / 7!
    * P(X=7) = (0.2231 * 17.0859)/5040
    * P(X=7) = 3.811/5040
    * P(X = 7) ≈ 0.000756