Question 1200520
**1. Define the Probability of Success**

* **Probability of Success (p):** 
    * The probability that a randomly selected manager uses social media to screen applications.
    * p = 72% = 0.72

**2. Define the Probability of Failure (q)**

* **Probability of Failure (q):** 
    * The probability that a randomly selected manager does *not* use social media to screen applications.
    * q = 1 - p = 1 - 0.72 = 0.28

**3. Use the Binomial Probability Formula**

* The probability of exactly k successes in n independent trials, where the probability of success in each trial is p, is given by the binomial probability formula:

   P(X = k) = (nCk) * p^k * q^(n-k) 

   where:
      * nCk = n! / (k! * (n-k)!) is the binomial coefficient 

**a) Probability that exactly 3 of the 5 managers use social media:**

* n = 5 (number of trials)
* k = 3 (number of successes)
* p = 0.72 
* q = 0.28

* P(X = 3) = (5C3) * (0.72)^3 * (0.28)^(5-3) 
           = (5! / (3! * 2!)) * (0.72)^3 * (0.28)²
           = 10 * 0.373248 * 0.0784
           = 0.2916

**Therefore, the probability that exactly 3 of the 5 managers use social media to screen applications is approximately 0.2916.**

**b) Probability that none of the 5 managers use social media:**

* n = 5
* k = 0 (no successes)

* P(X = 0) = (5C0) * (0.72)^0 * (0.28)^(5-0) 
           = 1 * 1 * (0.28)^5
           = 0.0014

**Therefore, the probability that none of the 5 managers use social media to screen applications is approximately 0.0014.**

**c) Probability that all 5 managers use social media:**

* n = 5
* k = 5 (all successes)

* P(X = 5) = (5C5) * (0.72)^5 * (0.28)^(5-5) 
           = 1 * (0.72)^5 * 1
           = 0.1935

**Therefore, the probability that all 5 managers use social media to screen applications is approximately 0.1935.**