Question 1200520: Out of 1000 managers, 72% say they use social media to screen applications. Out of 5 applications, what is the probability that exactly 3 of the managers use social media to screen
What is the probability that non will use social media
What is the probability that all 5 will use social media
Answer by GingerAle(43)   (Show Source):
You can put this solution on YOUR website! **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.**
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