Assume that when human resources

Question

Assume that when human resource managers are randomly​ selected, 43​% say job applicants should follow up within two weeks. If 6 human resource managers are randomly​ selected, find the probability that at least 2 of them say job applicants should follow up within two weeks.

Answer 1

Hi Amanda,

This is a binomial probability problem. There are a few criteria for that, but let’s keep it simple, Binomial means we have two potential outcomes, success and failure, which we can define however we deem appropriate. In this case, either the manager says applicants should follow up within two weeks or not, a binary outcome. Formula for binomial probability:

P=C(n,x)p^xq^1-x

Let’s break this down:

C implies combination—just means order of selection does not matter, not a variable

n=sample size

x=value you were given

p=probability of success

q=1-p=probability of failure

Now, before we can compute this, we need to compute C(n, x)

C(n,x)=n!/(x!(n-x)!

!=factorial=n*(n-1)*(n-2) I.e. 3*@*1, etc.

Here,

n=6

x=2

C(6,2)=6!/(2!*4!)

C(6,2)=15

Now, back to the original formula:

P=C(n,x)p^xq^n-x

C(n,x)=C(6,2)=15

x=2

n=6

p=0.43

q=1-p=0.57

Thus:

P(X=2)=15*0.43^2*0.57⁴

P(X=2)=0.293

But we’re not done. The question asked for the probability of at least two managers affirming the claim. We now have two options. We can compute the binomial probability for x=(3,4,5,6), but that’s a lot of work, Alternatively, we can do P(X=0) and P(X=1) and apply the complement rule, or as I call it, the 1 minus trick. This states that P(Success)=1-P(Failure). So, if we compute and add the probabilities P(x=0) and P(X=1), we can just subtract the sum from 1 to get the probability of at least two.

So:

P(X=1)=C(6,1)*0.43¹*0.57⁵

C(6,1)=6!/(5!*1!)=6

Note that 1!=1.

Now,

P(X=1)=6*0.43¹*0.57⁵

P(X=1)=0.155

Now, let’s compute P(X=0) I.e. no managers affirming claim:

P(X=0)=C(6,0)*0.43⁰*0.57⁶

C(6,0)=6!/(0!*6!)

C(6,0)=1

Note that 0!=1 by definition, and any number divided by itself is 1.

Thus:

P(X=0)=1*0.43⁰*0.57⁶

Note that any number raised to the 0 power is 1: n⁰=1. Thus:

P(X=0)=0.57⁶

P(X=0)=0.034

Now, we add this to P(X=1):

P(X<2)=0.034 + 0.155

P(X<2)=0.189

Now, we use the complement rule or 1 minus trick to get the probability we want:

P(X>=2)=1-0.189

P(X>=2)=0.811

I hope this helps.