MHB What is the probability that one is hired and one is not?

[elimeli]

I cannot get past this question:

Assume that 12 people, including the husband and wife pair, apply for 5 sales positions. People are hired at random. What is the probability that one is hired and one is not?

The sample space is C(12, 5). I tried finding first the probability that one is hired, which I suppose it's C(11,4). However, I am not sure how I am supposed to go about with finding the probability for not hiring one. I assumed C(11, 5), but then it wouldn't work if I try to multiply it with C(11,4) to find the event of both situations happening at the same time.

 

[Jameson]

I cannot get past this question:

Assume that 12 people, including the husband and wife pair, apply for 5 sales positions. People are hired at random. What is the probability that one is hired and one is not?

The sample space is C(12, 5). I tried finding first the probability that one is hired, which I suppose it's C(11,4). However, I am not sure how I am supposed to go about with finding the probability for not hiring one. I assumed C(11, 5), but then it wouldn't work if I try to multiply it with C(11,4) to find the event of both situations happening at the same time.

Hi elimeli,

You are on the right track but let's start with a simpler question. What is the probability of 1 person being hired for one of the positions?

 

[elimeli]

Hi elimeli,

You are on the right track but let's start with a simpler question. What is the probability of 1 person being hired for one of the positions?

Wouldn't it be C(11,4)? Because that would mean that somebody already took 1 job and 11 people would remain for the 4 vacant spots...?

 

[Jameson]

Let's put it like this.

If only one of the two is hired then the number of ways this can happen is $$\binom{2}{1}$$. What about the other remaining positions?

Hint: There are 10 people remaining. How many jobs remaining?

 

[Jameson]

Let's put it like this.

If only one of the two is hired then the number of ways this can happen is $$\binom{2}{1}$$. What about the other remaining positions?

Hint: There are 10 people remaining. How many jobs remaining?

Since some time has passed I'll go ahead and post my solution to this problem.

The number of ways one of the two can be hired is $$\binom{2}{1}$$. For the remaining people and positions, the number of ways to allocate them is $$\binom{10}{4}$$.

As the OP correctly stated the sample space is $$\binom{12}{5}$$.

Putting this all together the probability is $$\frac{\binom{2}{1}\binom{10}{4}}{\binom{12}{5}}=\frac{35}{66}$$