Scheduling Probability

[jerodfrowick]

Hi there, I have a real life example that I need some help with. It has to do with my schedule at work and I'm not quite sure how to word it so I'll just spell it out. One week, (Monday thru Sunday) I was supposed to be allowed to write my own schedule and I chose: Monday, Thursday, and Sunday.
The official schedule came out and I was assigned: Tuesday, Wednesday, Friday, and Saturday...the exact opposite. My question is, what is the probability of RANDOMLY receiving the exact opposite schedule from what I requested. I'm guessing around one in 20,000 or so but I have no idea how to actually figure this out. PLEASE HELP as I feel very slapped in the face and would love to have the actual probability at my disposal when I talk to the "higher-up's" in the company about this situation. Thanks a million

 

[galactus]

It's not nearly as rare as 1 in 20,000. There are only 7 days in the week.

You chose 3 days that were not random. In other words, you did not choose them at random but were chosen due to convenience, etc.

Now, let's say they chose 4 days out of 7 for you at random without regard to your choices and chose none of yours.

They would be choosing 4 from 7. This is called a combination.

Think of this as kind of like a small lottery. The probability of matching your days instead of matching numbered balls.

There are 35 ways to choose 4 things from 7. Written mathematically, it is \(\displaystyle \binom{7}{4}=\frac{7!}{4!(7-4)!}=\frac{7\cdot \not{6}\cdot 5\cdot \not{4}\cdot \not{3}\cdot \not{2}}{(\not{4}\cdot \not{3}\cdot \not{2})(\not{3}\cdot \not{2})}=35\)

So, the probability is 1/35.

Also, look at it this way as well. The first choice they choose 4 days, other than yours, from 7. That would be 4/7.

A day is not replaced, so the next day would be with probability 3/6. Then, 2/5. Then, 1/4.

\(\displaystyle \frac{4}{7}\cdot \frac{3}{6}\cdot \frac{2}{5}\cdot \frac{1}{4}=\frac{1}{35}\)

Actually, the probability they matched your days would be the same if chosen at random. \(\displaystyle \binom{7}{3}=35\).