Probability problem....

[MBH_14]

Jennifer, Dan, Larry, Linda, Mike, Angelique, and Vince go to a football game together and randomly sit in 7 adjacent chairs. What is the probability that Vince does not sit in the middle chair?

I got 6/7 as the answer. Is it that easy, or is there something wrong that i did? Thanks for any help!

 

[soroban]

Hello, MBH_14!

Your answer is correct . . . good work!


The probability that Vince does sit in the middle chair is \(\displaystyle \frac{1}{7}\).

Therefore, the probability that he does not is: \(\displaystyle \,\frac{6}{7}\).


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We can, of course, do it "the long way".


There are: \(\displaystyle \,7!\,=\,5040\) possible seating arrangement.

If Vince does not take the middle chair, he has \(\displaystyle 6\) choices.
For each of these, the other six can be seated in \(\displaystyle 6!\,=\,720\) ways.
Hence, there are: \(\displaystyle \,6\,\times\,720\:=\:4320\) ways.

Therefore: \(\displaystyle \,P({\text{Vince not in middle})\:=\:\frac{4320}{5040}\:=\:\frac{6}{7}\)