(Sorry not sure if this is basic or university level)
A diner has 16 two-seat booths.
26 guys walk in at random.
what is the probability that at least one booth is empty?
Is this right:
Total seating arrangements = P( 32, 26) = 3.655 * 10^32
Arrangements with 3 empty booths = 26! * P( 16, 3 ) = 1.355 * 10^30
Arrangements with 2 empty = P( 28, 26) * P( 16, 2 ) = 3.659 * 10^31
Arrangements with 1 empty = P( 30, 26) * 16 = 1.768 * 10^32
(seating arrangements with empty booths) / (total seating arrangements)
= ( 2.148 * 10^32 ) / ( 3.655 * 10^32 ) = 0.588
Originally Posted by billym
I think you have some problems with over-counting. Your arrangements with 1 empty booth have been included in the count of 2 empty booths, etc.
Are you familiar with the Principle of Inclusion / Exclusion, also known as the Sieve Principle? This might be a good place to apply it.
Okay so I was forgetting to subtract when then 2 empty seats circumstance yielded an event when the guys arranged themselves in a way that left another empty seat. And in the 1 empty seat circumstance there would be the events of both one extra empty seat and two extra empty seats that would need to be subtracted. (???)
Originally Posted by awkward
So: (16!/13!)*26! + (30!/4!) + (28!/2!)[(16!/14!) - 28)] = 4.472 * 10^31
(4.472 * 10^31) / (3.655 * 10^32) = 0.122
Does that look more reasonable? Am I doing anything right?
Can somebody please show me how to do this problem? I've tried it a billion ways and I keep getting a different answer. All it says is "focus on the empty seats."
Let’s reframe the problem. “How many ways can we put twenty-six identical balls into sixteen different boxes if the boxes can not hold more two balls?”
Originally Posted by billym
Let d stand for double occupancy, s stand for single occupancy and e stand for empty.
Now we have four cases;
To find the number of ways place the balls is the sum of those four cases.
For the second case: “how many ways can we arrange a string of 12 d’s, 2 s’s and 2 e’s?”
We notice that only one of those cases do we have no empty box.
My brain is so fried right now i only kind of understand. I get the chart bit. Is it a simple as the probability of having at least one empty box is 0.75?
What is the purpose of "For the second case: “how many ways can we arrange a string of 12 d’s, 2 s’s and 2 e’s?”
Could you actually write down the correct answer so I'll know when I've done it right?
The total number of ways to fill those boxes is .
The total number of ways to get the last row is
So the probability of having no empty booth is .
The to the question is
It will help if you go to this website Wolfram|Alpha
In the input window, type in this exact expression: expand (1+x+x^2)^16 (you can copy & paste)
Click the equals bar at the right-hand end of the input window.
In the expansion you will see the coefficient of is that is no accident.
Now ask it expand (x+x^2)^16.
See the coefficient of is that is also no accident.