Q2: A lunch counter has 16 stools in a single row. Currently, 5 stools are occup

Question

Q2: A lunch counter has 16 stools in a single row. Currently, 5 stools are occupied by customers. You notice that no two customers sit next to each other Assume that customers select stools at random (that is, all selections of 5 stools are equally likely to be occupied). (a) What is the probability that no two adjacent stools are occupied? (b) What is the probability that all 5 customers sit in adjacent stools (that is, there are no empty stools between any two customers)? HINT: A selection of 5 from 16 stools can be represented as an arrangement of 5 0's and 11 Es, such as EOEEOEEOEEOEDEEE or EEEOOEOEEDEEOEEE. In the second example, but not the first, there are two customers sitting next to each other. Then, an arrangement can be constructed in the following way. Make 'slots' on each side of the 11 E's: There are 12 slots (they should also be placed at the ends of the arangement). Then, the O's can be placed in the slots according to some relevant rule.

Explanation / Answer

You'll have 5 occupied seats and 11 empty seats which can be thought of as a string of 5 Os and 11 Es.

First let's count the total number of possible arrangements of 5 Os and 11 Es. To do that, you can just compute "16 choose 5" as the ways to pick which of the 5 seats are occupied. Note: This is the same as computing "16 choose 11" as the ways to pick which of the 11 seats are empty.
16C5 = (16 x 15 x 14 x 13 x 12) / (5 x 4 x 3 x 2 x 1)
= 4,368 ways

PART A:
We want to count the ways that we can have a valid seating arrangement with no one sitting together. Using the hint, imagine you had _ E_E_E_E_E_E_E_E_E_E_E_. There are 12 gaps between or beside each of these empty seats where you could put an occupied seat. But you can only put 1 occupied seat in these spots, not 2 or more. So we just take the 12 spaces between and choose the 5 places where we are going to put an occupied seat.
C(12,5) = (12 x 11 x 10 x 9 x 8) / (5 x 4 x 3 x 2 x 1)
= 792 ways

Thus the probability of this outcome is 792/4368 but that can be reduced to 33/182 (or about 18.13%)

PART B:
Again use the hint, but this time we have to put all 5 Os into a single space between the existing 11 empty seats. We have 12 gaps and we are picking 1.
12C1 = 12 ways

Thus the probability of this event is 12/4368 but that reduces to 1/364 (or about 0.27%)

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.