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B
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Difficulty:
95%
(hard)
Question Stats:
36% (02:30) correct
64%
(03:15)
wrong
based on 25
sessions
History
Date
Time
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Ramon will have to visit customers in Bayville on five days next month, but on no day after the 8th. All possible groups of five days are equally likely. A hotel in Bayville offers a room upgrade for stays of four nights or more, but Ramon will stay any given night at the hotel if and only if he has to visit a customer in Bayville the next morning. What is the probability that the hotel will offer Ramon a room upgrade next month?
(A) 3/14
(B) 2/7
(C) 3/8
(D) 5/14
(E) 3/7
(A) 3/14
(B) 2/7
(C) 3/8
(D) 5/14
(E) 3/7
25 Aug 2026, 00:07
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When will Ramon get the upgrade?
Case 1: If he stays for 4 consecutive nights, or
Case 2: if he stays for 5 consecutive nights
Total # of outcomes:
From 8 days, how many sets of 5 days are possible? 8C5 = 56.
Favorable # of outcomes:
Case 2: if he stays for 5 consecutive nights
Consider the 5 consecutive days as 1 block (X). Apart from X, we have 3 days among the 8 days (say O, O, O).
Number of ways Ramon could keep 5 of the 8 days consecutive --- is essentially --- all possible arrangements of X, O, O, O = 4!/3! = 4
We can visualize these easily ->
XOOO
OXOO
OOXO
OOOX
Where X is a block of 5 consecutive days.
Therefore,
# of favorable outcomes in Case 2 = 4
Case 1: If he stays for 4 consecutive nights
Consider the 4 consecutive days as 1 block (X). Consider the other day Ramon will have to make a visit as Y. Consider the other days as O,O,O.
We need to find the number of arrangements of X,Y,O,O,O such that X and Y are never together (Because if X and Y are together, that would make it 5 consecutive days, not 4).
- # of arrangements of X, Y, O, O, O = 5!/3! = 20
- # of arrangements of X, Y, O, O, O where X and Y are together = 4!/3! x 2! = 8
Logic: consider XY as 1 block called P.
P, O, O, O can be arranged in 4!/3! ways. But we cannot forget that within the block P, we could have XY or YX i.e., 2! arrangements.
Hence, 4!/3! x 2! = 8
Therefore - number of arrangements of X,Y,O,O,O such that X and Y are never together = 20 - 8 = 12
# of favorable outcomes in Case 1 = 12
# of favorable outcomes = 4 + 12 = 16
Hence,
Probability that he will be offered the upgrade:
16/56 = 2/7. Choice B.
---
Harsha
Case 1: If he stays for 4 consecutive nights, or
Case 2: if he stays for 5 consecutive nights
Total # of outcomes:
From 8 days, how many sets of 5 days are possible? 8C5 = 56.
Favorable # of outcomes:
Case 2: if he stays for 5 consecutive nights
Consider the 5 consecutive days as 1 block (X). Apart from X, we have 3 days among the 8 days (say O, O, O).
Number of ways Ramon could keep 5 of the 8 days consecutive --- is essentially --- all possible arrangements of X, O, O, O = 4!/3! = 4
We can visualize these easily ->
XOOO
OXOO
OOXO
OOOX
Where X is a block of 5 consecutive days.
Therefore,
# of favorable outcomes in Case 2 = 4
Case 1: If he stays for 4 consecutive nights
Consider the 4 consecutive days as 1 block (X). Consider the other day Ramon will have to make a visit as Y. Consider the other days as O,O,O.
We need to find the number of arrangements of X,Y,O,O,O such that X and Y are never together (Because if X and Y are together, that would make it 5 consecutive days, not 4).
- # of arrangements of X, Y, O, O, O = 5!/3! = 20
- # of arrangements of X, Y, O, O, O where X and Y are together = 4!/3! x 2! = 8
Logic: consider XY as 1 block called P.
P, O, O, O can be arranged in 4!/3! ways. But we cannot forget that within the block P, we could have XY or YX i.e., 2! arrangements.
Hence, 4!/3! x 2! = 8
Therefore - number of arrangements of X,Y,O,O,O such that X and Y are never together = 20 - 8 = 12
# of favorable outcomes in Case 1 = 12
# of favorable outcomes = 4 + 12 = 16
Hence,
Probability that he will be offered the upgrade:
16/56 = 2/7. Choice B.
---
Harsha
Moderator:
