### Brain Teasers

# Hotel Ade

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

Mad Ade has just inherited a hotel from his rich Auntie Climax.

The rooms are numbered from 101 to 550.

A room is chosen at random for Mad Ade's first guest.

What is the probability that room number starts with 1, 2 or 3 and ends with 4, 5 or 6?

The rooms are numbered from 101 to 550.

A room is chosen at random for Mad Ade's first guest.

What is the probability that room number starts with 1, 2 or 3 and ends with 4, 5 or 6?

### Answer

There are total 450 rooms.Out of which 299 room numbers start with either 1, 2 or 3.

Now out of those 299 rooms only 90 room numbers end with 4, 5 or 6.

So the probability is 90/450 i.e. 1/5 or 0.20

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## Comments

There are only 299 rooms beginning with 1, 2 or 3; there is no room #100.

Does this hotel happen to serve kababs for dinner?

299/450 * 3/10 = 897/4500 = 299/1500.

not quite 20% chance

not quite 20% chance

Egeon: the fact that he miscounted the number of rooms that met the first criterion, while a mistake, does not effect the answer because the improperly listed room, #100, does not meet the second criterion and thus is not one of the 90 rooms that meets both.

i think your answer is wrong. it is not 300 rooms which are started with 1,2, or 3. It is 299!

here the solution:

101-200 -> 200-101+1= 100 rooms

201-300 -> 300-201+1= 100 rooms

301-399 -> 399-301+1= 99 rooms

so there are 100+100+99= 299 rooms not 300 rooms (you count room number 400)

you answer still correct, because finding rooms which is started with 1,2,3 is no use..

here the solution:

101-200 -> 200-101+1= 100 rooms

201-300 -> 300-201+1= 100 rooms

301-399 -> 399-301+1= 99 rooms

so there are 100+100+99= 299 rooms not 300 rooms (you count room number 400)

you answer still correct, because finding rooms which is started with 1,2,3 is no use..

I changed 300 to 299 as others noticed above.

Nice one

...although if I wanted to be picky, I'd say that hotel rooms aren't numbered linearly like that, usually the first number is the floor, and the other numbers are the room on that floor (in this case that would be rooms 101-150, 201-250, etc). But I don't want to be picky, so I won't say that

...although if I wanted to be picky, I'd say that hotel rooms aren't numbered linearly like that, usually the first number is the floor, and the other numbers are the room on that floor (in this case that would be rooms 101-150, 201-250, etc). But I don't want to be picky, so I won't say that

JYMZY:

The first probability is indeed 299/450 (being assigned to a room that begins with 1, 2, or 3). However, your second probability -- 3/10 -- is incorrect. Let's break it down further.

P(starting with 1) = 99/450

P(starting with 2) = P(starting with 3) = 100/450

The next probability changes, from what you list, however.

P(ending with 4 or 5 or 6|started with 1) = 30/99

P(ending with 4 or 5 or 6|started with 2) = P(ending with 4 or 5 or 6|started with 3) = 30/100

Now, (99/450)*(30/99) = 1/15. Similarly, (100/450)*(30/100) = 1/15. And, 3*(1/15) = .2.

The first probability is indeed 299/450 (being assigned to a room that begins with 1, 2, or 3). However, your second probability -- 3/10 -- is incorrect. Let's break it down further.

P(starting with 1) = 99/450

P(starting with 2) = P(starting with 3) = 100/450

The next probability changes, from what you list, however.

P(ending with 4 or 5 or 6|started with 1) = 30/99

P(ending with 4 or 5 or 6|started with 2) = P(ending with 4 or 5 or 6|started with 3) = 30/100

Now, (99/450)*(30/99) = 1/15. Similarly, (100/450)*(30/100) = 1/15. And, 3*(1/15) = .2.

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