Which is Matthew's Favourite Restaurant ?

[Tomm]

Restaurant D.
Divide the number of times he went to each restaurant by the number of days worked at the nearby office per week to get a rate. D is the highest rate.

You've got it, Open heart. Congrats!

 

[Tigger45]

You've got it, Open heart. Congrats!

 

[Tomm]

...

 

[Meowzltov]

Since you haven't divided up the restaurants by office this time... the answer to question #1 is A. He went there 39 times.

The answer to question #2 is an hour since the only clue is "lunch HOUR".

All other information is irrelevant.

 

[brinny]

You've got it, Open heart. Congrats!

Congrats Open heart!!!

 

[Tomm]

Since you haven't divided up the restaurants by office this time... the answer to question #1 is A. He went there 39 times.

The answer to question #2 is an hour since the only clue is "lunch HOUR".

All other information is irrelevant.

Welcome again, Open Heart.
Sorry this time the puzzle is still based on the same map as last time. No change to the map.

 

[Meowzltov]

I no longer have the original maps, but I suspect that given the wide difference between A and the other restaurants, that A is still the right answer.

 

[Tomm]

I no longer have the original maps, but I suspect that given the wide difference between A and the other restaurants, that A is still the right answer.

The original map can be found on page 1 of this thread, see the first message.

 

[Hank77]

Recently, Matthew got 2 new part-time jobs.
The jobs are roster-based. He works at
each job in alternate weeks. In office 1, he works
in week #1, week #3, week #5, and so forth.
In office 2, he works in week #2, week #4, week #6,
and so forth.

In office 1, he works 3 days per week. In office 2, he
works only 2 days per week.

During lunch time, he always eats at a nearby restaurant.

During the past 8 weeks, he ate at restaurant A 17 times,
restaurant B 7 times, restaurant C 0 times, restaurant D
16 times and restaurant E 0 times. Below is a map of his
workplaces and nearby restaurants.

Now, the question is so far which is Matthew's favourite
restaurant ?

View attachment 164163

I have a hard time with statics. When I read this problem to begin with I see that Matthew worked 20 days in 8wks. and ate 40 lunches. Would I be correct in saying that, that does not matter or am reading something incorrectly?

 

[Tomm]

I have a hard time with statics. When I read this problem to begin with I see that Matthew worked 20 days in 8wks. and ate 40 lunches. Would I be correct in saying that, that does not matter or am reading something incorrectly?

No, he worked 55 days during the 11 weeks, and he ate 55 lunches at restaurants.

 

[Tomm]

....

 

[Tomm]

...

 

[Meowzltov]

1/2
It doesn't really matter what one child is. The probability of any child is the same.
Actually it is a slight bit more than 1/2. The ratio of conceived children is 1:2, but more male children die in utero.

 

[Tomm]

1/2
It doesn't really matter what one child is.

1/2......hmmmm.......

 

[Tomm]

1/2
It doesn't really matter what one child is. The probability of any child is the same.
Actually it is a slight bit more than 1/2. The ratio of conceived children is 1:2, but more male children die in utero.

Hmmmm.... 1/2 is wrong !
It does matter what one child is. This is conditional probability.
The answer is 1/3.

 

[Meowzltov]

Hmmmm.... 1/2 is wrong !
It does matter what one child is. This is conditional probability.
The answer is 1/3.

If I flip a coin, the odds of it coming up heads are 1/2. EVERY TIME. If it came up heads 5 times in a row before and I flip it again, the odds are still 1/2.

So in the same respect, it doesn't matter the sex of the previous child. The odds of conceiving either sex in ANY and EVERY pregnancy are always 1/2.

No worries! Your mistake is extremely common. I used to make it too. I dated a casino owner for a short period of time, and he told me I was making the Gamblers error. It took me years of mulling it over to finally understand WHY he was right.

 

[Meowzltov]

ooops

 

[Tomm]

Yes, I understand the Gambler's fallacy, I studied statistics (and still do).

Actually, there might be a miscommunication. The original
puzzle is not what you have in mind, it goes like this:
There are 2 children, one of them is a boy, what is the probability
that the other is also a boy ?

This is different from what you have in mind, which is:
The first child is a boy, what's the probability that the second is also a
boy ?

They are different. The first one is conditional probability since the probability
depends on a condition. The answer is 1/3.

The answer for the second case is 1/2.

 

[Meowzltov]

Yes, I understand the Gambler's fallacy, I studied statistics (and still do).

Actually, there might be a miscommunication. The original
puzzle is not what you have in mind, it goes like this:
There are 2 children, one of them is a boy, what is the probability
that the other is also a boy ?

This is different from what you have in mind, which is:
The first child is a boy, what's the probability that the second is also a
boy ?

They are different. The first one is conditional probability since the probability
depends on a condition. The answer is 1/3.

The answer for the second case is 1/2.

I don't see it. I admit I've never taken statistics, but I do have a good mind. There are only two possibilities for the second child: a boy or a girl, thus the chances of it being a boy is 1/2.

 

[Tomm]

Please read the puzzle again before reading on. The two puzzles are different in their wordings.

In this puzzle, in its sample space, there are 4 possible combinations:

1) Boy, Boy
2) Boy, Girl
3) Girl, Girl
4) Girl, Boy

Therefore, the answer is 1/3, not 1/2. It's a typical conditional probability problem.
=========================================================
In the other puzzle, the possible combinations are:
1) Boy, Boy
2) Boy, Girl

Therefore, the answer is 1/2.