Lift persons problem

[pratikhalder]

A lift starts from the ground floor with 7 passengers and stops at 10 floors. Find the probability that exactly three of them leave at 2nd floor.

 

[lev888]

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[Steven G]

I guess that you did not read the posting guidelines. This is a free math help forum not a free homework service forum.
You need to show us the work you have done so far so we know what type of help you need. As far as a helper solving this problem for you that will not happen. We want to help you which means we want you to solve your own problem with hints from forum helpers.

Hint: Think of there being two types of floors- 2nd floor and others floors. See what you can do with this hint.

 

[Deleted member 4993]

A lift starts from the ground floor with 7 passengers and stops at 10 floors. Find the probability that exactly three of them leave at 2nd floor.

Please post - preferably a photocopy - of the EXACT problem (along with your work).

As posted, it is not clear that the lift even stops at the second floor!

 

[pratikhalder]

what I have done so far -
1 out of 10 floors can be selected in 10c1ways = 10 ways
3 out of 7 persons can be selected in 7c3 ways = 35 ways
Favorable outcomes =350
total outcomes = 10^7
Required probability = 35/10^6

 

[Steven G]

As posted, it is not clear that the lift even stops at the second floor!

Why don't I have the ability to see such things! You'd think after being on this forum fo 6 years I would have improved in this area. Yuck!!

 

[pratikhalder]

The problem and my work as requested.

 

[Steven G]

what I have done so far -
1 out of 10 floors can be selected in 10c1ways = 10 ways
3 out of 7 persons can be selected in 7c3 ways = 35 ways
Favorable outcomes =350
total outcomes = 10^7
Required probability = 35/10^6

Favorable outcomes = 350. Can you list them?
I am bit confused. There are 7 people on a lift and you want to count the number of ways that 3 of them can get off on the 2nd floor. What does the answer to that question have to do with the number of floors. If there were 20 floors are you saying that changes the number of ways you can choose 3 people out of 7 to exit the lift on the 2nd floor. Sorry, but I do not buy this!

 

[pratikhalder]

Yes, I am getting confused. That is why I asked for help/solution. I have solved another part of the question and I think that part is correct (please find attached document in my previous post).

 

[pka]

The problem and my work as requested.

Thank you for posting the original. The question clearly says the lift stops at ten floors.
Now the is a pure occupancy problem. That is, how many get off not who.
This like counting the ways to place seven identical marbles into ten different slots.
The answer total possible is $\dbinom{16}{7}=11440$ ways those seven people can exit the lift.
For part (i) there are $\dbinom{10}{7}=120$ ways those seven can exist on different floors.
For (iii) we need to know how many ways that four can exist on the other nine floors. What is that?

 

[Steven G]

Is my hint in post #3 valid?

 

[Deleted member 4993]

Now the is a pure occupancy problem. That is, how many get off not who.
This like counting the ways to place seven identical marbles into ten different slots.
The answer total possible is $\dbinom{16}{7}=11440$ ways those seven people can exit the lift.
For part (i) there are $\dbinom{10}{7}=120$ ways those seven can exist on different floors.
For (iii) we need to know how many ways that four can exist on the other nine floors. What is that?

The question clearly says the lift stops at ten floors.

The lift could be operating in an office-building with 15 floors - and stops at 10 floors without stopping at 2nd. floor.

I was trying to point out the need for "explicit" assumption/s.