not sure if binomial? (1 Viewer)

[Dumbarse]

four rats are in a maze, and there a five exits. each rat is equally likely to exit any maze, so the probability of a rat exiting by a certain exit is 1/5.

1) what is the probabilty of all four rats leaving the maze by the same exit?

2)what is the probability of at least 2 rats leaving by the same exit?

 

[BlackJack]

Assuming you're talking about a certain exit, and not about any exit:
1) should be (1/5 )^4, as all rats go for this one out of five
2) 1 - (4/5)^4 - (4/5)^3 * 1/5, the conjugate of one & no rats leaving by this exit

If you're talking about any exit:

1) multiply (1/5 )^4 by 5 to cater fo all five exits...
re-edit: this is 1/125...

2) Still use the conjugate, but since there mustn't be two rats using the same exit:
1- 5/5 * 4/5 * 3/5 *2/5
edit: (first rats get five, second can only choose four out of five exits, etcetera...)
re-edit:

 

[Lazarus]

Question 1
P(four) = 1 * 1/5 * 1/5 * 1/5
= 1/125

interpretation: It doesn't matter which exit the first rat leaves by. Each subsequent rat then has a one in five chance of leaving by that same exit.

Question 2
P(at least two) = 1 - P(none or one)
= 1 - [P(none) + P(one)]
= 1 - [(1 * 4/5 * 3/5 * 2/5) + (1 * 1/5 * 3/5 * 2/5)]
= 1 - [24/125 + 6/125]
= 1 - 30/125
= 19/25
= 0.76

I think that's right.


Edit: Oh hmm BlackJack posted before me, now I'm going to have to double-check this. Damn.

 

[Dumbarse]

wrong !!

 

[Dumbarse]

wrong !! :mad1:

 

[Dumbarse]

no wait laz your right, the fact that its only 1/5^3 for the first one because it is dependant on whatever exit the first one goes to

good stuff

 

[Dumbarse]

for the second one the answer is

1 - ( 1/5^3 + 4c3*1/5^2 *4/5^3)

 

[Dumbarse]

hold up

ahh i did it this arvo and i forgot now
its something like that

 

[Lazarus]

Hmmmm... are you sure my answer for the second question isn't correct?

With probability, you shouldn't need to enumerate the combinations...

 

[Dumbarse]

ohh wait, yeh mines wrong,
i did when it was AT MOST 2 go out the same exit
which is
1- (1/5*3 + 4c3 * 1/5^3 * 4/5^2)

yours is right

 

[BlackJack]

I'm still cloudy about the second question... explain please...
P( 1+ rats in the same exit as first) = 1- P(no rats in same exit) ?

 

[kaseita]

you sure that's right laz?
cause you've done
p(at least 2 )= 1 - [p(1 rat each exit) + p(2 rat one exit)]
cause p[none] either means no rat goes through that exit, which doesn't make sense, or means that one rat goes in each exit.
so p[one] would be that one two rats go through the same exit?

 

[spice girl]

Idunno whether anyone's got an answer here yet, here's mine:

P( >=2 rats same exit) = 1 - P(rats go different exits)

= 1 - (5*4*3*2) / 5^4

= 1 - 24/125

= 101/125

 

[Simmo]

hey didnt he say the prob for A rat to exit out of a CERTAIN exit is 1/5 then allothers 1/5 after that...so it should be 1/625

 

[Lazarus]

Originally posted by Simmo
hey didnt he say the prob for A rat to exit out of a CERTAIN exit is 1/5 then allothers 1/5 after that...so it should be 1/625

The probability of one rat leaving by a certain exit is 1/5, but the probability of four rats leaving by the same exit (irrespective of which exit it is) is 1/5³.

Originally posted by kaseita
cause you've done
p(at least 2 )= 1 - [p(1 rat each exit) + p(2 rat one exit)]
cause p[none] either means no rat goes through that exit, which doesn't make sense, or means that one rat goes in each exit.
so p[one] would be that one two rats go through the same exit?

Yes. Let me rewrite my first line a bit more clearly:

P(at least 2 rats leaving by the same exit) = 1 - [P(no rats leaving by the same exit) + P(one rat leaving by the same exit)]

where:
no rats leaving by the same exit = all rats leaving through different exits
one rat leaving by the same exit = two rats going through the same exit

You have to be careful how you write it, or you'll confuse the inversion. Most students know P(at least one) = 1 - P(none), but it's often blindly memorised without an actual understanding of why this is so.

P(a group of events occurring) = 1 - P(all other events occurring)
P(all events except the '0' event occurring) = 1 - P(the '0' event occurring)
P(at least one) = 1 - P(none)

P(all events except the '0' and '1' event occurring) = 1 - P(the '0' event OR the '1' event occurring)
P(at least two) = 1 - [P(none) + P(one)]

Does that make sense?

This question is probably easier if you rephrase it.

"What is the probability of at least 2 rats leaving by the same exit?"
We know that the exit through which the first rat leaves doesn't matter.
So it's really asking,
"What is the probability of at least 1 rat leaving by the exit the first rat went through?"
Which is also the same as
"What is the probability of at least 1 rat leaving by a certain exit?"

P(at least one) = 1 - P(none)
P(at least one) = 1 - [5/5 * 4/5 * 3/5 * 2/5]
P(at least one) = 101/125

And spice girl is correct. I just proved myself wrong. Yeah, well, annoying when you do that. The reason I was wrong is that P(one) is really two rats leaving by the same exit and so isn't one of the complementary events - my answer would have given the probability of at least three rats leaving by the same exit. Oh yeah and BlackJack solved it too; well done.

 

[BlackJack]

Originally posted by spice girl
Idunno whether anyone's got an answer here yet, here's mine:

P( >=2 rats same exit) = 1 - P(rats go different exits)

= 1 - (5*4*3*2) / 5^4

= 1 - 24/125

= 101/125

Phew... so I did get it right... I thought I needed to revise the whole binomial stuff again...
I did all the basics of probability from a book on logic puzzles, like, 8 or 9 years ago, so I did something else while the teacher was going on about the topic...

 

[Dumbarse]

yeh true thats right
good stuff