Question

Quick probability question, just want to check my approach, please don't post the answer.

Answer 1

Supose you have 5 bins and 5 balls. Each of the balls has probability of 1/5 of landing in the bin when thrown. So, suppose you try to throw all five balls into the bins. What is the probability that at least one of the bins is empty?
So, what I'm doing is:
P(at least 1 empty) = 1 - P(none empty)
P(none empty, meaning a ball in each of the bins) = 1/5*5/5 * 1/5*4/5 * 1/5*3/5 * 1/5*2/5 * 1/5*1/5

Answer 2

Does this look right?

Answer 3

I got 5/5*4/5*3/5*2/5*1/5

Answer 4

well 1 minus that

Answer 5

Guys, but there's a probability of 1/5 of a ball even landing in the bin

Answer 6

P(none empty)=P(all full)

Answer 7

oh, they aren't ordered lol

Answer 8

Interpreting the problem is a pain in the bum... hang on ill attach a screenshot of the actual problem

Answer 9

I think I"m right about it though, 5/5 chances to get into a empty, then 4/5 chances to get into an empty, then 3/5 and so on and so forth

Answer 10

so it's 1-5!/5^5

Answer 11

I mean disregard the numbers, I just changed them around a bit, in order to not be accused of cheating on my hw - but yea it's the question in the screenshot.

Answer 12

ya, but same logic... try PA, the mod there will help

Answer 13

PK halp, I need a third opinion.

Answer 14

That "1/4 probability" sentence could also mean that a ball is equally likely to fall in any of the boxes - no biases, that is.

Answer 15

Like when we do problems about fair coins, they may specify something like the chance of landing a head is 1/2.

Answer 16

Yea, I mean in that case it would have to be 1/4!

Answer 17

Half the time I have trouble interpreting the problem

Answer 18

I know, it's a pain in the posterior.

Plus it uses the phrase "any particular box", so it does seem like this means nothing but that a ball is equally likely to fall in all four boxes.

Answer 19

yeah that's 80% of probability

Answer 20

So Inky's answer seems fair enough to me.

Answer 21

Okay fine that explains it - I originally took that approach, but then thought I had to do something about that 1/4 probability.