Question

probability - there is a container with 100 marbles: 30 green, 60 red and 10 blue. Withdrawing simultaneously and randomly two marbles, what is the probability that at least one is blue?

Answer 1

I have the problem solved here only with fractions, but I don't know how it works

Answer 2

Easiest way is to find the probability that none are blue and subtract that value from "1", since that is the complement of "at least one is blue".

P(none blue) = P(first pick is not blue) x P(2nd pick is not blue)

That p(2nd pick is not blue) is dependent on the first pick because there will be one less marble on that second pick (and one less that is not blue)

P(none blue) = (90/100) x (89/99)

P(at least one is blue) = 1 - (90/100) x (89/99)

Answer 3

right, same here, I'm a bit confused with the terms at least, at most, can you explain how to deal with them?

Answer 4

np. Which term? Before you answer that, a comment. Notice that I used "first pick and second pick". That would appear to go against picking 2 simultaneously, but it does not go against that. That first pick is from 90 "non-blues", where there are a total at that time of 100 marbles. That second pick is from 89 "non-blues", where there are only 99 marbles left at that point (after the first pick).

Answer 5

yes, I would ask you this "simultaneously" thing, isn't it the same as picking one then other one?

Answer 6

Picking one and then the other is the same thing as picking both together in this case. If you pick both together (blindly), you can imagine putting your hand in a bag without looking and putting one in your hand and then another, without looking at the first. Here's another way to do the problem that will clarify this:

The total # of ways you can pick 2 marbles that are non-blue (together, same time) from that group of 90 is:

90C2

The total number of ways you can pick any 2 marbles from the whole lot of 100 is:

100C2

If you do:

1 - (90C2)/(100C2)

you get the same answer.

Answer 7

ok, about the 'at least' and 'at most' I have to use the complement to solve?

Answer 8

right, thanks!

Answer 9

That's the first part and actually the easiest. You are looking for the probability of "at least 1". That means you would have to add P(1 blue) and P(2 blue). If you figured each out individually, you would have to do 2 probabilities and add them. That's inefficient when you stop and realize that the only other possibility is P(zero blue) and:

P(zero blue) + P(1 blue) + P(2 blue) = 1

So, just figure out the P(zero blue) and subtract from 1.

Answer 10

This is because you picking only 2 marbles. Imagine if you were picking a lot more. You wouldn't want to add up:

P(1) + P(2) + P(3) + P(4) + etc.

Answer 11

All good now, @lucaz ?

Answer 12

yeah, thanks =]

Answer 13

uw! Good luck to you in all of your studies and thx for the recognition! @lucaz