Probability question?

Question

N = number of balls initially in a basket.
x = number of balls picked from the basket, each in turn, without replacing them.
Probability of picking a particular ball out of a full basket = 1/N.

Question:

1. There are N balls in the basket to begin with.

2. You pick "x" balls

3. You then put all the balls back, and again pick "x" balls

What is the probability that you have picked, in step 3, a ball you did not pick in step 2?

Thanks a lot for any answers.

Answer 1

It's perhaps easier to think of this in terms of the probability of picking all the same balls the second time around... your answer would just be 1 minus that probability of picking the exact same set of balls the next time around. Since the probability of picking a ball doesn't change whether you picked it the first time or not, this is equivalent to the probability of picking x particular balls out of the total N. And we can assume a uniform distribution since it's given that the probability of picking each ball is 1/N.

So the probability of picking x particular balls would be the product of picking each of the x balls. The probability of taking one of the x balls would be x/N. The probability of taking another of the x balls would be (x-1)/(N-1), because we now have one less ball in the bag (boy that sounds bad!). The probability of taking the final of the x balls after picking out the others would then be 1/(N-x+1). The probability then winds up being...

--- x/N * (x-1)/(N-1) * (x-2)/(N-2) * ... * 1/(N-x+1)

But remember you're looking for the chance of taking a different ball the second time around, so what you're looking for is...

--- 1 - [x/N * (x-1)/(N-1) * (x-2)/(N-2) * ... * 1/(N-x+1)]

Note that this assumes that x is at least 1 and not more then N. If x=1 then the answer would simply be 1-(1/N), or (N-1/N). If x=N, then the answer would be 0 (you picked all the balls the first time around and so couldn't pick a different ball the second time around).

Answer 2

1

Answer 3

i've got faith my argument might desire to not be nicely won. I say the prospect is 50% permit a ??, permit b ?? and randomly decide on the values for a and b. As already referred to, for a ? 0, P( a < b²) = a million, it rather is trivial. basically particularly much less trivial is the thought P(a < 0 ) = a million/2 and subsequently P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what happens whilst a > 0 For a > 0, on a similar time because it rather is easy to tutor there's a non 0 risk for a finite b, the decrease, the prospect is 0. a < b² is resembling asserting 0 < a < b², remember we are basically finding at a > 0. If this a finite era on an endless line. The risk that a is an component to this era is 0. P( a < b² | a > 0) = 0 As such we've a complete risk P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 remember, it fairly is through countless instruments. no count what form of era you draw on paper or on a working laptop or computing device you will come across a finite risk that seems to physique of recommendations a million. yet it rather is with the aid of the finite random variety turbines on the computing device and if we had this question asked with finite values there could be a a answer extra advantageous than 50%. i don't mean to be condescending, yet please clarify why utilising the Gaussian to approximate a uniform distribution is a robust thought? are not countless numbers exciting. Cantor whilst mad working with them! :)

Answer 4

A=(n choose x) is the number of possible pickings
B=((n-x) choose x) is the number of picking removing the first set

P = B/A ... assuming n > 2x and also since you said each ball is equally likely.

Answer 5

It's a function of how many balls you have and how many you take out. If X and N are the same numner then the probability is 100%

probability = N/X * 100%

Answer 6

1/N = N / x ???

Answer 7

Depends on the color of the balls. Red balls always get picked, whereas blue do not.

Answer 8

it is one in three, also stated 1:3, or 33 1/3% which also is .333