Stats - Probability of redrawing a ball (draw, replace)

[Michael805]

Homework Statement

There are balls numbered 1 through n in a box. Suppose that a boy successively draws a ball from the box, each time replacing the one drawn before drawing another. This continues until the boy draws a ball that he has previously drawn before. Let X denote the number of draws, and compute its probability density function.

Homework Equations

Probability density function

The Attempt at a Solution

I was thinking at first this would simply be 1/n, but I'm unsure how to account for putting the ball back then calculating how many attempts it will take before drawing a previously drawn one.

 

[Ray Vickson]

Homework Statement

There are balls numbered 1 through n in a box. Suppose that a boy successively draws a ball from the box, each time replacing the one drawn before drawing another. This continues until the boy draws a ball that he has previously drawn before. Let X denote the number of draws, and compute its probability density function.

Homework Equations

Probability density function

The Attempt at a Solution

I was thinking at first this would simply be 1/n, but I'm unsure how to account for putting the ball back then calculating how many attempts it will take before drawing a previously drawn one.

Event {X=2} occurs if the second ball drawn matches the first one drawn, and the probability of this is 1/n. Event {X=3} occurs if the second ball fails to match the first but the third one matches one of the first two. The probability the second does not match the first is (n-1)/n, while the probability the third one matches one of the first two is 2/n. You can put these two pieces of information together to determine Pr{X = 3}. Continue in this way to get Pr{X=k} for any k.

RGV

 

[Michael805]

So if I'm understanding this right, the probability of drawing a ball that was previously drawn would be (X-1)/n, where X is the number of draws? This makes sense to me, but it's just not quite fully clicking for some reason.

 

[Ray Vickson]

So if I'm understanding this right, the probability of drawing a ball that was previously drawn would be (X-1)/n, where X is the number of draws? This makes sense to me, but it's just not quite fully clicking for some reason.

The only way you could be making draw X is if the first (X-1) numbers are all different. So, the chance that draw X gives you a repetition is (X-1)/N, because each individual number has chance 1/N of being drawn, and any of the current (X-1) would give you a match. What is it about this argument that leaves you confused?

RGV