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

In summary: 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?
  • #1
Michael805
5
0

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.
 
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  • #2
Michael805 said:

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
 
  • #3
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.
 
  • #4
Michael805 said:
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
 

1. What is the probability of redrawing a red ball?

The probability of redrawing a red ball depends on the number of red balls in the pool of balls and the total number of balls in the pool. For example, if there are 10 red balls and 50 total balls, the probability of redrawing a red ball would be 10/50 or 20%.

2. Can the probability of redrawing a ball change after each draw?

Yes, the probability of redrawing a ball can change after each draw if the ball is not replaced back into the pool. With each draw, the total number of balls decreases, affecting the probability of redrawing a specific color.

3. How does the probability of redrawing a ball change if the ball is replaced?

If the ball is replaced after each draw, the probability of redrawing a specific color remains the same. This is because the total number of balls in the pool does not change, and the probability is calculated based on the number of balls of a specific color over the total number of balls.

4. What is the probability of not redrawing a red ball?

The probability of not redrawing a red ball is the complementary event of redrawing a red ball, and it can be calculated by subtracting the probability of redrawing a red ball from 1. For example, if the probability of redrawing a red ball is 20%, the probability of not redrawing a red ball would be 1-0.2 = 0.8 or 80%.

5. How does the number of balls in the pool affect the probability of redrawing a ball?

The number of balls in the pool directly affects the probability of redrawing a specific color. The more balls there are in the pool, the lower the probability of redrawing a specific color. For example, if there are 100 balls in the pool with 10 red balls, the probability of redrawing a red ball is 10/100 or 10%. However, if there are only 50 balls in the pool with 10 red balls, the probability of redrawing a red ball increases to 10/50 or 20%.

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