Drawing tickets from box - probability

[goldfronts1]

Homework Statement

A box contains 8 tickets. Two are marked 1, two are marked 2, two marked 3, and two marked 4. Tickets are drawn at random from the box without replacement until a number appears that has appeared before. Let X be the number of draws that are made. Make a table to display the probability distribution of X.

The Attempt at a Solution

Unsure of where to begin.

 

[Dick]

Start at the beginning. The probability of stopping at the first draw is 0. The probability of stopping at the second draw is 1/7 (why?). The probability of stopping at the third draw is the probability you haven't already stopped before (6/7) times the probability of picking a duplicate to one of the two cards you are already holding out the six remaining. Etc.

 

[system downs]

If you have studied in-depth physics and you believe in determinism, then you know that probability is the accuracy of a prediction at random, and not % of what will happen

 

[Dick]

If you have studied in-depth physics and you believe in determinism, then you know that probability is the accuracy of a prediction at random, and not % of what will happen

Is that supposed to mean something? This is a math problem.

 

[goldfronts1]

The probability of stopping at the first draw is 0. (because you have only drawn one ticket). The probability of stopping at the second draw is 1/7 (because out of the 7 remaining cards their is only 1 possibility for a duplicate of the first). The probability of stopping at the third draw is the probability you haven't already stopped before (6/7) (Can you explain the 6/7 to me) times the probability of picking a duplicate to one of the two cards you are already holding out the six remaining.(would this be 2/6)?

So would the probability of stopping at the fourth draw be (5/6)?
And the probability of picking a duplicate to one of the three cards I already have out of the 5 remaining be (1/5)?

Thanks

 

[Dick]

To get to the third draw you have to have not stopped before. The probability you haven't stopped yet is 1-1/7=6/7. And yes, the probability you get a duplicate at the third draw is 2/6=1/3. The probability of stopping at the fourth draw is, again, the probability you haven't stopped in the first three, times the probability of getting a match to one of the first three cards out of the remaining five. I think you are getting this.

 

[goldfronts1]

The probability of stopping at the fourth draw is, again, the probability you haven't stopped in the first three (1-6/7)?, times the probability of getting a match to one of the first three cards out of the remaining five (3/5)?. I'm am a little confused on how to calculate the probability I haven't stopped in the previous draws. Is it always the complement rule of P(not A) = 1-P(A)?

 

[Dick]

What are you finally getting for the probability of stopping at the third? Sure, the probability of stopping at draw n is 1-(the sum of the probabilities of stopping at the previous draws). All of these probabilities of stopping are mutually exclusive.

 

[goldfronts1]

The probability of stopping at the third draw is 2/6? So the probability of stopping at the fourth draw is (1-2/6)? So this times the probability of getting a match to one of the first three cards out of the remaining five is (3/5)? But (2/3) * (3/5) is 2/5. But the answer in the back of the books says 12/35.

 

[Dick]

The probability of stopping at the third draw is 2/6? So the probability of stopping at the fourth draw is (1-2/6)? So this times the probability of getting a match to one of the first three cards out of the remaining five is (3/5)? But (2/3) * (3/5) is 2/5. But the answer in the back of the books says 12/35.

No. The probability of stopping at the third draw is (6/7)*(1/3). So the probability of making it to a fourth draw is (1-(1/7)-(6/7)*(1/3)) and the probability of getting a match then is 3/5. Multiply those and you'll get 12/35.

 

[goldfronts1]

Hold on. I think I got it. But how are getting the probability of getting a match to one of three previous cards out of the remaining 5? You have (1/3). I say it is (3/5) because there are three chances of getting a match to anyone of the three cards..

 

[Dick]

I have a 1/3=2/6 for the third draw probability of matching (as you said). For the fourth draw it's 3/5 (as you said). Now factor in the probability of even making it to a third or fourth draw. It's 1-the probability of quitting earlier.

 

[goldfronts1]

Oh man! I feel so silly. I got it now. It's always 1-the probability of quitting earlier. Oh man!. Thank you for all of your help.

 

[Dick]

I knew you'd get it. You were so close.