# Distribution table, drawing tickets w/o replacement

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• Oct 4th 2009, 09:35 AM
Dfowj
Distribution table, drawing tickets w/o replacement
Hey guys,

I recently encountered this problem while studying for a test, and im having trouble understanding how to solve it.

A box contains 8 tickets. Two are marked 1, two marked 2, two marked 3, 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.

Now, i dont need the actual table, i just need to know how the book comes up with the answer P(X = 2) = 5/35, P(X = 3) = 10/35 (it only gives a partial answer, those are the only two it gives.)

From my understanding, after drawing once from the box (and thus taking one out), the probability of the second draw being the match to the first would be 1/7. So the sequence of draws would be (1/8)*(1/7).

1/56 != 5/35...

What am i doing wrong here?
• Oct 5th 2009, 02:23 AM
The Second Solution
Quote:

Originally Posted by Dfowj
Hey guys,

I recently encountered this problem while studying for a test, and im having trouble understanding how to solve it.

A box contains 8 tickets. Two are marked 1, two marked 2, two marked 3, 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.

Now, i dont need the actual table, i just need to know how the book comes up with the answer P(X = 2) = 5/35, P(X = 3) = 10/35 (it only gives a partial answer, those are the only two it gives.)

From my understanding, after drawing once from the box (and thus taking one out), the probability of the second draw being the match to the first would be 1/7. So the sequence of draws would be (1/8)*(1/7).

1/56 != 3/35...

What am i doing wrong here?

Try drawing a tree diagram to see how to calculate the probabilities.
• Oct 5th 2009, 02:29 AM
Grandad
Hello Dfowj
Quote:

Originally Posted by Dfowj
Hey guys,

I recently encountered this problem while studying for a test, and im having trouble understanding how to solve it.

A box contains 8 tickets. Two are marked 1, two marked 2, two marked 3, 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.

Now, i dont need the actual table, i just need to know how the book comes up with the answer P(X = 2) = 5/35, P(X = 3) = 10/35 (it only gives a partial answer, those are the only two it gives.)

From my understanding, after drawing once from the box (and thus taking one out), the probability of the second draw being the match to the first would be 1/7. So the sequence of draws would be (1/8)*(1/7).

1/56 != 3/35...

What am i doing wrong here?

Why have you multiplied by $\tfrac18$? It doesn't matter which number is drawn first time. The probability that the second matches it is $\tfrac17 = \tfrac{5}{35}$. (Not $\tfrac{3}{35}$, incidentally.)

The probability that the first two are different, on the other hand, is $\tfrac67$, and if this has occurred, there are then $2$ cards out of the remaining $6$ that will give a match. So $p(x=3) = \tfrac67\times\tfrac26 = \tfrac27=\tfrac{10}{35}$.

Grandad