Finding the probability distribution function

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lucrob

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The question is as follows:

Ten identically shaped discs are in a bag, two of them are black and 8 of them are white. The discs are drawn from the bag at random and are not replaced. Let G be the number of discs drawn until the first black one is drawn.
(i) Find the probability distribution function f(g) = P(G = g)

I know that P(G = 1) = 8/10 x 2/9 then P(G = 2) = 7/9 x 2/8 and so on. Tried to make a few different formulae based on this idea but all didn't match the answer key. Not too sure where to go from here. Any help would be greatly appreciated as this question is really bothering me. Thanks.
 
lucrob said:
The question is as follows:

Ten identically shaped discs are in a bag, two of them are black and 8 of them are white. The discs are drawn from the bag at random and are not replaced. Let G be the number of discs drawn until the first black one is drawn.
(i) Find the probability distribution function f(g) = P(G = g)

I know that P(G = 1) = 8/10 x 2/9 then P(G = 2) = 7/9 x 2/8 and so on. Tried to make a few different formulae based on this idea but all didn't match the answer key. Not too sure where to go from here. Any help would be greatly appreciated as this question is really bothering me. Thanks.
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First, we need to clarify the problem. Does "the number of discs drawn until the first black one is drawn" mean "before the first black one is drawn", so that the event "first drawn is black" is G=0, or is "through the first black one", so that the event "first drawn is black" is G=1? Your work suggests you are taking it as the former, since your P(G=1) appears to be P(WB). But you don't mention P(G=0), and your P(G=2) seems to ignore the first draw.

Second, it will be helpful if you show us what the answer key shows, so we can be sure what form they want the answer to be in; this will also help to make sure what the problem is intended to mean.

Finally, I'd like to see your actual answer for comparison to that, in case you just made a small mistake, or did correct work with a wrong interpretation of the problem, or didn't go far enough, or something like that.
 
Dr.Peterson said:
First, we need to clarify the problem. Does "the number of discs drawn until the first black one is drawn" mean "before the first black one is drawn", so that the event "first drawn is black" is G=0, or is "through the first black one", so that the event "first drawn is black" is G=1? Your work suggests you are taking it as the former, since your P(G=1) appears to be P(WB). But you don't mention P(G=0), and your P(G=2) seems to ignore the first draw.

Second, it will be helpful if you show us what the answer key shows, so we can be sure what form they want the answer to be in; this will also help to make sure what the problem is intended to mean.

Finally, I'd like to see your actual answer for comparison to that, in case you just made a small mistake, or did correct work with a wrong interpretation of the problem, or didn't go far enough, or something like that.
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1. the question suggests that it is until the first black one is drawn.

2. the answer for this question is: 10-g/45

3. I tried to come up with different answers but all of them didn't work:
2/(10-g) x (8-g/10-g) & (1/10) x (8/10)^g-1

thanks
 
lucrob said:
1. the question suggests that it is until the first black one is drawn.
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Well, as quoted, it literally says "until"; the question is what that means. Does it include the black one, or not? (This is like ambiguity of the word "between".)

lucrob said:
2. the answer for this question is: 10-g/45
Click to expand...
That seems too simple; I didn't expect the distribution to be a linear function (without having actually tried solving it).

The claim, then, would be that P(G=0) = 10. That's nonsense! So I'll suppose you meant (10-g)/45. Then P(G=0) = 10/45. But clearly the probability that the first draw is black is 2/10; and that's what this formula gives for f(1), so they must be taking it (as I would unless told otherwise) as inclusive: G is the number of draws until and including the first black one, so its smallest value is 1.

So now we seem to know the intended interpretation. When you said, "P(G = 1) = 8/10 x 2/9", you were apparently taking G = 1 to mean there are two draws, the first white and the second black. Give it another try.
 
When I tried to solve this problem, my thinking was like that

I could draw a black disk from the first try, so i would say that when [MATH]g = 1[/MATH], [MATH]p(1) = \frac{2}{10}[/MATH]
And I think that this problem could be also solved using Negative Binomial Distribution, but I did not try because I am lazy.
 
If you are unwilling to try then why should any of the helpers here be willing to try?
 
Jomo said:
If you are unwilling to try then why should any of the helpers here be willing to try?
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Lucrob did try, and showed some work (post #3). We just need to see another try, which may be stuck in moderation (or not).

You are responding to nasi112, who also made a start, but as a helper at the moment is not obligated to try every conceivable method. I get lazy that way too (letting the OP do the work). I also get lazy sometimes about looking to the top to find out who the OP is.
 
Dr.Peterson said:
Lucrob did try, and showed some work (post #3). We just need to see another try, which may be stuck in moderation (or not).

You are responding to nasi112, who also made a start, but as a helper at the moment is not obligated to try every conceivable method. I get lazy that way too (letting the OP do the work). I also get lazy sometimes about looking to the top to find out who the OP is.
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Thanks for pointing that out!
 
Yes, [MATH]P(G=g) = \frac{(10-g)}{45}[/MATH] (g=1 to 9, the only possible values for g)

[MATH]P(G=1)=\frac{2}{10}[/MATH] clearly
[MATH]P(G=2)=\frac{8}{10}\times \frac{2}{9}[/MATH][MATH]P(G=3)=\frac{8}{10}\times \frac{7}{9} \times \frac{2}{8}[/MATH] as this requires that you get white first, white second and black third.

Continue this 'table', writing the multiplications like this and then try to get a general expression for the [MATH]n^\text{th}[/MATH] term (or [MATH]g^\text{th}[/MATH] term), by inspection, cancelling terms or proving it using factorial notation to get an expression which cancels down to the required expression.
 
Just for fun.

[MATH]f(g) = \frac{\frac{2!}{(2 - 1)!} \cdot \frac{8!}{(8 - [g - 1])!}}{\frac{10!}{(10 - g)!}}[/MATH]

[MATH]\sum_{g=1}^{9} \frac{\frac{2!}{(2 - 1)!} \cdot \frac{8!}{(8 - [g - 1])!}}{\frac{10!}{(10 - g)!}} = 1[/MATH]
 
Dr.Peterson said:
Lucrob did try, and showed some work (post #3). We just need to see another try, which may be stuck in moderation (or not).

You are responding to nasi112, who also made a start, but as a helper at the moment is not obligated to try every conceivable method. I get lazy that way too (letting the OP do the work). I also get lazy sometimes about looking to the top to find out who the OP is.
Click to expand...

Ok, I believe I figured it out. When you multiply 8/10 x 2/9 you end up with 8/45 which is P(G = 1). Every multiplication cumulative multiplication after that would make it even smaller. Additionally, Another way of writing 2/10 which is P(G = 0) is 9/45. I did a table to better explain this below:

P(G = 0) = 9/45
P(G = 1) = 8/45
P(G = 2) = 7/45
P(G = 3) = 6/45
...etc

From this it can be deduced that the probability distribution formula for this is:
9-G/45

Although, the answer says its 10-G/45 so I'm not too sure what to do after this step.
 
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