Conditional probability(black & white ball game)

[Cylab]

A Box contains black balls and white balls, with 1/2 of each. Now we add black balls, with 1/8 of total. What is probability of drawing 1 black, 2 black and 3 black balls respectively (no replication)?

Analysis:
Drawing one black ball : 1/8*1/2+1/2 = 9/16.
Thinking of conditional probability, we simply consider what is left after first one drawn and multiply its probability to the first one. However, here, we only know probability of black ball(9/16). Please shed some light, what is probability of the second ball drawn is black, given first one is black ball? , and so on...

Thanks for your attention.

 

[Sourabh N]

I don't think you have the probability of drawing one ball correct. Assume there were $n$ balls in the box initially and proceed.

 

[Cylab]

In other words, maybe we can also say; Pr(Black) = 9/16, and Pr(White)=7/16 in the box. Then what is probability of drawing two black balls (no replication)?
　　Analysis:
, the probability of drawing 1st black ball is 9/16, I just do not know how to calculate of drawing the 2nd black one, given 1st one is black ball. It should be: Pr(B1&B2) = Pr(B1) * Pr(B2|B1)

Please share your ideas! or hint, anything will be appreciated extremely.

 

[Ray Vickson]

A Box contains black balls and white balls, with 1/2 of each. Now we add black balls, with 1/8 of total. What is probability of drawing 1 black, 2 black and 3 black balls respectively (no replication)?

Analysis:
Drawing one black ball : 1/8*1/2+1/2 = 9/16.
Thinking of conditional probability, we simply consider what is left after first one drawn and multiply its probability to the first one. However, here, we only know probability of black ball(9/16). Please shed some light, what is probability of the second ball drawn is black, given first one is black ball? , and so on...

Thanks for your attention.

The probability of drawing black on the first draw does not depend on the number of balls, but that is not true for drawing the second or third black ball; that is, the probabilities of getting black on the second and on the third draw WILL depend on the number of balls you start with. Given the description, the only possible black/white numbers (B,W) are (5,4), (10,8), (15,12), etc. In each of these cases you can work out the answer to your question, but the answers will be a little bit different for different (B,W) combinations.

RGV

 

[rcgldr]

Given the description, the only possible black/white numbers (B,W) are (5,4), (10,8), (14,12)

(14,12) should be (15,12). The b/w numbers will be a multiple of (5,4).

 

[Ray Vickson]

(14,12) should be (15,12). The b/w numbers will be a multiple of (5,4).

Right: that was a typo, now fixed.

RGV

 

[Cylab]

Could you please explain how did you come up with such rule?
The Pr(B)=9/16 . But the multiple of (5,4), it is 5/9, 10/18, 15/27 .
Did I fail to understand something?

 

[Ray Vickson]

Could you please explain how did you come up with such rule?
The Pr(B)=9/16 . But the multiple of (5,4), it is 5/9, 10/18, 15/27 .
Did I fail to understand something?

Could you please explain what you mean by "such rule"? What rule?

A lot depends on your interpretation of the question "What is probability of drawing 1 black, 2 black and 3 black balls respectively (no replication)?"

Does this mean the following three questions? (1) draw 1 ball; what is the probability it is black? (2) Draw two balls; what is the probability they are both black? (3) Draw three balls; what is the probability all three are black? In (1) the probability would be 5/9 for any allowed (B,W) combination. However, the answers to (2) and (3) would vary among the allowed combinations.

Or, does the question mean: we draw three balls. What are the probabilities for (i) exactly 1 black ball; (ii) exactly two black balls; or (iii) all three are black. Obviously, the probability of one black in this case would not be 5/9, but would now depend on the applicable (B,W) values.

RGV

 

[Cylab]

Let me clarify the question!
A Box contains black balls and white balls, with 1/2 of each. Now we add another black balls into the box (that already has black and white ball with same probabilities), with 1/8 of total (which means if the box has 40(black)+40(white), then we add 10 black ball into it).
Now after the ADDING, the Pr(B)=9/16 (it is easy to calculate, right?). Now the question is what is probability of drawing 2nd black ball, given 1st one is black?

 

[Ray Vickson]

Let me clarify the question!
A Box contains black balls and white balls, with 1/2 of each. Now we add another black balls into the box (that already has black and white ball with same probabilities), with 1/8 of total (which means if the box has 40(black)+40(white), then we add 10 black ball into it).
Now after the ADDING, the Pr(B)=9/16 (it is easy to calculate, right?). Now the question is what is probability of drawing 2nd black ball, given 1st one is black?

If there are originally 4 black and 4 white, after adding there will be 5 black and 4 white, so P(B) = 5/9 on the first draw.

If there are originally 8 black and 8 white, after adding there will be 10 black and 8 white, so P(B) = 10/18 = 5/9 on the first draw.

For ANY multiple of (B,W) = (5,4), the probability P(B) = 5/9 on the first draw. However, for the second and higher draws the probabilities of additional blacks will vary between the different cases, as I have already said more than once.

RGV