- #1
ChrisVer
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Suppose you have a box with 10 balls, the 4 are red, the 5 are blue and the 1 is white.
The probability of dragging each ball is [itex]0.4~,~0.5~,~0.1[/itex] respectively.
Now if I ask what is the probability of getting 3 balls in the order: red -> blue -> white, what is the right answer if balls are replaced/not replaced?
I have some problem in understanding the logic behind my approach to answer this...
first I thought: [itex]P_{tot} = P(R) P(B) P(W) [/itex] This gives the probability of dragging a red, a blue and a white ball... It doesn't seem right for ordered draggings since it also counts the dragging of : blue->red->white.
If the balls are replaced, then the dragging doesn't affect the probabilities, and so the correct probability of the successive drawing would be:
[itex]P(RBW)= \frac{P(R) P(B) P(W)}{3 !}= \frac{0.02}{6} = \frac{1}{300}[/itex]
Since the written [itex]P_{tot} = P(RBW)+ P(RWB)+ P(BRW) + P(BWR) + P(WBR) + P(WRB)[/itex] (permutations of 3).
Then I thought of using the conditional probability...I define two events: [itex]A_1,A_2[/itex]. The [itex]A_1[/itex] is the event that I drag a blue ball given that I had dragged a red ball before: [itex]A_1 = (B|R)[/itex] and the probability is:
[itex] P(A_1)= P(B|R)= \frac{P(B \cap R)}{P(R)}= \frac{P(B) P(R)}{P(R)} = P(B)[/itex]
Then [itex]A_2[/itex] is dragging a white ball, given that [itex]A_1[/itex] is true:
[itex]P(A_2)= P(W|A_1)= \frac{P (W \cap A_1)}{P(A_1)} = P(W)[/itex]
and so the [itex]P(RBW)=P(W)=0.1 \ne \frac{1}{300}[/itex]
I used that [itex]P(x \cap y) = P(x) P(y)[/itex] since they are statistically independent events.
So what is wrong in either of these approaches? They obviously don't give the same result...However, logically they make sense to me![Frown :frown: :frown:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
The probability of dragging each ball is [itex]0.4~,~0.5~,~0.1[/itex] respectively.
Now if I ask what is the probability of getting 3 balls in the order: red -> blue -> white, what is the right answer if balls are replaced/not replaced?
I have some problem in understanding the logic behind my approach to answer this...
first I thought: [itex]P_{tot} = P(R) P(B) P(W) [/itex] This gives the probability of dragging a red, a blue and a white ball... It doesn't seem right for ordered draggings since it also counts the dragging of : blue->red->white.
If the balls are replaced, then the dragging doesn't affect the probabilities, and so the correct probability of the successive drawing would be:
[itex]P(RBW)= \frac{P(R) P(B) P(W)}{3 !}= \frac{0.02}{6} = \frac{1}{300}[/itex]
Since the written [itex]P_{tot} = P(RBW)+ P(RWB)+ P(BRW) + P(BWR) + P(WBR) + P(WRB)[/itex] (permutations of 3).
Then I thought of using the conditional probability...I define two events: [itex]A_1,A_2[/itex]. The [itex]A_1[/itex] is the event that I drag a blue ball given that I had dragged a red ball before: [itex]A_1 = (B|R)[/itex] and the probability is:
[itex] P(A_1)= P(B|R)= \frac{P(B \cap R)}{P(R)}= \frac{P(B) P(R)}{P(R)} = P(B)[/itex]
Then [itex]A_2[/itex] is dragging a white ball, given that [itex]A_1[/itex] is true:
[itex]P(A_2)= P(W|A_1)= \frac{P (W \cap A_1)}{P(A_1)} = P(W)[/itex]
and so the [itex]P(RBW)=P(W)=0.1 \ne \frac{1}{300}[/itex]
I used that [itex]P(x \cap y) = P(x) P(y)[/itex] since they are statistically independent events.
So what is wrong in either of these approaches? They obviously don't give the same result...However, logically they make sense to me
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