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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
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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