Probability of third ball drawnDiscarding 2 balls without color known - probablity 3rd ball is redProbability to draw a white ball from the third box.initial probability , 2rd drawn is red ballProbability of a certain ball drawn from one box given that other balls were drawnWhat of the probability of drawing a blue ball in this example?Show the probability that the first red ball drawn is the $(k+1)$th ball drawn equals $binomr+b-k-1r-1/binomr+bb$What is the probability that the drawn ball is redTough Probability Question.Does drawn count effect probability?After $K$ balls are drawn, what is the probability that next ball drawn is green.
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Probability of third ball drawn
Discarding 2 balls without color known - probablity 3rd ball is redProbability to draw a white ball from the third box.initial probability , 2rd drawn is red ballProbability of a certain ball drawn from one box given that other balls were drawnWhat of the probability of drawing a blue ball in this example?Show the probability that the first red ball drawn is the $(k+1)$th ball drawn equals $binomr+b-k-1r-1/binomr+bb$What is the probability that the drawn ball is redTough Probability Question.Does drawn count effect probability?After $K$ balls are drawn, what is the probability that next ball drawn is green.
$begingroup$
Having a box containing 15 red and 7 blue balls. We want to draw 3 balls at random by these conditions on each draw:
- If the ball is red you set it aside
- If the ball is blue put it back in
What is probability that the third draw is blue? (If you get a blue ball it counts as a draw even though you put it back in the box.)
Can I say by these rules?
We have three balls drawn the total possibility is the sum of:
- Both the first two balls are red
- Both the first two balls are blue
- The first ball is red and the second is blue
- The first ball is blue and the second is red.
Then for each case I can calculate the probability of blue ball at third draw.
$
P=frac1522*frac1421*frac720+frac722*frac722*frac722+frac722*frac1522*frac721+frac1522*frac721*frac721
$
probability balls-in-bins
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Having a box containing 15 red and 7 blue balls. We want to draw 3 balls at random by these conditions on each draw:
- If the ball is red you set it aside
- If the ball is blue put it back in
What is probability that the third draw is blue? (If you get a blue ball it counts as a draw even though you put it back in the box.)
Can I say by these rules?
We have three balls drawn the total possibility is the sum of:
- Both the first two balls are red
- Both the first two balls are blue
- The first ball is red and the second is blue
- The first ball is blue and the second is red.
Then for each case I can calculate the probability of blue ball at third draw.
$
P=frac1522*frac1421*frac720+frac722*frac722*frac722+frac722*frac1522*frac721+frac1522*frac721*frac721
$
probability balls-in-bins
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
If I understand what you're saying, this seems right.
$endgroup$
– Brian Tung
Mar 13 at 7:20
$begingroup$
Seems right. Combinations are b * (bb + br+rb+rr)
$endgroup$
– Michael Paris
Mar 13 at 7:21
add a comment |
$begingroup$
Having a box containing 15 red and 7 blue balls. We want to draw 3 balls at random by these conditions on each draw:
- If the ball is red you set it aside
- If the ball is blue put it back in
What is probability that the third draw is blue? (If you get a blue ball it counts as a draw even though you put it back in the box.)
Can I say by these rules?
We have three balls drawn the total possibility is the sum of:
- Both the first two balls are red
- Both the first two balls are blue
- The first ball is red and the second is blue
- The first ball is blue and the second is red.
Then for each case I can calculate the probability of blue ball at third draw.
$
P=frac1522*frac1421*frac720+frac722*frac722*frac722+frac722*frac1522*frac721+frac1522*frac721*frac721
$
probability balls-in-bins
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Having a box containing 15 red and 7 blue balls. We want to draw 3 balls at random by these conditions on each draw:
- If the ball is red you set it aside
- If the ball is blue put it back in
What is probability that the third draw is blue? (If you get a blue ball it counts as a draw even though you put it back in the box.)
Can I say by these rules?
We have three balls drawn the total possibility is the sum of:
- Both the first two balls are red
- Both the first two balls are blue
- The first ball is red and the second is blue
- The first ball is blue and the second is red.
Then for each case I can calculate the probability of blue ball at third draw.
$
P=frac1522*frac1421*frac720+frac722*frac722*frac722+frac722*frac1522*frac721+frac1522*frac721*frac721
$
probability balls-in-bins
probability balls-in-bins
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 13 at 9:55
Cettt
1,888622
1,888622
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asked Mar 13 at 7:15
Ali J.Ali J.
484
484
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
$begingroup$
If I understand what you're saying, this seems right.
$endgroup$
– Brian Tung
Mar 13 at 7:20
$begingroup$
Seems right. Combinations are b * (bb + br+rb+rr)
$endgroup$
– Michael Paris
Mar 13 at 7:21
add a comment |
1
$begingroup$
If I understand what you're saying, this seems right.
$endgroup$
– Brian Tung
Mar 13 at 7:20
$begingroup$
Seems right. Combinations are b * (bb + br+rb+rr)
$endgroup$
– Michael Paris
Mar 13 at 7:21
1
1
$begingroup$
If I understand what you're saying, this seems right.
$endgroup$
– Brian Tung
Mar 13 at 7:20
$begingroup$
If I understand what you're saying, this seems right.
$endgroup$
– Brian Tung
Mar 13 at 7:20
$begingroup$
Seems right. Combinations are b * (bb + br+rb+rr)
$endgroup$
– Michael Paris
Mar 13 at 7:21
$begingroup$
Seems right. Combinations are b * (bb + br+rb+rr)
$endgroup$
– Michael Paris
Mar 13 at 7:21
add a comment |
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1
$begingroup$
If I understand what you're saying, this seems right.
$endgroup$
– Brian Tung
Mar 13 at 7:20
$begingroup$
Seems right. Combinations are b * (bb + br+rb+rr)
$endgroup$
– Michael Paris
Mar 13 at 7:21