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.













4












$begingroup$


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
$










share|cite|improve this question









New contributor




Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$







  • 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















4












$begingroup$


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
$










share|cite|improve this question









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













4












4








4





$begingroup$


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
$










share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited Mar 13 at 9:55









Cettt

1,888622




1,888622






New contributor




Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Mar 13 at 7:15









Ali J.Ali J.

484




484




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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.
Check out our Code of Conduct.







  • 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




    $begingroup$
    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










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