Conditional probability (urns and balls)












5














From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears.
What is the probability that:
1) second white ball appears on the second step
2) second white ball appears on the third step
3) second white ball appears on the k-th step



My solution is the following.
The probability of the second white ball is
$$Pr(X=2)=frac{6}{18}frac{5}{17}$$



The probability of the third white ball is
$$Pr(X=3)=frac{6}{18}frac{12}{17}frac{5}{16}+ frac{12}{18}frac{6}{17}frac{5}{16} = frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$



Therefore, the probability
$$Pr(X=k)=frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$



Could somebody, please, check my solution, especially the third part. Thank you.






probability balls-in-bins







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  • In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
    – N. F. Taussig
    Nov 27 '18 at 19:34












  • Yes, the calculation looks fine to me.
    – DavidPM
    Nov 27 '18 at 19:38










  • Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
    – Makhayama
    Nov 27 '18 at 19:40
















5














From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears.
What is the probability that:
1) second white ball appears on the second step
2) second white ball appears on the third step
3) second white ball appears on the k-th step



My solution is the following.
The probability of the second white ball is
$$Pr(X=2)=frac{6}{18}frac{5}{17}$$



The probability of the third white ball is
$$Pr(X=3)=frac{6}{18}frac{12}{17}frac{5}{16}+ frac{12}{18}frac{6}{17}frac{5}{16} = frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$



Therefore, the probability
$$Pr(X=k)=frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$



Could somebody, please, check my solution, especially the third part. Thank you.






probability balls-in-bins







share|cite|improve this question
























  • In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
    – N. F. Taussig
    Nov 27 '18 at 19:34












  • Yes, the calculation looks fine to me.
    – DavidPM
    Nov 27 '18 at 19:38










  • Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
    – Makhayama
    Nov 27 '18 at 19:40














5












5








5


1





From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears.
What is the probability that:
1) second white ball appears on the second step
2) second white ball appears on the third step
3) second white ball appears on the k-th step



My solution is the following.
The probability of the second white ball is
$$Pr(X=2)=frac{6}{18}frac{5}{17}$$



The probability of the third white ball is
$$Pr(X=3)=frac{6}{18}frac{12}{17}frac{5}{16}+ frac{12}{18}frac{6}{17}frac{5}{16} = frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$



Therefore, the probability
$$Pr(X=k)=frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$



Could somebody, please, check my solution, especially the third part. Thank you.






probability balls-in-bins







share|cite|improve this question















From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears.
What is the probability that:
1) second white ball appears on the second step
2) second white ball appears on the third step
3) second white ball appears on the k-th step



My solution is the following.
The probability of the second white ball is
$$Pr(X=2)=frac{6}{18}frac{5}{17}$$



The probability of the third white ball is
$$Pr(X=3)=frac{6}{18}frac{12}{17}frac{5}{16}+ frac{12}{18}frac{6}{17}frac{5}{16} = frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$



Therefore, the probability
$$Pr(X=k)=frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$



Could somebody, please, check my solution, especially the third part. Thank you.






probability balls-in-bins




probability balls-in-bins






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edited Nov 28 '18 at 11:07

























asked Nov 27 '18 at 19:26









Makhayama

262




262












  • In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
    – N. F. Taussig
    Nov 27 '18 at 19:34












  • Yes, the calculation looks fine to me.
    – DavidPM
    Nov 27 '18 at 19:38










  • Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
    – Makhayama
    Nov 27 '18 at 19:40


















  • In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
    – N. F. Taussig
    Nov 27 '18 at 19:34












  • Yes, the calculation looks fine to me.
    – DavidPM
    Nov 27 '18 at 19:38










  • Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
    – Makhayama
    Nov 27 '18 at 19:40
















In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
– N. F. Taussig
Nov 27 '18 at 19:34






In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
– N. F. Taussig
Nov 27 '18 at 19:34














Yes, the calculation looks fine to me.
– DavidPM
Nov 27 '18 at 19:38




Yes, the calculation looks fine to me.
– DavidPM
Nov 27 '18 at 19:38












Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
– Makhayama
Nov 27 '18 at 19:40




Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
– Makhayama
Nov 27 '18 at 19:40










1 Answer
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I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.






share|cite|improve this answer

















  • 1




    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • @GrahamKemp An elegant alternative approach.
    – callculus
    Nov 28 '18 at 2:39











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I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.






share|cite|improve this answer

















  • 1




    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • @GrahamKemp An elegant alternative approach.
    – callculus
    Nov 28 '18 at 2:39
















1














I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.






share|cite|improve this answer

















  • 1




    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • @GrahamKemp An elegant alternative approach.
    – callculus
    Nov 28 '18 at 2:39














1












1








1






I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.






share|cite|improve this answer












I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 '18 at 21:02









callculus

17.8k31427




17.8k31427








  • 1




    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • @GrahamKemp An elegant alternative approach.
    – callculus
    Nov 28 '18 at 2:39














  • 1




    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • @GrahamKemp An elegant alternative approach.
    – callculus
    Nov 28 '18 at 2:39








1




1




Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
– Graham Kemp
Nov 28 '18 at 1:56






Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
– Graham Kemp
Nov 28 '18 at 1:56














@GrahamKemp An elegant alternative approach.
– callculus
Nov 28 '18 at 2:39




@GrahamKemp An elegant alternative approach.
– callculus
Nov 28 '18 at 2:39


















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