Conditional probability (urns and balls)
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
asked Nov 27 '18 at 19:26
Makhayama
262
add a comment |
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
asked Nov 27 '18 at 19:26
Makhayama
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
add a comment |
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
asked Nov 27 '18 at 19:26
Makhayama
262
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
asked Nov 27 '18 at 19:26
Makhayama
262
asked Nov 27 '18 at 19:26
Makhayama
262
edited Nov 28 '18 at 11:07
asked Nov 27 '18 at 19:26
Makhayama
262
asked Nov 27 '18 at 19:26
Makhayama
262
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
add a comment |
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
add a comment |
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.
answered Nov 27 '18 at 21:02
callculus
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
add a comment |
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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.
answered Nov 27 '18 at 21:02
callculus
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
add a comment |
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.
answered Nov 27 '18 at 21:02
callculus
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
add a comment |
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.
answered Nov 27 '18 at 21:02
callculus
17.8k31427
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.
answered Nov 27 '18 at 21:02
callculus
17.8k31427
answered Nov 27 '18 at 21:02
callculus
17.8k31427
answered Nov 27 '18 at 21:02
callculus
17.8k31427
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
add a comment |
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
add a comment |
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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