Probablity of drawing all the red balls while blue and green are still left
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3
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Suppose that a box contains 10 red balls, 20 green balls, and 30 blue balls.
Suppose also that balls are drawn from the box one at a time at random.
What is the probability that all the red balls are drawn before the blue or green balls are themselves exhausted. What is the probability that as the last red ball is drawn, there remains at least one blue and one green left in the box.
The answer I was given is $dfrac{7}{12}$ and a general equation is:
$$
dfrac{b g}{1-b}+dfrac{b g}{1-g}
$$ where
$$
g=dfrac{20}{60},b=dfrac{30}{60}
$$
but why?
probability probability-theory
add a comment |
up vote
3
down vote
favorite
Suppose that a box contains 10 red balls, 20 green balls, and 30 blue balls.
Suppose also that balls are drawn from the box one at a time at random.
What is the probability that all the red balls are drawn before the blue or green balls are themselves exhausted. What is the probability that as the last red ball is drawn, there remains at least one blue and one green left in the box.
The answer I was given is $dfrac{7}{12}$ and a general equation is:
$$
dfrac{b g}{1-b}+dfrac{b g}{1-g}
$$ where
$$
g=dfrac{20}{60},b=dfrac{30}{60}
$$
but why?
probability probability-theory
2
Your claimed answer makes no sense. Even if we draw only one ball out of the $60$ total, the probability that this ball is red will be only $10/60 = 1/6$, which is much smaller than $7/12$; and the probability of getting more than one red ball for multiple draws without replacement cannot possibly exceed $1/6$.
– heropup
Jun 19 '14 at 6:02
1
Perhaps he means the probability that all 10 red balls are selected before all 20 greens or all 30 blues? Not that the probability of that happening is necessarily 7/12.
– Avraham
Jun 19 '14 at 6:05
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose that a box contains 10 red balls, 20 green balls, and 30 blue balls.
Suppose also that balls are drawn from the box one at a time at random.
What is the probability that all the red balls are drawn before the blue or green balls are themselves exhausted. What is the probability that as the last red ball is drawn, there remains at least one blue and one green left in the box.
The answer I was given is $dfrac{7}{12}$ and a general equation is:
$$
dfrac{b g}{1-b}+dfrac{b g}{1-g}
$$ where
$$
g=dfrac{20}{60},b=dfrac{30}{60}
$$
but why?
probability probability-theory
Suppose that a box contains 10 red balls, 20 green balls, and 30 blue balls.
Suppose also that balls are drawn from the box one at a time at random.
What is the probability that all the red balls are drawn before the blue or green balls are themselves exhausted. What is the probability that as the last red ball is drawn, there remains at least one blue and one green left in the box.
The answer I was given is $dfrac{7}{12}$ and a general equation is:
$$
dfrac{b g}{1-b}+dfrac{b g}{1-g}
$$ where
$$
g=dfrac{20}{60},b=dfrac{30}{60}
$$
but why?
probability probability-theory
probability probability-theory
edited Jun 19 '14 at 15:22
Avraham
2,5071131
2,5071131
asked Jun 19 '14 at 5:55
LCFactorization
1,107618
1,107618
2
Your claimed answer makes no sense. Even if we draw only one ball out of the $60$ total, the probability that this ball is red will be only $10/60 = 1/6$, which is much smaller than $7/12$; and the probability of getting more than one red ball for multiple draws without replacement cannot possibly exceed $1/6$.
– heropup
Jun 19 '14 at 6:02
1
Perhaps he means the probability that all 10 red balls are selected before all 20 greens or all 30 blues? Not that the probability of that happening is necessarily 7/12.
– Avraham
Jun 19 '14 at 6:05
add a comment |
2
Your claimed answer makes no sense. Even if we draw only one ball out of the $60$ total, the probability that this ball is red will be only $10/60 = 1/6$, which is much smaller than $7/12$; and the probability of getting more than one red ball for multiple draws without replacement cannot possibly exceed $1/6$.
– heropup
Jun 19 '14 at 6:02
1
Perhaps he means the probability that all 10 red balls are selected before all 20 greens or all 30 blues? Not that the probability of that happening is necessarily 7/12.
– Avraham
Jun 19 '14 at 6:05
2
2
Your claimed answer makes no sense. Even if we draw only one ball out of the $60$ total, the probability that this ball is red will be only $10/60 = 1/6$, which is much smaller than $7/12$; and the probability of getting more than one red ball for multiple draws without replacement cannot possibly exceed $1/6$.
– heropup
Jun 19 '14 at 6:02
Your claimed answer makes no sense. Even if we draw only one ball out of the $60$ total, the probability that this ball is red will be only $10/60 = 1/6$, which is much smaller than $7/12$; and the probability of getting more than one red ball for multiple draws without replacement cannot possibly exceed $1/6$.
– heropup
Jun 19 '14 at 6:02
1
1
Perhaps he means the probability that all 10 red balls are selected before all 20 greens or all 30 blues? Not that the probability of that happening is necessarily 7/12.
– Avraham
Jun 19 '14 at 6:05
Perhaps he means the probability that all 10 red balls are selected before all 20 greens or all 30 blues? Not that the probability of that happening is necessarily 7/12.
– Avraham
Jun 19 '14 at 6:05
add a comment |
3 Answers
3
active
oldest
votes
up vote
12
down vote
accepted
Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn.
The probability that the last ball is blue and that the last green comes after the last red is $dfrac{30}{10+20+30}times dfrac{20}{10+20} =dfrac{1}{3}$ or more generally $b times dfrac{g}{1-b}$.
The probability that the last ball is green and that the last blue comes after the last red is $dfrac{20}{10+20+30}times dfrac{30}{10+30} =dfrac{1}{4}$ or more generally $g times dfrac{b}{1-g}$.
So the probability that all reds are drawn before the final blue and final green are drawn is $dfrac{1}{3}+dfrac{1}{4}=dfrac{7}{12}$ or more generally $ dfrac{bg}{1-b} + dfrac{bg}{1-g}$.
add a comment |
up vote
3
down vote
Edit
The following gives the probability of draw the all red balls before any ball of other colour. It is not the asked answer and it is due to a misunderstanding.
There is only one possibility to choose all $10$ red balls if we draw $10$ ball. The number of possible ways to draw $10$ balls is $binom{60}{10}.$ Thus the probability is
$$frac{1}{binom{60}{10}}approx 1.3263 cdot 10^{-11}.$$
this is not the right answer.
– LCFactorization
Jun 19 '14 at 7:58
1
It is the answer to a different question, answering "what is the probability that the last red ball is drawn before any blue balls or green balls are drawn". The original question is closer to "what is the probability that the last red ball is drawn before the last blue ball and last green ball are drawn".
– Henry
Jun 19 '14 at 8:41
You are right, I misunderstood the question.
– mfl
Jun 19 '14 at 12:25
I edited the question to make it clearer.
– Avraham
Jun 19 '14 at 16:30
add a comment |
up vote
1
down vote
You might be able to prove it by induction. Let $P(R,B,G)$ be the probability that reds are the first to go given $R$ reds, $B$ blues and $G$ greens. It is easy to check your formula for $P(1,1,1)$. Note also that $P(R,B,0)=0$.
Now suppose your formula is true for $P(R,B,G)$ whenever $R+B+Gle N$. Then $$P(R+1,B,G)=frac{R+1}{R+1+B+G}P(R,B,G)+frac{B}{R+1+B+G}P(R+1,B-1,G)+frac{G}{R+1+B+G}P(R+1,B,G-1)$$
and now you can check the formula is true when $R+B+G=N+1$.
Your "note that $P(R, B, 0) = 0$" should start all alarms ringing: as OP isn't interested in blue/green balls, their colors are irrelevant, and $P(R, B, G) = P(R, B + G, 0)$
– vonbrand
Jun 19 '14 at 8:01
1
No, I think that if you have run out of green balls, then you didn't run out of red first. In the same way, I think $P(0,B,G)=1$
– Empy2
Jun 19 '14 at 9:32
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn.
The probability that the last ball is blue and that the last green comes after the last red is $dfrac{30}{10+20+30}times dfrac{20}{10+20} =dfrac{1}{3}$ or more generally $b times dfrac{g}{1-b}$.
The probability that the last ball is green and that the last blue comes after the last red is $dfrac{20}{10+20+30}times dfrac{30}{10+30} =dfrac{1}{4}$ or more generally $g times dfrac{b}{1-g}$.
So the probability that all reds are drawn before the final blue and final green are drawn is $dfrac{1}{3}+dfrac{1}{4}=dfrac{7}{12}$ or more generally $ dfrac{bg}{1-b} + dfrac{bg}{1-g}$.
add a comment |
up vote
12
down vote
accepted
Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn.
The probability that the last ball is blue and that the last green comes after the last red is $dfrac{30}{10+20+30}times dfrac{20}{10+20} =dfrac{1}{3}$ or more generally $b times dfrac{g}{1-b}$.
The probability that the last ball is green and that the last blue comes after the last red is $dfrac{20}{10+20+30}times dfrac{30}{10+30} =dfrac{1}{4}$ or more generally $g times dfrac{b}{1-g}$.
So the probability that all reds are drawn before the final blue and final green are drawn is $dfrac{1}{3}+dfrac{1}{4}=dfrac{7}{12}$ or more generally $ dfrac{bg}{1-b} + dfrac{bg}{1-g}$.
add a comment |
up vote
12
down vote
accepted
up vote
12
down vote
accepted
Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn.
The probability that the last ball is blue and that the last green comes after the last red is $dfrac{30}{10+20+30}times dfrac{20}{10+20} =dfrac{1}{3}$ or more generally $b times dfrac{g}{1-b}$.
The probability that the last ball is green and that the last blue comes after the last red is $dfrac{20}{10+20+30}times dfrac{30}{10+30} =dfrac{1}{4}$ or more generally $g times dfrac{b}{1-g}$.
So the probability that all reds are drawn before the final blue and final green are drawn is $dfrac{1}{3}+dfrac{1}{4}=dfrac{7}{12}$ or more generally $ dfrac{bg}{1-b} + dfrac{bg}{1-g}$.
Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn.
The probability that the last ball is blue and that the last green comes after the last red is $dfrac{30}{10+20+30}times dfrac{20}{10+20} =dfrac{1}{3}$ or more generally $b times dfrac{g}{1-b}$.
The probability that the last ball is green and that the last blue comes after the last red is $dfrac{20}{10+20+30}times dfrac{30}{10+30} =dfrac{1}{4}$ or more generally $g times dfrac{b}{1-g}$.
So the probability that all reds are drawn before the final blue and final green are drawn is $dfrac{1}{3}+dfrac{1}{4}=dfrac{7}{12}$ or more generally $ dfrac{bg}{1-b} + dfrac{bg}{1-g}$.
answered Jun 19 '14 at 8:29
Henry
97k474154
97k474154
add a comment |
add a comment |
up vote
3
down vote
Edit
The following gives the probability of draw the all red balls before any ball of other colour. It is not the asked answer and it is due to a misunderstanding.
There is only one possibility to choose all $10$ red balls if we draw $10$ ball. The number of possible ways to draw $10$ balls is $binom{60}{10}.$ Thus the probability is
$$frac{1}{binom{60}{10}}approx 1.3263 cdot 10^{-11}.$$
this is not the right answer.
– LCFactorization
Jun 19 '14 at 7:58
1
It is the answer to a different question, answering "what is the probability that the last red ball is drawn before any blue balls or green balls are drawn". The original question is closer to "what is the probability that the last red ball is drawn before the last blue ball and last green ball are drawn".
– Henry
Jun 19 '14 at 8:41
You are right, I misunderstood the question.
– mfl
Jun 19 '14 at 12:25
I edited the question to make it clearer.
– Avraham
Jun 19 '14 at 16:30
add a comment |
up vote
3
down vote
Edit
The following gives the probability of draw the all red balls before any ball of other colour. It is not the asked answer and it is due to a misunderstanding.
There is only one possibility to choose all $10$ red balls if we draw $10$ ball. The number of possible ways to draw $10$ balls is $binom{60}{10}.$ Thus the probability is
$$frac{1}{binom{60}{10}}approx 1.3263 cdot 10^{-11}.$$
this is not the right answer.
– LCFactorization
Jun 19 '14 at 7:58
1
It is the answer to a different question, answering "what is the probability that the last red ball is drawn before any blue balls or green balls are drawn". The original question is closer to "what is the probability that the last red ball is drawn before the last blue ball and last green ball are drawn".
– Henry
Jun 19 '14 at 8:41
You are right, I misunderstood the question.
– mfl
Jun 19 '14 at 12:25
I edited the question to make it clearer.
– Avraham
Jun 19 '14 at 16:30
add a comment |
up vote
3
down vote
up vote
3
down vote
Edit
The following gives the probability of draw the all red balls before any ball of other colour. It is not the asked answer and it is due to a misunderstanding.
There is only one possibility to choose all $10$ red balls if we draw $10$ ball. The number of possible ways to draw $10$ balls is $binom{60}{10}.$ Thus the probability is
$$frac{1}{binom{60}{10}}approx 1.3263 cdot 10^{-11}.$$
Edit
The following gives the probability of draw the all red balls before any ball of other colour. It is not the asked answer and it is due to a misunderstanding.
There is only one possibility to choose all $10$ red balls if we draw $10$ ball. The number of possible ways to draw $10$ balls is $binom{60}{10}.$ Thus the probability is
$$frac{1}{binom{60}{10}}approx 1.3263 cdot 10^{-11}.$$
edited Jun 19 '14 at 12:28
answered Jun 19 '14 at 6:09
mfl
26k12141
26k12141
this is not the right answer.
– LCFactorization
Jun 19 '14 at 7:58
1
It is the answer to a different question, answering "what is the probability that the last red ball is drawn before any blue balls or green balls are drawn". The original question is closer to "what is the probability that the last red ball is drawn before the last blue ball and last green ball are drawn".
– Henry
Jun 19 '14 at 8:41
You are right, I misunderstood the question.
– mfl
Jun 19 '14 at 12:25
I edited the question to make it clearer.
– Avraham
Jun 19 '14 at 16:30
add a comment |
this is not the right answer.
– LCFactorization
Jun 19 '14 at 7:58
1
It is the answer to a different question, answering "what is the probability that the last red ball is drawn before any blue balls or green balls are drawn". The original question is closer to "what is the probability that the last red ball is drawn before the last blue ball and last green ball are drawn".
– Henry
Jun 19 '14 at 8:41
You are right, I misunderstood the question.
– mfl
Jun 19 '14 at 12:25
I edited the question to make it clearer.
– Avraham
Jun 19 '14 at 16:30
this is not the right answer.
– LCFactorization
Jun 19 '14 at 7:58
this is not the right answer.
– LCFactorization
Jun 19 '14 at 7:58
1
1
It is the answer to a different question, answering "what is the probability that the last red ball is drawn before any blue balls or green balls are drawn". The original question is closer to "what is the probability that the last red ball is drawn before the last blue ball and last green ball are drawn".
– Henry
Jun 19 '14 at 8:41
It is the answer to a different question, answering "what is the probability that the last red ball is drawn before any blue balls or green balls are drawn". The original question is closer to "what is the probability that the last red ball is drawn before the last blue ball and last green ball are drawn".
– Henry
Jun 19 '14 at 8:41
You are right, I misunderstood the question.
– mfl
Jun 19 '14 at 12:25
You are right, I misunderstood the question.
– mfl
Jun 19 '14 at 12:25
I edited the question to make it clearer.
– Avraham
Jun 19 '14 at 16:30
I edited the question to make it clearer.
– Avraham
Jun 19 '14 at 16:30
add a comment |
up vote
1
down vote
You might be able to prove it by induction. Let $P(R,B,G)$ be the probability that reds are the first to go given $R$ reds, $B$ blues and $G$ greens. It is easy to check your formula for $P(1,1,1)$. Note also that $P(R,B,0)=0$.
Now suppose your formula is true for $P(R,B,G)$ whenever $R+B+Gle N$. Then $$P(R+1,B,G)=frac{R+1}{R+1+B+G}P(R,B,G)+frac{B}{R+1+B+G}P(R+1,B-1,G)+frac{G}{R+1+B+G}P(R+1,B,G-1)$$
and now you can check the formula is true when $R+B+G=N+1$.
Your "note that $P(R, B, 0) = 0$" should start all alarms ringing: as OP isn't interested in blue/green balls, their colors are irrelevant, and $P(R, B, G) = P(R, B + G, 0)$
– vonbrand
Jun 19 '14 at 8:01
1
No, I think that if you have run out of green balls, then you didn't run out of red first. In the same way, I think $P(0,B,G)=1$
– Empy2
Jun 19 '14 at 9:32
add a comment |
up vote
1
down vote
You might be able to prove it by induction. Let $P(R,B,G)$ be the probability that reds are the first to go given $R$ reds, $B$ blues and $G$ greens. It is easy to check your formula for $P(1,1,1)$. Note also that $P(R,B,0)=0$.
Now suppose your formula is true for $P(R,B,G)$ whenever $R+B+Gle N$. Then $$P(R+1,B,G)=frac{R+1}{R+1+B+G}P(R,B,G)+frac{B}{R+1+B+G}P(R+1,B-1,G)+frac{G}{R+1+B+G}P(R+1,B,G-1)$$
and now you can check the formula is true when $R+B+G=N+1$.
Your "note that $P(R, B, 0) = 0$" should start all alarms ringing: as OP isn't interested in blue/green balls, their colors are irrelevant, and $P(R, B, G) = P(R, B + G, 0)$
– vonbrand
Jun 19 '14 at 8:01
1
No, I think that if you have run out of green balls, then you didn't run out of red first. In the same way, I think $P(0,B,G)=1$
– Empy2
Jun 19 '14 at 9:32
add a comment |
up vote
1
down vote
up vote
1
down vote
You might be able to prove it by induction. Let $P(R,B,G)$ be the probability that reds are the first to go given $R$ reds, $B$ blues and $G$ greens. It is easy to check your formula for $P(1,1,1)$. Note also that $P(R,B,0)=0$.
Now suppose your formula is true for $P(R,B,G)$ whenever $R+B+Gle N$. Then $$P(R+1,B,G)=frac{R+1}{R+1+B+G}P(R,B,G)+frac{B}{R+1+B+G}P(R+1,B-1,G)+frac{G}{R+1+B+G}P(R+1,B,G-1)$$
and now you can check the formula is true when $R+B+G=N+1$.
You might be able to prove it by induction. Let $P(R,B,G)$ be the probability that reds are the first to go given $R$ reds, $B$ blues and $G$ greens. It is easy to check your formula for $P(1,1,1)$. Note also that $P(R,B,0)=0$.
Now suppose your formula is true for $P(R,B,G)$ whenever $R+B+Gle N$. Then $$P(R+1,B,G)=frac{R+1}{R+1+B+G}P(R,B,G)+frac{B}{R+1+B+G}P(R+1,B-1,G)+frac{G}{R+1+B+G}P(R+1,B,G-1)$$
and now you can check the formula is true when $R+B+G=N+1$.
edited Jun 19 '14 at 6:45
answered Jun 19 '14 at 6:25
Empy2
33.3k12261
33.3k12261
Your "note that $P(R, B, 0) = 0$" should start all alarms ringing: as OP isn't interested in blue/green balls, their colors are irrelevant, and $P(R, B, G) = P(R, B + G, 0)$
– vonbrand
Jun 19 '14 at 8:01
1
No, I think that if you have run out of green balls, then you didn't run out of red first. In the same way, I think $P(0,B,G)=1$
– Empy2
Jun 19 '14 at 9:32
add a comment |
Your "note that $P(R, B, 0) = 0$" should start all alarms ringing: as OP isn't interested in blue/green balls, their colors are irrelevant, and $P(R, B, G) = P(R, B + G, 0)$
– vonbrand
Jun 19 '14 at 8:01
1
No, I think that if you have run out of green balls, then you didn't run out of red first. In the same way, I think $P(0,B,G)=1$
– Empy2
Jun 19 '14 at 9:32
Your "note that $P(R, B, 0) = 0$" should start all alarms ringing: as OP isn't interested in blue/green balls, their colors are irrelevant, and $P(R, B, G) = P(R, B + G, 0)$
– vonbrand
Jun 19 '14 at 8:01
Your "note that $P(R, B, 0) = 0$" should start all alarms ringing: as OP isn't interested in blue/green balls, their colors are irrelevant, and $P(R, B, G) = P(R, B + G, 0)$
– vonbrand
Jun 19 '14 at 8:01
1
1
No, I think that if you have run out of green balls, then you didn't run out of red first. In the same way, I think $P(0,B,G)=1$
– Empy2
Jun 19 '14 at 9:32
No, I think that if you have run out of green balls, then you didn't run out of red first. In the same way, I think $P(0,B,G)=1$
– Empy2
Jun 19 '14 at 9:32
add a comment |
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2
Your claimed answer makes no sense. Even if we draw only one ball out of the $60$ total, the probability that this ball is red will be only $10/60 = 1/6$, which is much smaller than $7/12$; and the probability of getting more than one red ball for multiple draws without replacement cannot possibly exceed $1/6$.
– heropup
Jun 19 '14 at 6:02
1
Perhaps he means the probability that all 10 red balls are selected before all 20 greens or all 30 blues? Not that the probability of that happening is necessarily 7/12.
– Avraham
Jun 19 '14 at 6:05