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Probability of drawing unknown color at the beginning and at the end of nth draw
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How many balls will I have to draw?Probability of drawing at least one red and at least one green ball.Probability of drawing at least 1 red, 1 blue, 1 green, 1 white, 1 black, and 1 grey when drawing 8 balls from a pool of 30?Probability of drawing balls without replacement in first and last drawMean number of balls drawnProbability of drawing the same color marble twice in a bag of $10$ red, $10$ orange, $10$ green.Probability of draw 1000 balls of the same color from 2000Probability of drawing balls of unknown coloursProbability of drawing same ball until draw 4 or 'greater'?Probability of Drawing Two Balls of Different Color in Second Draw.
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I am not sure how to handle the unknown color situation. If the color of the first ball was known then I would have tried in a way (showing below, not sure I am correct but that's my initial approach.
Actual question:
You have 4 red balls, 8 green balls, and 10 purple balls. You draw a ball from the bag, record its color. Continue drawing without replacement until you get a ball that matches the color of the first ball. What is the probability that it takes exactly 3 additional draws (not including draw 0) to get a matching ball?
Say I knew the color and it was red then (again, I could be wrong):
pr = (8+10)/(4+8+10-1) * (8+10-1)/(4+8+10-2) * (3/(4+8+10-3))
I really appreciate any direction/explanation that anyone can provide
probability statistics
asked Mar 26 at 0:25
Musa BiraloMusa Biralo
31
$endgroup$
add a comment |
$begingroup$
I am not sure how to handle the unknown color situation. If the color of the first ball was known then I would have tried in a way (showing below, not sure I am correct but that's my initial approach.
Actual question:
You have 4 red balls, 8 green balls, and 10 purple balls. You draw a ball from the bag, record its color. Continue drawing without replacement until you get a ball that matches the color of the first ball. What is the probability that it takes exactly 3 additional draws (not including draw 0) to get a matching ball?
Say I knew the color and it was red then (again, I could be wrong):
pr = (8+10)/(4+8+10-1) * (8+10-1)/(4+8+10-2) * (3/(4+8+10-3))
I really appreciate any direction/explanation that anyone can provide
probability statistics
asked Mar 26 at 0:25
Musa BiraloMusa Biralo
31
$endgroup$
$begingroup$
Hint: I guess you can do similar computations for green and purple, right?
$endgroup$
– Ertxiem
Mar 26 at 0:35
add a comment |
$begingroup$
I am not sure how to handle the unknown color situation. If the color of the first ball was known then I would have tried in a way (showing below, not sure I am correct but that's my initial approach.
Actual question:
You have 4 red balls, 8 green balls, and 10 purple balls. You draw a ball from the bag, record its color. Continue drawing without replacement until you get a ball that matches the color of the first ball. What is the probability that it takes exactly 3 additional draws (not including draw 0) to get a matching ball?
Say I knew the color and it was red then (again, I could be wrong):
pr = (8+10)/(4+8+10-1) * (8+10-1)/(4+8+10-2) * (3/(4+8+10-3))
I really appreciate any direction/explanation that anyone can provide
probability statistics
asked Mar 26 at 0:25
Musa BiraloMusa Biralo
31
$endgroup$
I am not sure how to handle the unknown color situation. If the color of the first ball was known then I would have tried in a way (showing below, not sure I am correct but that's my initial approach.
Actual question:
You have 4 red balls, 8 green balls, and 10 purple balls. You draw a ball from the bag, record its color. Continue drawing without replacement until you get a ball that matches the color of the first ball. What is the probability that it takes exactly 3 additional draws (not including draw 0) to get a matching ball?
Say I knew the color and it was red then (again, I could be wrong):
pr = (8+10)/(4+8+10-1) * (8+10-1)/(4+8+10-2) * (3/(4+8+10-3))
I really appreciate any direction/explanation that anyone can provide
probability statistics
probability statistics
asked Mar 26 at 0:25
Musa BiraloMusa Biralo
31
asked Mar 26 at 0:25
Musa BiraloMusa Biralo
31
asked Mar 26 at 0:25
Musa BiraloMusa Biralo
31
asked Mar 26 at 0:25
Musa BiraloMusa Biralo
31
asked Mar 26 at 0:25
Musa BiraloMusa Biralo
31
31
$begingroup$
Hint: I guess you can do similar computations for green and purple, right?
$endgroup$
– Ertxiem
Mar 26 at 0:35
add a comment |
$begingroup$
Hint: I guess you can do similar computations for green and purple, right?
$endgroup$
– Ertxiem
Mar 26 at 0:35
$begingroup$
Hint: I guess you can do similar computations for green and purple, right?
$endgroup$
– Ertxiem
Mar 26 at 0:35
$begingroup$
Hint: I guess you can do similar computations for green and purple, right?
$endgroup$
– Ertxiem
Mar 26 at 0:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It may not be efficient but:
The probability of the first ball being Red is $frac 4{22}$. If the first ball is red the probability of the next two not being red is $frac {18*17}{21*20}$ and if that happens the probability of the fourth ball being red is $frac 3{19}$. So the probability of the first ball being red and my getting a second red ball exactly three balls later is $frac 4{22}frac {18*17}{21*20}frac 3{19}$.
Doing the same for if the first ball is green we can $frac 8{22}$ and then $frac {14*13}{21*20}$ and $frac 7{19}$. So the prob of that is $frac {8*14*13*7}{22*...*19}$.
And for the first ball being purple we have $frac {10}{22}frac {12*11}{21*20}frac 9{19}$
So the probability is
$frac {4*18*17*3}{22*...*19} + frac{8*14*13*7}{22*..*19}+ frac {10*12*11*9}{22*...*19}$.
answered Mar 26 at 0:37
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
Suppose the first ball was red. This occurs with probability $frac{4}{22}$ as there are $4$ red balls out of $4+8+10=22$ balls total.
Given that the first ball was red, in order for the fourth ball to be the first re-occurrence of a red ball, that means that the next two draws must both be non-red. As such the second ball being non-red given that the first is red will occur with probability $frac{18}{21}$, and given that the first is red and second is non-red the third ball also being non-red will occur with probability $frac{17}{20}$. Finally, given the first is red and the next two are non-red the probability that the fourth ball is red will be $frac{3}{19}$
This gives the probability that the first is red and the first re-occurrence of red occurs on the fourth draw as being $frac{4}{22}timesfrac{18}{21}timesfrac{17}{20}timesfrac{3}{19}$
Continue in the same fashion for if the first ball happened to be green or purple and add the results.
answered Mar 26 at 0:38
JMoravitzJMoravitz
49.4k44091
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
It may not be efficient but:
The probability of the first ball being Red is $frac 4{22}$. If the first ball is red the probability of the next two not being red is $frac {18*17}{21*20}$ and if that happens the probability of the fourth ball being red is $frac 3{19}$. So the probability of the first ball being red and my getting a second red ball exactly three balls later is $frac 4{22}frac {18*17}{21*20}frac 3{19}$.
Doing the same for if the first ball is green we can $frac 8{22}$ and then $frac {14*13}{21*20}$ and $frac 7{19}$. So the prob of that is $frac {8*14*13*7}{22*...*19}$.
And for the first ball being purple we have $frac {10}{22}frac {12*11}{21*20}frac 9{19}$
So the probability is
$frac {4*18*17*3}{22*...*19} + frac{8*14*13*7}{22*..*19}+ frac {10*12*11*9}{22*...*19}$.
answered Mar 26 at 0:37
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
It may not be efficient but:
The probability of the first ball being Red is $frac 4{22}$. If the first ball is red the probability of the next two not being red is $frac {18*17}{21*20}$ and if that happens the probability of the fourth ball being red is $frac 3{19}$. So the probability of the first ball being red and my getting a second red ball exactly three balls later is $frac 4{22}frac {18*17}{21*20}frac 3{19}$.
Doing the same for if the first ball is green we can $frac 8{22}$ and then $frac {14*13}{21*20}$ and $frac 7{19}$. So the prob of that is $frac {8*14*13*7}{22*...*19}$.
And for the first ball being purple we have $frac {10}{22}frac {12*11}{21*20}frac 9{19}$
So the probability is
$frac {4*18*17*3}{22*...*19} + frac{8*14*13*7}{22*..*19}+ frac {10*12*11*9}{22*...*19}$.
answered Mar 26 at 0:37
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
It may not be efficient but:
The probability of the first ball being Red is $frac 4{22}$. If the first ball is red the probability of the next two not being red is $frac {18*17}{21*20}$ and if that happens the probability of the fourth ball being red is $frac 3{19}$. So the probability of the first ball being red and my getting a second red ball exactly three balls later is $frac 4{22}frac {18*17}{21*20}frac 3{19}$.
Doing the same for if the first ball is green we can $frac 8{22}$ and then $frac {14*13}{21*20}$ and $frac 7{19}$. So the prob of that is $frac {8*14*13*7}{22*...*19}$.
And for the first ball being purple we have $frac {10}{22}frac {12*11}{21*20}frac 9{19}$
So the probability is
$frac {4*18*17*3}{22*...*19} + frac{8*14*13*7}{22*..*19}+ frac {10*12*11*9}{22*...*19}$.
answered Mar 26 at 0:37
fleabloodfleablood
1
$endgroup$
It may not be efficient but:
The probability of the first ball being Red is $frac 4{22}$. If the first ball is red the probability of the next two not being red is $frac {18*17}{21*20}$ and if that happens the probability of the fourth ball being red is $frac 3{19}$. So the probability of the first ball being red and my getting a second red ball exactly three balls later is $frac 4{22}frac {18*17}{21*20}frac 3{19}$.
Doing the same for if the first ball is green we can $frac 8{22}$ and then $frac {14*13}{21*20}$ and $frac 7{19}$. So the prob of that is $frac {8*14*13*7}{22*...*19}$.
And for the first ball being purple we have $frac {10}{22}frac {12*11}{21*20}frac 9{19}$
So the probability is
$frac {4*18*17*3}{22*...*19} + frac{8*14*13*7}{22*..*19}+ frac {10*12*11*9}{22*...*19}$.
answered Mar 26 at 0:37
fleabloodfleablood
1
answered Mar 26 at 0:37
fleabloodfleablood
1
answered Mar 26 at 0:37
fleabloodfleablood
1
answered Mar 26 at 0:37
fleabloodfleablood
1
1
add a comment |
add a comment |
$begingroup$
Suppose the first ball was red. This occurs with probability $frac{4}{22}$ as there are $4$ red balls out of $4+8+10=22$ balls total.
Given that the first ball was red, in order for the fourth ball to be the first re-occurrence of a red ball, that means that the next two draws must both be non-red. As such the second ball being non-red given that the first is red will occur with probability $frac{18}{21}$, and given that the first is red and second is non-red the third ball also being non-red will occur with probability $frac{17}{20}$. Finally, given the first is red and the next two are non-red the probability that the fourth ball is red will be $frac{3}{19}$
This gives the probability that the first is red and the first re-occurrence of red occurs on the fourth draw as being $frac{4}{22}timesfrac{18}{21}timesfrac{17}{20}timesfrac{3}{19}$
Continue in the same fashion for if the first ball happened to be green or purple and add the results.
answered Mar 26 at 0:38
JMoravitzJMoravitz
49.4k44091
$endgroup$
add a comment |
$begingroup$
Suppose the first ball was red. This occurs with probability $frac{4}{22}$ as there are $4$ red balls out of $4+8+10=22$ balls total.
Given that the first ball was red, in order for the fourth ball to be the first re-occurrence of a red ball, that means that the next two draws must both be non-red. As such the second ball being non-red given that the first is red will occur with probability $frac{18}{21}$, and given that the first is red and second is non-red the third ball also being non-red will occur with probability $frac{17}{20}$. Finally, given the first is red and the next two are non-red the probability that the fourth ball is red will be $frac{3}{19}$
This gives the probability that the first is red and the first re-occurrence of red occurs on the fourth draw as being $frac{4}{22}timesfrac{18}{21}timesfrac{17}{20}timesfrac{3}{19}$
Continue in the same fashion for if the first ball happened to be green or purple and add the results.
answered Mar 26 at 0:38
JMoravitzJMoravitz
49.4k44091
$endgroup$
add a comment |
$begingroup$
Suppose the first ball was red. This occurs with probability $frac{4}{22}$ as there are $4$ red balls out of $4+8+10=22$ balls total.
Given that the first ball was red, in order for the fourth ball to be the first re-occurrence of a red ball, that means that the next two draws must both be non-red. As such the second ball being non-red given that the first is red will occur with probability $frac{18}{21}$, and given that the first is red and second is non-red the third ball also being non-red will occur with probability $frac{17}{20}$. Finally, given the first is red and the next two are non-red the probability that the fourth ball is red will be $frac{3}{19}$
This gives the probability that the first is red and the first re-occurrence of red occurs on the fourth draw as being $frac{4}{22}timesfrac{18}{21}timesfrac{17}{20}timesfrac{3}{19}$
Continue in the same fashion for if the first ball happened to be green or purple and add the results.
answered Mar 26 at 0:38
JMoravitzJMoravitz
49.4k44091
$endgroup$
Suppose the first ball was red. This occurs with probability $frac{4}{22}$ as there are $4$ red balls out of $4+8+10=22$ balls total.
Given that the first ball was red, in order for the fourth ball to be the first re-occurrence of a red ball, that means that the next two draws must both be non-red. As such the second ball being non-red given that the first is red will occur with probability $frac{18}{21}$, and given that the first is red and second is non-red the third ball also being non-red will occur with probability $frac{17}{20}$. Finally, given the first is red and the next two are non-red the probability that the fourth ball is red will be $frac{3}{19}$
This gives the probability that the first is red and the first re-occurrence of red occurs on the fourth draw as being $frac{4}{22}timesfrac{18}{21}timesfrac{17}{20}timesfrac{3}{19}$
Continue in the same fashion for if the first ball happened to be green or purple and add the results.
answered Mar 26 at 0:38
JMoravitzJMoravitz
49.4k44091
answered Mar 26 at 0:38
JMoravitzJMoravitz
49.4k44091
answered Mar 26 at 0:38
JMoravitzJMoravitz
49.4k44091
answered Mar 26 at 0:38
JMoravitzJMoravitz
49.4k44091
49.4k44091
add a comment |
add a comment |
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$begingroup$
Hint: I guess you can do similar computations for green and purple, right?
$endgroup$
– Ertxiem
Mar 26 at 0:35