Mixing
Probability of a specific outcome while taking balls out of a bag
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You have a bag with 8 white and 4 red balls. You pick out 5 balls from the bag without replacement. What is the probability of getting 2 red and 3 white balls?
I've a doubt with a specific approach in solving this problem.
Method 1:
Since I don't care about the ordering of the specific outcome, I can find the probability as 8C3*4C2/12C5 = 14/33
Method 2:
Lets take a particular ordering of the desired outcome.
WRWRW
The probability of this outcome would be (8*4*7*3*6)/(12*11*10*9*8)
If you observe the numerator, it remains the same for any ordering of the red and white balls. Hence any ordering of 2 red and 3 white balls has equal probability of occurrence. Now total such orderings of 2 red and 3 white balls are 5!/(3!2!) = 10
Hence the final probability is (8*4*7*3*6)/(12*11*10*9*8)*10 = 14/33
Method 3:
Whenever I take out 5 balls, I can the following outcomes. I've listed the type of outcome, the total arrangements possible for every outcome, and total ways of getting x red and y white balls from that bag, i.e.
#. type of outcome ; possible arrangements ; no. of ways from the bag
- 0R5W ; 1 ; 8C5 = 56
- 1R4W ; 5 ; 8C4*4C1 = 280
- 2R3W ; 10 ; 8C3*4C2 = 336
- 3R2W ; 10 ; 8C2*4C3 = 112
- 4R1W ; 1 ; 8C1*4C4 = 8
Now, if I take the final probability as (No. of ways of #3) / (No. of ways for all #) that gives the correct answer i.e. 14/33.
BUT, for lets say, 4R1W, there are 8 ways to get 4R and 1W out of the bag, and for each of those ways, the 4R1W balls could be arranged in 5 ways. So total no. of outcomes where you have 4R1W balls should be 5*8 = 40. Which is basically 8C1*4C4*5!/(4!1!)
So should the probability be the ratio of product of 2nd and 3rd columns, and not just the 3rd column? Why is this incorrect?
probability
asked 18 hours ago
qwerty_uiop
412
add a comment |
up vote
0
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favorite
You have a bag with 8 white and 4 red balls. You pick out 5 balls from the bag without replacement. What is the probability of getting 2 red and 3 white balls?
I've a doubt with a specific approach in solving this problem.
Method 1:
Since I don't care about the ordering of the specific outcome, I can find the probability as 8C3*4C2/12C5 = 14/33
Method 2:
Lets take a particular ordering of the desired outcome.
WRWRW
The probability of this outcome would be (8*4*7*3*6)/(12*11*10*9*8)
If you observe the numerator, it remains the same for any ordering of the red and white balls. Hence any ordering of 2 red and 3 white balls has equal probability of occurrence. Now total such orderings of 2 red and 3 white balls are 5!/(3!2!) = 10
Hence the final probability is (8*4*7*3*6)/(12*11*10*9*8)*10 = 14/33
Method 3:
Whenever I take out 5 balls, I can the following outcomes. I've listed the type of outcome, the total arrangements possible for every outcome, and total ways of getting x red and y white balls from that bag, i.e.
#. type of outcome ; possible arrangements ; no. of ways from the bag
- 0R5W ; 1 ; 8C5 = 56
- 1R4W ; 5 ; 8C4*4C1 = 280
- 2R3W ; 10 ; 8C3*4C2 = 336
- 3R2W ; 10 ; 8C2*4C3 = 112
- 4R1W ; 1 ; 8C1*4C4 = 8
Now, if I take the final probability as (No. of ways of #3) / (No. of ways for all #) that gives the correct answer i.e. 14/33.
BUT, for lets say, 4R1W, there are 8 ways to get 4R and 1W out of the bag, and for each of those ways, the 4R1W balls could be arranged in 5 ways. So total no. of outcomes where you have 4R1W balls should be 5*8 = 40. Which is basically 8C1*4C4*5!/(4!1!)
So should the probability be the ratio of product of 2nd and 3rd columns, and not just the 3rd column? Why is this incorrect?
probability
asked 18 hours ago
qwerty_uiop
412
Are we assuming that all white balls are different and all red balls are different ( like for examples all balls are differently numbered)? If that's the case then I can see why my method 3 would be wrong.
– qwerty_uiop
17 hours ago
Yes, we are. That is indicated by the fact that we discern $binom{12}5$ different outcomes.
– drhab
17 hours ago
add a comment |
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up vote
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down vote
favorite
You have a bag with 8 white and 4 red balls. You pick out 5 balls from the bag without replacement. What is the probability of getting 2 red and 3 white balls?
I've a doubt with a specific approach in solving this problem.
Method 1:
Since I don't care about the ordering of the specific outcome, I can find the probability as 8C3*4C2/12C5 = 14/33
Method 2:
Lets take a particular ordering of the desired outcome.
WRWRW
The probability of this outcome would be (8*4*7*3*6)/(12*11*10*9*8)
If you observe the numerator, it remains the same for any ordering of the red and white balls. Hence any ordering of 2 red and 3 white balls has equal probability of occurrence. Now total such orderings of 2 red and 3 white balls are 5!/(3!2!) = 10
Hence the final probability is (8*4*7*3*6)/(12*11*10*9*8)*10 = 14/33
Method 3:
Whenever I take out 5 balls, I can the following outcomes. I've listed the type of outcome, the total arrangements possible for every outcome, and total ways of getting x red and y white balls from that bag, i.e.
#. type of outcome ; possible arrangements ; no. of ways from the bag
- 0R5W ; 1 ; 8C5 = 56
- 1R4W ; 5 ; 8C4*4C1 = 280
- 2R3W ; 10 ; 8C3*4C2 = 336
- 3R2W ; 10 ; 8C2*4C3 = 112
- 4R1W ; 1 ; 8C1*4C4 = 8
Now, if I take the final probability as (No. of ways of #3) / (No. of ways for all #) that gives the correct answer i.e. 14/33.
BUT, for lets say, 4R1W, there are 8 ways to get 4R and 1W out of the bag, and for each of those ways, the 4R1W balls could be arranged in 5 ways. So total no. of outcomes where you have 4R1W balls should be 5*8 = 40. Which is basically 8C1*4C4*5!/(4!1!)
So should the probability be the ratio of product of 2nd and 3rd columns, and not just the 3rd column? Why is this incorrect?
probability
asked 18 hours ago
qwerty_uiop
412
You have a bag with 8 white and 4 red balls. You pick out 5 balls from the bag without replacement. What is the probability of getting 2 red and 3 white balls?
I've a doubt with a specific approach in solving this problem.
Method 1:
Since I don't care about the ordering of the specific outcome, I can find the probability as 8C3*4C2/12C5 = 14/33
Method 2:
Lets take a particular ordering of the desired outcome.
WRWRW
The probability of this outcome would be (8*4*7*3*6)/(12*11*10*9*8)
If you observe the numerator, it remains the same for any ordering of the red and white balls. Hence any ordering of 2 red and 3 white balls has equal probability of occurrence. Now total such orderings of 2 red and 3 white balls are 5!/(3!2!) = 10
Hence the final probability is (8*4*7*3*6)/(12*11*10*9*8)*10 = 14/33
Method 3:
Whenever I take out 5 balls, I can the following outcomes. I've listed the type of outcome, the total arrangements possible for every outcome, and total ways of getting x red and y white balls from that bag, i.e.
#. type of outcome ; possible arrangements ; no. of ways from the bag
- 0R5W ; 1 ; 8C5 = 56
- 1R4W ; 5 ; 8C4*4C1 = 280
- 2R3W ; 10 ; 8C3*4C2 = 336
- 3R2W ; 10 ; 8C2*4C3 = 112
- 4R1W ; 1 ; 8C1*4C4 = 8
Now, if I take the final probability as (No. of ways of #3) / (No. of ways for all #) that gives the correct answer i.e. 14/33.
BUT, for lets say, 4R1W, there are 8 ways to get 4R and 1W out of the bag, and for each of those ways, the 4R1W balls could be arranged in 5 ways. So total no. of outcomes where you have 4R1W balls should be 5*8 = 40. Which is basically 8C1*4C4*5!/(4!1!)
So should the probability be the ratio of product of 2nd and 3rd columns, and not just the 3rd column? Why is this incorrect?
probability
probability
asked 18 hours ago
qwerty_uiop
412
asked 18 hours ago
qwerty_uiop
412
asked 18 hours ago
qwerty_uiop
412
asked 18 hours ago
qwerty_uiop
412
asked 18 hours ago
qwerty_uiop
412
412
Are we assuming that all white balls are different and all red balls are different ( like for examples all balls are differently numbered)? If that's the case then I can see why my method 3 would be wrong.
– qwerty_uiop
17 hours ago
Yes, we are. That is indicated by the fact that we discern $binom{12}5$ different outcomes.
– drhab
17 hours ago
add a comment |
Are we assuming that all white balls are different and all red balls are different ( like for examples all balls are differently numbered)? If that's the case then I can see why my method 3 would be wrong.
– qwerty_uiop
17 hours ago
Yes, we are. That is indicated by the fact that we discern $binom{12}5$ different outcomes.
– drhab
17 hours ago
Are we assuming that all white balls are different and all red balls are different ( like for examples all balls are differently numbered)? If that's the case then I can see why my method 3 would be wrong.
– qwerty_uiop
17 hours ago
Are we assuming that all white balls are different and all red balls are different ( like for examples all balls are differently numbered)? If that's the case then I can see why my method 3 would be wrong.
– qwerty_uiop
17 hours ago
Yes, we are. That is indicated by the fact that we discern $binom{12}5$ different outcomes.
– drhab
17 hours ago
Yes, we are. That is indicated by the fact that we discern $binom{12}5$ different outcomes.
– drhab
17 hours ago
add a comment |
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Are we assuming that all white balls are different and all red balls are different ( like for examples all balls are differently numbered)? If that's the case then I can see why my method 3 would be wrong.
– qwerty_uiop
17 hours ago
Yes, we are. That is indicated by the fact that we discern $binom{12}5$ different outcomes.
– drhab
17 hours ago