Mixing
Lottery Probability Confusion
$begingroup$
(A) Estimate the probability of getting $4$ out of $6$ numbers correct in a lottery system that uses $49$ white balls numbered $1$ through $49.$
(B) If a red "powerball" is chosen from a group of $10$ balls numbered $0-9$ is added to part A (after picking the $6$ from the group of $49$),
what are the odds of getting $4$ out of $6$ numbers correct and then the correct powerball?
Thank you for your help.
PS. I got (a) $.000969$ or $1$ out of $1032.4$
and (b) $.000097$ or $1$ out of $10324.$
But I have A as marked wrong on my paper.
This is what I did:
Ways to get $4$ of $6$ numbers correct $C(6,4)=largefrac{6!}{4!2!}$
Ways to get $2$ incorrect $C(43,2)=903.$
All possible combinations to win $C(49,6)=13983816.$
Then Probability of get $4$ out of $6$ of the white balls (there being as of yet no red ball)
$large frac{(C(6,4)* C(43,2)}{C(49,6)}$
For (B) I worked like this:
Probability is $large frac{C(6,4)*C(43,2)*C(1,1)}{C49,6)*C(10,1)}$
with $C(1,1)=1$ and $C(10,1)=10.$
probability
edited Jan 31 at 11:37
Gaby Alfonso
1,1931418
asked Apr 30 '14 at 23:44
watanakewatanake
385
$endgroup$
add a comment |
$begingroup$
(A) Estimate the probability of getting $4$ out of $6$ numbers correct in a lottery system that uses $49$ white balls numbered $1$ through $49.$
(B) If a red "powerball" is chosen from a group of $10$ balls numbered $0-9$ is added to part A (after picking the $6$ from the group of $49$),
what are the odds of getting $4$ out of $6$ numbers correct and then the correct powerball?
Thank you for your help.
PS. I got (a) $.000969$ or $1$ out of $1032.4$
and (b) $.000097$ or $1$ out of $10324.$
But I have A as marked wrong on my paper.
This is what I did:
Ways to get $4$ of $6$ numbers correct $C(6,4)=largefrac{6!}{4!2!}$
Ways to get $2$ incorrect $C(43,2)=903.$
All possible combinations to win $C(49,6)=13983816.$
Then Probability of get $4$ out of $6$ of the white balls (there being as of yet no red ball)
$large frac{(C(6,4)* C(43,2)}{C(49,6)}$
For (B) I worked like this:
Probability is $large frac{C(6,4)*C(43,2)*C(1,1)}{C49,6)*C(10,1)}$
with $C(1,1)=1$ and $C(10,1)=10.$
probability
edited Jan 31 at 11:37
Gaby Alfonso
1,1931418
asked Apr 30 '14 at 23:44
watanakewatanake
385
$endgroup$
$begingroup$
The second is one-tenth of the first, so cannot be $0.00097$. Just a typo.
$endgroup$
– André Nicolas
May 1 '14 at 0:06
$begingroup$
Note that the numbers like $binom{6}{4}$ are not probabilities,they are counts.
$endgroup$
– André Nicolas
May 1 '14 at 0:11
add a comment |
$begingroup$
(A) Estimate the probability of getting $4$ out of $6$ numbers correct in a lottery system that uses $49$ white balls numbered $1$ through $49.$
(B) If a red "powerball" is chosen from a group of $10$ balls numbered $0-9$ is added to part A (after picking the $6$ from the group of $49$),
what are the odds of getting $4$ out of $6$ numbers correct and then the correct powerball?
Thank you for your help.
PS. I got (a) $.000969$ or $1$ out of $1032.4$
and (b) $.000097$ or $1$ out of $10324.$
But I have A as marked wrong on my paper.
This is what I did:
Ways to get $4$ of $6$ numbers correct $C(6,4)=largefrac{6!}{4!2!}$
Ways to get $2$ incorrect $C(43,2)=903.$
All possible combinations to win $C(49,6)=13983816.$
Then Probability of get $4$ out of $6$ of the white balls (there being as of yet no red ball)
$large frac{(C(6,4)* C(43,2)}{C(49,6)}$
For (B) I worked like this:
Probability is $large frac{C(6,4)*C(43,2)*C(1,1)}{C49,6)*C(10,1)}$
with $C(1,1)=1$ and $C(10,1)=10.$
probability
edited Jan 31 at 11:37
Gaby Alfonso
1,1931418
asked Apr 30 '14 at 23:44
watanakewatanake
385
$endgroup$
(A) Estimate the probability of getting $4$ out of $6$ numbers correct in a lottery system that uses $49$ white balls numbered $1$ through $49.$
(B) If a red "powerball" is chosen from a group of $10$ balls numbered $0-9$ is added to part A (after picking the $6$ from the group of $49$),
what are the odds of getting $4$ out of $6$ numbers correct and then the correct powerball?
Thank you for your help.
PS. I got (a) $.000969$ or $1$ out of $1032.4$
and (b) $.000097$ or $1$ out of $10324.$
But I have A as marked wrong on my paper.
This is what I did:
Ways to get $4$ of $6$ numbers correct $C(6,4)=largefrac{6!}{4!2!}$
Ways to get $2$ incorrect $C(43,2)=903.$
All possible combinations to win $C(49,6)=13983816.$
Then Probability of get $4$ out of $6$ of the white balls (there being as of yet no red ball)
$large frac{(C(6,4)* C(43,2)}{C(49,6)}$
For (B) I worked like this:
Probability is $large frac{C(6,4)*C(43,2)*C(1,1)}{C49,6)*C(10,1)}$
with $C(1,1)=1$ and $C(10,1)=10.$
probability
probability
edited Jan 31 at 11:37
Gaby Alfonso
1,1931418
asked Apr 30 '14 at 23:44
watanakewatanake
385
edited Jan 31 at 11:37
Gaby Alfonso
1,1931418
asked Apr 30 '14 at 23:44
watanakewatanake
385
edited Jan 31 at 11:37
Gaby Alfonso
1,1931418
edited Jan 31 at 11:37
Gaby Alfonso
1,1931418
edited Jan 31 at 11:37
Gaby Alfonso
1,1931418
1,1931418
asked Apr 30 '14 at 23:44
watanakewatanake
385
asked Apr 30 '14 at 23:44
watanakewatanake
385
asked Apr 30 '14 at 23:44
watanakewatanake
385
385
$begingroup$
The second is one-tenth of the first, so cannot be $0.00097$. Just a typo.
$endgroup$
– André Nicolas
May 1 '14 at 0:06
$begingroup$
Note that the numbers like $binom{6}{4}$ are not probabilities,they are counts.
$endgroup$
– André Nicolas
May 1 '14 at 0:11
add a comment |
$begingroup$
The second is one-tenth of the first, so cannot be $0.00097$. Just a typo.
$endgroup$
– André Nicolas
May 1 '14 at 0:06
$begingroup$
Note that the numbers like $binom{6}{4}$ are not probabilities,they are counts.
$endgroup$
– André Nicolas
May 1 '14 at 0:11
$begingroup$
The second is one-tenth of the first, so cannot be $0.00097$. Just a typo.
$endgroup$
– André Nicolas
May 1 '14 at 0:06
$begingroup$
The second is one-tenth of the first, so cannot be $0.00097$. Just a typo.
$endgroup$
– André Nicolas
May 1 '14 at 0:06
$begingroup$
Note that the numbers like $binom{6}{4}$ are not probabilities,they are counts.
$endgroup$
– André Nicolas
May 1 '14 at 0:11
$begingroup$
Note that the numbers like $binom{6}{4}$ are not probabilities,they are counts.
$endgroup$
– André Nicolas
May 1 '14 at 0:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your calculation of A is fine, but B the decimal should have one more zero. The chance of B is one tenth the chance of A, as your expression says.
answered May 1 '14 at 0:23
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
I am confused because the decimal already is move over a tenth. I am very sorry, but I think I am missing something here. Thank you very much for your help
$endgroup$
– watanake
May 1 '14 at 0:24
$begingroup$
I see three zeros before the 9 in both cases. The conversion between $0.00097$ and $1$ in $10324$ is not correct.
$endgroup$
– Ross Millikan
May 1 '14 at 0:25
$begingroup$
Thank you, I mistyped that. So is it now correct that you can see?
$endgroup$
– watanake
May 1 '14 at 0:31
$begingroup$
Yes, I agree now.
$endgroup$
– Ross Millikan
May 1 '14 at 0:33
$begingroup$
I appreciate very much this help. Thank you.
$endgroup$
– watanake
May 1 '14 at 0:34
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your calculation of A is fine, but B the decimal should have one more zero. The chance of B is one tenth the chance of A, as your expression says.
answered May 1 '14 at 0:23
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
I am confused because the decimal already is move over a tenth. I am very sorry, but I think I am missing something here. Thank you very much for your help
$endgroup$
– watanake
May 1 '14 at 0:24
$begingroup$
I see three zeros before the 9 in both cases. The conversion between $0.00097$ and $1$ in $10324$ is not correct.
$endgroup$
– Ross Millikan
May 1 '14 at 0:25
$begingroup$
Thank you, I mistyped that. So is it now correct that you can see?
$endgroup$
– watanake
May 1 '14 at 0:31
$begingroup$
Yes, I agree now.
$endgroup$
– Ross Millikan
May 1 '14 at 0:33
$begingroup$
I appreciate very much this help. Thank you.
$endgroup$
– watanake
May 1 '14 at 0:34
add a comment |
$begingroup$
Your calculation of A is fine, but B the decimal should have one more zero. The chance of B is one tenth the chance of A, as your expression says.
answered May 1 '14 at 0:23
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
I am confused because the decimal already is move over a tenth. I am very sorry, but I think I am missing something here. Thank you very much for your help
$endgroup$
– watanake
May 1 '14 at 0:24
$begingroup$
I see three zeros before the 9 in both cases. The conversion between $0.00097$ and $1$ in $10324$ is not correct.
$endgroup$
– Ross Millikan
May 1 '14 at 0:25
$begingroup$
Thank you, I mistyped that. So is it now correct that you can see?
$endgroup$
– watanake
May 1 '14 at 0:31
$begingroup$
Yes, I agree now.
$endgroup$
– Ross Millikan
May 1 '14 at 0:33
$begingroup$
I appreciate very much this help. Thank you.
$endgroup$
– watanake
May 1 '14 at 0:34
add a comment |
$begingroup$
Your calculation of A is fine, but B the decimal should have one more zero. The chance of B is one tenth the chance of A, as your expression says.
answered May 1 '14 at 0:23
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Your calculation of A is fine, but B the decimal should have one more zero. The chance of B is one tenth the chance of A, as your expression says.
answered May 1 '14 at 0:23
Ross MillikanRoss Millikan
301k24200375
answered May 1 '14 at 0:23
Ross MillikanRoss Millikan
301k24200375
answered May 1 '14 at 0:23
Ross MillikanRoss Millikan
301k24200375
answered May 1 '14 at 0:23
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
I am confused because the decimal already is move over a tenth. I am very sorry, but I think I am missing something here. Thank you very much for your help
$endgroup$
– watanake
May 1 '14 at 0:24
$begingroup$
I see three zeros before the 9 in both cases. The conversion between $0.00097$ and $1$ in $10324$ is not correct.
$endgroup$
– Ross Millikan
May 1 '14 at 0:25
$begingroup$
Thank you, I mistyped that. So is it now correct that you can see?
$endgroup$
– watanake
May 1 '14 at 0:31
$begingroup$
Yes, I agree now.
$endgroup$
– Ross Millikan
May 1 '14 at 0:33
$begingroup$
I appreciate very much this help. Thank you.
$endgroup$
– watanake
May 1 '14 at 0:34
add a comment |
$begingroup$
I am confused because the decimal already is move over a tenth. I am very sorry, but I think I am missing something here. Thank you very much for your help
$endgroup$
– watanake
May 1 '14 at 0:24
$begingroup$
I see three zeros before the 9 in both cases. The conversion between $0.00097$ and $1$ in $10324$ is not correct.
$endgroup$
– Ross Millikan
May 1 '14 at 0:25
$begingroup$
Thank you, I mistyped that. So is it now correct that you can see?
$endgroup$
– watanake
May 1 '14 at 0:31
$begingroup$
Yes, I agree now.
$endgroup$
– Ross Millikan
May 1 '14 at 0:33
$begingroup$
I appreciate very much this help. Thank you.
$endgroup$
– watanake
May 1 '14 at 0:34
$begingroup$
I am confused because the decimal already is move over a tenth. I am very sorry, but I think I am missing something here. Thank you very much for your help
$endgroup$
– watanake
May 1 '14 at 0:24
$begingroup$
I am confused because the decimal already is move over a tenth. I am very sorry, but I think I am missing something here. Thank you very much for your help
$endgroup$
– watanake
May 1 '14 at 0:24
$begingroup$
I see three zeros before the 9 in both cases. The conversion between $0.00097$ and $1$ in $10324$ is not correct.
$endgroup$
– Ross Millikan
May 1 '14 at 0:25
$begingroup$
I see three zeros before the 9 in both cases. The conversion between $0.00097$ and $1$ in $10324$ is not correct.
$endgroup$
– Ross Millikan
May 1 '14 at 0:25
$begingroup$
Thank you, I mistyped that. So is it now correct that you can see?
$endgroup$
– watanake
May 1 '14 at 0:31
$begingroup$
Thank you, I mistyped that. So is it now correct that you can see?
$endgroup$
– watanake
May 1 '14 at 0:31
$begingroup$
Yes, I agree now.
$endgroup$
– Ross Millikan
May 1 '14 at 0:33
$begingroup$
Yes, I agree now.
$endgroup$
– Ross Millikan
May 1 '14 at 0:33
$begingroup$
I appreciate very much this help. Thank you.
$endgroup$
– watanake
May 1 '14 at 0:34
$begingroup$
I appreciate very much this help. Thank you.
$endgroup$
– watanake
May 1 '14 at 0:34
add a comment |
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$begingroup$
The second is one-tenth of the first, so cannot be $0.00097$. Just a typo.
$endgroup$
– André Nicolas
May 1 '14 at 0:06
$begingroup$
Note that the numbers like $binom{6}{4}$ are not probabilities,they are counts.
$endgroup$
– André Nicolas
May 1 '14 at 0:11