Probability lottery question with multiple variables to account for
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Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetition allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if at least 7 of your numbers are there in the 11 chosen by the Lottery Commission. What is the probability of winning the lottery?
I just wanted to check if what I am thinking is correct or not.
Sample size = $80 choose 8$
Winning Events = $11 choose 7$ * $73choose 1$ $~$ $bigg($i.e. $11choose7$ if at least 7 number matches and $73choose 1$ for $8^{th}$ number can be anything $bigg)$
Therefore, the probability is given by:
$$ P = frac{{11 choose 7} {73 choose 1} }{80 choose 8}$$
probability lotteries
asked Dec 6 '18 at 17:11
zealzeal
716
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add a comment |
$begingroup$
Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetition allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if at least 7 of your numbers are there in the 11 chosen by the Lottery Commission. What is the probability of winning the lottery?
I just wanted to check if what I am thinking is correct or not.
Sample size = $80 choose 8$
Winning Events = $11 choose 7$ * $73choose 1$ $~$ $bigg($i.e. $11choose7$ if at least 7 number matches and $73choose 1$ for $8^{th}$ number can be anything $bigg)$
Therefore, the probability is given by:
$$ P = frac{{11 choose 7} {73 choose 1} }{80 choose 8}$$
probability lotteries
asked Dec 6 '18 at 17:11
zealzeal
716
$endgroup$
add a comment |
$begingroup$
Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetition allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if at least 7 of your numbers are there in the 11 chosen by the Lottery Commission. What is the probability of winning the lottery?
I just wanted to check if what I am thinking is correct or not.
Sample size = $80 choose 8$
Winning Events = $11 choose 7$ * $73choose 1$ $~$ $bigg($i.e. $11choose7$ if at least 7 number matches and $73choose 1$ for $8^{th}$ number can be anything $bigg)$
Therefore, the probability is given by:
$$ P = frac{{11 choose 7} {73 choose 1} }{80 choose 8}$$
probability lotteries
asked Dec 6 '18 at 17:11
zealzeal
716
$endgroup$
Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetition allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if at least 7 of your numbers are there in the 11 chosen by the Lottery Commission. What is the probability of winning the lottery?
I just wanted to check if what I am thinking is correct or not.
Sample size = $80 choose 8$
Winning Events = $11 choose 7$ * $73choose 1$ $~$ $bigg($i.e. $11choose7$ if at least 7 number matches and $73choose 1$ for $8^{th}$ number can be anything $bigg)$
Therefore, the probability is given by:
$$ P = frac{{11 choose 7} {73 choose 1} }{80 choose 8}$$
probability lotteries
probability lotteries
asked Dec 6 '18 at 17:11
zealzeal
716
asked Dec 6 '18 at 17:11
zealzeal
716
asked Dec 6 '18 at 17:11
zealzeal
716
asked Dec 6 '18 at 17:11
zealzeal
716
asked Dec 6 '18 at 17:11
zealzeal
716
716
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$
answered Dec 6 '18 at 18:36
greedoidgreedoid
38.7k114797
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$begingroup$
I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
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– zeal
Dec 6 '18 at 19:12
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Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
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– greedoid
Dec 6 '18 at 19:28
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I hope I made clear this :)
$endgroup$
– greedoid
Dec 6 '18 at 19:29
$begingroup$
thank you very much that is clear now.
$endgroup$
– zeal
Dec 7 '18 at 2:09
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$
answered Dec 6 '18 at 18:36
greedoidgreedoid
38.7k114797
$endgroup$
$begingroup$
I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
$endgroup$
– zeal
Dec 6 '18 at 19:12
$begingroup$
Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
$endgroup$
– greedoid
Dec 6 '18 at 19:28
$begingroup$
I hope I made clear this :)
$endgroup$
– greedoid
Dec 6 '18 at 19:29
$begingroup$
thank you very much that is clear now.
$endgroup$
– zeal
Dec 7 '18 at 2:09
add a comment |
$begingroup$
Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$
answered Dec 6 '18 at 18:36
greedoidgreedoid
38.7k114797
$endgroup$
$begingroup$
I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
$endgroup$
– zeal
Dec 6 '18 at 19:12
$begingroup$
Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
$endgroup$
– greedoid
Dec 6 '18 at 19:28
$begingroup$
I hope I made clear this :)
$endgroup$
– greedoid
Dec 6 '18 at 19:29
$begingroup$
thank you very much that is clear now.
$endgroup$
– zeal
Dec 7 '18 at 2:09
add a comment |
$begingroup$
Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$
answered Dec 6 '18 at 18:36
greedoidgreedoid
38.7k114797
$endgroup$
Sample size is OK but Winning Events = $${11 choose 7} * {69choose 1} +{11 choose 8} * {69choose 0} $$
answered Dec 6 '18 at 18:36
greedoidgreedoid
38.7k114797
answered Dec 6 '18 at 18:36
greedoidgreedoid
38.7k114797
answered Dec 6 '18 at 18:36
greedoidgreedoid
38.7k114797
answered Dec 6 '18 at 18:36
greedoidgreedoid
38.7k114797
38.7k114797
$begingroup$
I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
$endgroup$
– zeal
Dec 6 '18 at 19:12
$begingroup$
Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
$endgroup$
– greedoid
Dec 6 '18 at 19:28
$begingroup$
I hope I made clear this :)
$endgroup$
– greedoid
Dec 6 '18 at 19:29
$begingroup$
thank you very much that is clear now.
$endgroup$
– zeal
Dec 7 '18 at 2:09
add a comment |
$begingroup$
I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
$endgroup$
– zeal
Dec 6 '18 at 19:12
$begingroup$
Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
$endgroup$
– greedoid
Dec 6 '18 at 19:28
$begingroup$
I hope I made clear this :)
$endgroup$
– greedoid
Dec 6 '18 at 19:29
$begingroup$
thank you very much that is clear now.
$endgroup$
– zeal
Dec 7 '18 at 2:09
$begingroup$
I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
$endgroup$
– zeal
Dec 6 '18 at 19:12
$begingroup$
I thought that way but then I thought this is more correct $frac{{11 choose 7} {73 choose 1}}{80 choose 8}$ as $73 choose 1$ will also include for the fact when all the 8th number also matches. As winning is when at least 7 numbers match that means 8th number can be anything- either from the 11 number chosen by lottery commision or any other number which is included in rest of 73 numbers. Please tell me where I am going wrong.
$endgroup$
– zeal
Dec 6 '18 at 19:12
$begingroup$
Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
$endgroup$
– greedoid
Dec 6 '18 at 19:28
$begingroup$
Yes, but in this case you are overcounting. Say commison numbers are 1 to 11. Then as you counted, the first 7 numbers are from that 11 and 8 number could be also, say 10. But now you have also this situation if you choose first 1,2,...,6 and 10 among first 11 and then 8-th number is 7.
$endgroup$
– greedoid
Dec 6 '18 at 19:28
$begingroup$
I hope I made clear this :)
$endgroup$
– greedoid
Dec 6 '18 at 19:29
$begingroup$
I hope I made clear this :)
$endgroup$
– greedoid
Dec 6 '18 at 19:29
$begingroup$
thank you very much that is clear now.
$endgroup$
– zeal
Dec 7 '18 at 2:09
$begingroup$
thank you very much that is clear now.
$endgroup$
– zeal
Dec 7 '18 at 2:09
add a comment |
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