Game of Keno: What is the percent chance that a player selects exactly 3 winning numbers?

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In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3,4, or 5 of the 20 winning numbers.
May anyone give me a hint to solve this problem, please? Thanks!






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  • 1) Count how many combinations of 20 numbers can be drawn from a set of 80 numbers. 2) You have 20 winning numbers and 60 losing numbers. Count how many combinations can be formed with $n$ winning numbers and $20-n$ losing numbers. 3) Divide what you get at step 2 by what you got at step 1.
    – nicola
    Aug 31 at 8:04














up vote
0
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In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3,4, or 5 of the 20 winning numbers.
May anyone give me a hint to solve this problem, please? Thanks!






probability







share|cite|improve this question





















  • 1) Count how many combinations of 20 numbers can be drawn from a set of 80 numbers. 2) You have 20 winning numbers and 60 losing numbers. Count how many combinations can be formed with $n$ winning numbers and $20-n$ losing numbers. 3) Divide what you get at step 2 by what you got at step 1.
    – nicola
    Aug 31 at 8:04












up vote
0
down vote

favorite









up vote
0
down vote

favorite











In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3,4, or 5 of the 20 winning numbers.
May anyone give me a hint to solve this problem, please? Thanks!






probability







share|cite|improve this question













In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3,4, or 5 of the 20 winning numbers.
May anyone give me a hint to solve this problem, please? Thanks!






probability




probability






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asked Aug 31 at 8:00









Willy

122




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  • 1) Count how many combinations of 20 numbers can be drawn from a set of 80 numbers. 2) You have 20 winning numbers and 60 losing numbers. Count how many combinations can be formed with $n$ winning numbers and $20-n$ losing numbers. 3) Divide what you get at step 2 by what you got at step 1.
    – nicola
    Aug 31 at 8:04
















  • 1) Count how many combinations of 20 numbers can be drawn from a set of 80 numbers. 2) You have 20 winning numbers and 60 losing numbers. Count how many combinations can be formed with $n$ winning numbers and $20-n$ losing numbers. 3) Divide what you get at step 2 by what you got at step 1.
    – nicola
    Aug 31 at 8:04















1) Count how many combinations of 20 numbers can be drawn from a set of 80 numbers. 2) You have 20 winning numbers and 60 losing numbers. Count how many combinations can be formed with $n$ winning numbers and $20-n$ losing numbers. 3) Divide what you get at step 2 by what you got at step 1.
– nicola
Aug 31 at 8:04




1) Count how many combinations of 20 numbers can be drawn from a set of 80 numbers. 2) You have 20 winning numbers and 60 losing numbers. Count how many combinations can be formed with $n$ winning numbers and $20-n$ losing numbers. 3) Divide what you get at step 2 by what you got at step 1.
– nicola
Aug 31 at 8:04










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The number of ways to select 20 out of 80 is $binom8020$. The player needs to select 3,4, or 5 of these AND the rest from the remaining 60 numbers: $binom203 times binom6017 + binom204 times binom6016 +binom205 times binom6015$. Can you sort out the rest?






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  • Thanks all! The percent chance: C(20,3) . C(60,17) / C(80,20) = 12.48%
    – Willy
    Sep 1 at 22:48










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1 Answer
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1 Answer
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active

oldest

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active

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active

oldest

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up vote
1
down vote



accepted










The number of ways to select 20 out of 80 is $binom8020$. The player needs to select 3,4, or 5 of these AND the rest from the remaining 60 numbers: $binom203 times binom6017 + binom204 times binom6016 +binom205 times binom6015$. Can you sort out the rest?






share|cite|improve this answer




















  • Thanks all! The percent chance: C(20,3) . C(60,17) / C(80,20) = 12.48%
    – Willy
    Sep 1 at 22:48














up vote
1
down vote



accepted










The number of ways to select 20 out of 80 is $binom8020$. The player needs to select 3,4, or 5 of these AND the rest from the remaining 60 numbers: $binom203 times binom6017 + binom204 times binom6016 +binom205 times binom6015$. Can you sort out the rest?






share|cite|improve this answer




















  • Thanks all! The percent chance: C(20,3) . C(60,17) / C(80,20) = 12.48%
    – Willy
    Sep 1 at 22:48












up vote
1
down vote



accepted







up vote
1
down vote



accepted






The number of ways to select 20 out of 80 is $binom8020$. The player needs to select 3,4, or 5 of these AND the rest from the remaining 60 numbers: $binom203 times binom6017 + binom204 times binom6016 +binom205 times binom6015$. Can you sort out the rest?






share|cite|improve this answer












The number of ways to select 20 out of 80 is $binom8020$. The player needs to select 3,4, or 5 of these AND the rest from the remaining 60 numbers: $binom203 times binom6017 + binom204 times binom6016 +binom205 times binom6015$. Can you sort out the rest?







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answered Aug 31 at 8:06









Alex

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  • Thanks all! The percent chance: C(20,3) . C(60,17) / C(80,20) = 12.48%
    – Willy
    Sep 1 at 22:48
















  • Thanks all! The percent chance: C(20,3) . C(60,17) / C(80,20) = 12.48%
    – Willy
    Sep 1 at 22:48















Thanks all! The percent chance: C(20,3) . C(60,17) / C(80,20) = 12.48%
– Willy
Sep 1 at 22:48




Thanks all! The percent chance: C(20,3) . C(60,17) / C(80,20) = 12.48%
– Willy
Sep 1 at 22:48

















 

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