Mixing
What is the probabillity of winning at ball drawing?
$begingroup$
If balls are numerated 1 to 48, and 35 balls are drawn, what is the chance we have 6 numbers ? Order of balls is not important.
It's local bingo and I was wondering what are the chance of winning.
probability combinatorics
edited Feb 2 at 23:39
Henno Brandsma
116k349127
asked Feb 1 at 12:33
Dinko PeharDinko Pehar
1033
$endgroup$
add a comment |
$begingroup$
If balls are numerated 1 to 48, and 35 balls are drawn, what is the chance we have 6 numbers ? Order of balls is not important.
It's local bingo and I was wondering what are the chance of winning.
probability combinatorics
edited Feb 2 at 23:39
Henno Brandsma
116k349127
asked Feb 1 at 12:33
Dinko PeharDinko Pehar
1033
$endgroup$
$begingroup$
Do you mean six particular numbers?
$endgroup$
– jvdhooft
Feb 1 at 12:35
$begingroup$
I mean any of 6 balls of 36 drawn in interval [1, 48] .
$endgroup$
– Dinko Pehar
Feb 1 at 12:37
$begingroup$
With or without replacing?
$endgroup$
– drhab
Feb 1 at 12:43
$begingroup$
@dhrab: Apparently it is bingo, so without replacement
$endgroup$
– Henry
Feb 1 at 12:45
add a comment |
$begingroup$
If balls are numerated 1 to 48, and 35 balls are drawn, what is the chance we have 6 numbers ? Order of balls is not important.
It's local bingo and I was wondering what are the chance of winning.
probability combinatorics
edited Feb 2 at 23:39
Henno Brandsma
116k349127
asked Feb 1 at 12:33
Dinko PeharDinko Pehar
1033
$endgroup$
If balls are numerated 1 to 48, and 35 balls are drawn, what is the chance we have 6 numbers ? Order of balls is not important.
It's local bingo and I was wondering what are the chance of winning.
probability combinatorics
probability combinatorics
edited Feb 2 at 23:39
Henno Brandsma
116k349127
asked Feb 1 at 12:33
Dinko PeharDinko Pehar
1033
edited Feb 2 at 23:39
Henno Brandsma
116k349127
asked Feb 1 at 12:33
Dinko PeharDinko Pehar
1033
edited Feb 2 at 23:39
Henno Brandsma
116k349127
edited Feb 2 at 23:39
Henno Brandsma
116k349127
edited Feb 2 at 23:39
Henno Brandsma
116k349127
116k349127
asked Feb 1 at 12:33
Dinko PeharDinko Pehar
1033
asked Feb 1 at 12:33
Dinko PeharDinko Pehar
1033
asked Feb 1 at 12:33
Dinko PeharDinko Pehar
1033
1033
$begingroup$
Do you mean six particular numbers?
$endgroup$
– jvdhooft
Feb 1 at 12:35
$begingroup$
I mean any of 6 balls of 36 drawn in interval [1, 48] .
$endgroup$
– Dinko Pehar
Feb 1 at 12:37
$begingroup$
With or without replacing?
$endgroup$
– drhab
Feb 1 at 12:43
$begingroup$
@dhrab: Apparently it is bingo, so without replacement
$endgroup$
– Henry
Feb 1 at 12:45
add a comment |
$begingroup$
Do you mean six particular numbers?
$endgroup$
– jvdhooft
Feb 1 at 12:35
$begingroup$
I mean any of 6 balls of 36 drawn in interval [1, 48] .
$endgroup$
– Dinko Pehar
Feb 1 at 12:37
$begingroup$
With or without replacing?
$endgroup$
– drhab
Feb 1 at 12:43
$begingroup$
@dhrab: Apparently it is bingo, so without replacement
$endgroup$
– Henry
Feb 1 at 12:45
$begingroup$
Do you mean six particular numbers?
$endgroup$
– jvdhooft
Feb 1 at 12:35
$begingroup$
Do you mean six particular numbers?
$endgroup$
– jvdhooft
Feb 1 at 12:35
$begingroup$
I mean any of 6 balls of 36 drawn in interval [1, 48] .
$endgroup$
– Dinko Pehar
Feb 1 at 12:37
$begingroup$
I mean any of 6 balls of 36 drawn in interval [1, 48] .
$endgroup$
– Dinko Pehar
Feb 1 at 12:37
$begingroup$
With or without replacing?
$endgroup$
– drhab
Feb 1 at 12:43
$begingroup$
With or without replacing?
$endgroup$
– drhab
Feb 1 at 12:43
$begingroup$
@dhrab: Apparently it is bingo, so without replacement
$endgroup$
– Henry
Feb 1 at 12:45
$begingroup$
@dhrab: Apparently it is bingo, so without replacement
$endgroup$
– Henry
Feb 1 at 12:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There are ${48 choose 35}$ equally likely sets of $35$ selected numbers
Of these, ${48-6 choose 35-6}$ contain all six desired numbers
So the chance that all six desired numbers are selected in the $35$ is $dfrac{42 choose 29}{48 choose 35}approx 0.13227$
answered Feb 1 at 12:53
HenryHenry
101k482170
$endgroup$
$begingroup$
Is it 13 % or 0.13 % ? I didn't get the last part
$endgroup$
– Dinko Pehar
Feb 1 at 13:39
$begingroup$
@DinkoPehar $13%$ or $frac{18445}{139449}$
$endgroup$
– Henry
Feb 1 at 14:06
add a comment |
$begingroup$
Hint:
Probably to be solved with hypergeometric distribution.
Turn things around and select randomly $6$ distinct numbers.
Then what is the probability that all $6$ numbers do not exceed $35$?
(on how many ways can you select $6$ numbers in $[1,35]$ and on how many ways can you select $6$ numbers in $[1,48]$?)
answered Feb 1 at 12:43
drhabdrhab
104k545136
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are ${48 choose 35}$ equally likely sets of $35$ selected numbers
Of these, ${48-6 choose 35-6}$ contain all six desired numbers
So the chance that all six desired numbers are selected in the $35$ is $dfrac{42 choose 29}{48 choose 35}approx 0.13227$
answered Feb 1 at 12:53
HenryHenry
101k482170
$endgroup$
$begingroup$
Is it 13 % or 0.13 % ? I didn't get the last part
$endgroup$
– Dinko Pehar
Feb 1 at 13:39
$begingroup$
@DinkoPehar $13%$ or $frac{18445}{139449}$
$endgroup$
– Henry
Feb 1 at 14:06
add a comment |
$begingroup$
There are ${48 choose 35}$ equally likely sets of $35$ selected numbers
Of these, ${48-6 choose 35-6}$ contain all six desired numbers
So the chance that all six desired numbers are selected in the $35$ is $dfrac{42 choose 29}{48 choose 35}approx 0.13227$
answered Feb 1 at 12:53
HenryHenry
101k482170
$endgroup$
$begingroup$
Is it 13 % or 0.13 % ? I didn't get the last part
$endgroup$
– Dinko Pehar
Feb 1 at 13:39
$begingroup$
@DinkoPehar $13%$ or $frac{18445}{139449}$
$endgroup$
– Henry
Feb 1 at 14:06
add a comment |
$begingroup$
There are ${48 choose 35}$ equally likely sets of $35$ selected numbers
Of these, ${48-6 choose 35-6}$ contain all six desired numbers
So the chance that all six desired numbers are selected in the $35$ is $dfrac{42 choose 29}{48 choose 35}approx 0.13227$
answered Feb 1 at 12:53
HenryHenry
101k482170
$endgroup$
There are ${48 choose 35}$ equally likely sets of $35$ selected numbers
Of these, ${48-6 choose 35-6}$ contain all six desired numbers
So the chance that all six desired numbers are selected in the $35$ is $dfrac{42 choose 29}{48 choose 35}approx 0.13227$
answered Feb 1 at 12:53
HenryHenry
101k482170
answered Feb 1 at 12:53
HenryHenry
101k482170
answered Feb 1 at 12:53
HenryHenry
101k482170
answered Feb 1 at 12:53
HenryHenry
101k482170
101k482170
$begingroup$
Is it 13 % or 0.13 % ? I didn't get the last part
$endgroup$
– Dinko Pehar
Feb 1 at 13:39
$begingroup$
@DinkoPehar $13%$ or $frac{18445}{139449}$
$endgroup$
– Henry
Feb 1 at 14:06
add a comment |
$begingroup$
Is it 13 % or 0.13 % ? I didn't get the last part
$endgroup$
– Dinko Pehar
Feb 1 at 13:39
$begingroup$
@DinkoPehar $13%$ or $frac{18445}{139449}$
$endgroup$
– Henry
Feb 1 at 14:06
$begingroup$
Is it 13 % or 0.13 % ? I didn't get the last part
$endgroup$
– Dinko Pehar
Feb 1 at 13:39
$begingroup$
Is it 13 % or 0.13 % ? I didn't get the last part
$endgroup$
– Dinko Pehar
Feb 1 at 13:39
$begingroup$
@DinkoPehar $13%$ or $frac{18445}{139449}$
$endgroup$
– Henry
Feb 1 at 14:06
$begingroup$
@DinkoPehar $13%$ or $frac{18445}{139449}$
$endgroup$
– Henry
Feb 1 at 14:06
add a comment |
$begingroup$
Hint:
Probably to be solved with hypergeometric distribution.
Turn things around and select randomly $6$ distinct numbers.
Then what is the probability that all $6$ numbers do not exceed $35$?
(on how many ways can you select $6$ numbers in $[1,35]$ and on how many ways can you select $6$ numbers in $[1,48]$?)
answered Feb 1 at 12:43
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Hint:
Probably to be solved with hypergeometric distribution.
Turn things around and select randomly $6$ distinct numbers.
Then what is the probability that all $6$ numbers do not exceed $35$?
(on how many ways can you select $6$ numbers in $[1,35]$ and on how many ways can you select $6$ numbers in $[1,48]$?)
answered Feb 1 at 12:43
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Hint:
Probably to be solved with hypergeometric distribution.
Turn things around and select randomly $6$ distinct numbers.
Then what is the probability that all $6$ numbers do not exceed $35$?
(on how many ways can you select $6$ numbers in $[1,35]$ and on how many ways can you select $6$ numbers in $[1,48]$?)
answered Feb 1 at 12:43
drhabdrhab
104k545136
$endgroup$
Hint:
Probably to be solved with hypergeometric distribution.
Turn things around and select randomly $6$ distinct numbers.
Then what is the probability that all $6$ numbers do not exceed $35$?
(on how many ways can you select $6$ numbers in $[1,35]$ and on how many ways can you select $6$ numbers in $[1,48]$?)
answered Feb 1 at 12:43
drhabdrhab
104k545136
edited Feb 1 at 12:49
answered Feb 1 at 12:43
drhabdrhab
104k545136
answered Feb 1 at 12:43
drhabdrhab
104k545136
answered Feb 1 at 12:43
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
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$begingroup$
Do you mean six particular numbers?
$endgroup$
– jvdhooft
Feb 1 at 12:35
$begingroup$
I mean any of 6 balls of 36 drawn in interval [1, 48] .
$endgroup$
– Dinko Pehar
Feb 1 at 12:37
$begingroup$
With or without replacing?
$endgroup$
– drhab
Feb 1 at 12:43
$begingroup$
@dhrab: Apparently it is bingo, so without replacement
$endgroup$
– Henry
Feb 1 at 12:45