Mixing
There are 20 people. Among the 12 months in the year there are 4 months containing exactly 2 birthdays and 4...
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There are 20 people. Find the probability that among the 12 months in the year there are 4 months containing exactly 2 birthdays and 4 containing exactly 3 birthdays?
I am unable to get from the problem as to whether the 4 months talked about are overlapping or disjoint? If they are disjoint then what about the remaining 4 months... Suppose the remaining 4 months have 0,1,2,3,...14,15 birthdays then we can possibly use multinomial theorem as
x1+x2+x3+y1+y2+ki=20 ;i=0,1,2,...14,15
Total no of ways becomes (say A): 19C4+19C5+19C6+...+19C13+19C14
Total no outcomes(D): 12^20
Required probability= 12C4×8C4×A/D
But I am not getting the answer. It is given as 1.0604 × 10^-3
probability
asked Jan 24 at 16:35
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
add a comment |
$begingroup$
There are 20 people. Find the probability that among the 12 months in the year there are 4 months containing exactly 2 birthdays and 4 containing exactly 3 birthdays?
I am unable to get from the problem as to whether the 4 months talked about are overlapping or disjoint? If they are disjoint then what about the remaining 4 months... Suppose the remaining 4 months have 0,1,2,3,...14,15 birthdays then we can possibly use multinomial theorem as
x1+x2+x3+y1+y2+ki=20 ;i=0,1,2,...14,15
Total no of ways becomes (say A): 19C4+19C5+19C6+...+19C13+19C14
Total no outcomes(D): 12^20
Required probability= 12C4×8C4×A/D
But I am not getting the answer. It is given as 1.0604 × 10^-3
probability
asked Jan 24 at 16:35
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
$begingroup$
There are $4$ month that each contain exactly $3$ birthdays, and $4$ other months that each contain exactly $2$ birthdays. That accounts for all $20$ birthdays; the other months contain none.
$endgroup$
– saulspatz
Jan 24 at 17:00
$begingroup$
Ohh ok. Thanks. The problem has the "each" term missing here
$endgroup$
– Abhishek Ghosh
Jan 24 at 17:02
add a comment |
$begingroup$
There are 20 people. Find the probability that among the 12 months in the year there are 4 months containing exactly 2 birthdays and 4 containing exactly 3 birthdays?
I am unable to get from the problem as to whether the 4 months talked about are overlapping or disjoint? If they are disjoint then what about the remaining 4 months... Suppose the remaining 4 months have 0,1,2,3,...14,15 birthdays then we can possibly use multinomial theorem as
x1+x2+x3+y1+y2+ki=20 ;i=0,1,2,...14,15
Total no of ways becomes (say A): 19C4+19C5+19C6+...+19C13+19C14
Total no outcomes(D): 12^20
Required probability= 12C4×8C4×A/D
But I am not getting the answer. It is given as 1.0604 × 10^-3
probability
asked Jan 24 at 16:35
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
There are 20 people. Find the probability that among the 12 months in the year there are 4 months containing exactly 2 birthdays and 4 containing exactly 3 birthdays?
I am unable to get from the problem as to whether the 4 months talked about are overlapping or disjoint? If they are disjoint then what about the remaining 4 months... Suppose the remaining 4 months have 0,1,2,3,...14,15 birthdays then we can possibly use multinomial theorem as
x1+x2+x3+y1+y2+ki=20 ;i=0,1,2,...14,15
Total no of ways becomes (say A): 19C4+19C5+19C6+...+19C13+19C14
Total no outcomes(D): 12^20
Required probability= 12C4×8C4×A/D
But I am not getting the answer. It is given as 1.0604 × 10^-3
probability
probability
asked Jan 24 at 16:35
Abhishek GhoshAbhishek Ghosh
878
asked Jan 24 at 16:35
Abhishek GhoshAbhishek Ghosh
878
asked Jan 24 at 16:35
Abhishek GhoshAbhishek Ghosh
878
asked Jan 24 at 16:35
Abhishek GhoshAbhishek Ghosh
878
asked Jan 24 at 16:35
Abhishek GhoshAbhishek Ghosh
878
878
$begingroup$
There are $4$ month that each contain exactly $3$ birthdays, and $4$ other months that each contain exactly $2$ birthdays. That accounts for all $20$ birthdays; the other months contain none.
$endgroup$
– saulspatz
Jan 24 at 17:00
$begingroup$
Ohh ok. Thanks. The problem has the "each" term missing here
$endgroup$
– Abhishek Ghosh
Jan 24 at 17:02
add a comment |
$begingroup$
There are $4$ month that each contain exactly $3$ birthdays, and $4$ other months that each contain exactly $2$ birthdays. That accounts for all $20$ birthdays; the other months contain none.
$endgroup$
– saulspatz
Jan 24 at 17:00
$begingroup$
Ohh ok. Thanks. The problem has the "each" term missing here
$endgroup$
– Abhishek Ghosh
Jan 24 at 17:02
$begingroup$
There are $4$ month that each contain exactly $3$ birthdays, and $4$ other months that each contain exactly $2$ birthdays. That accounts for all $20$ birthdays; the other months contain none.
$endgroup$
– saulspatz
Jan 24 at 17:00
$begingroup$
There are $4$ month that each contain exactly $3$ birthdays, and $4$ other months that each contain exactly $2$ birthdays. That accounts for all $20$ birthdays; the other months contain none.
$endgroup$
– saulspatz
Jan 24 at 17:00
$begingroup$
Ohh ok. Thanks. The problem has the "each" term missing here
$endgroup$
– Abhishek Ghosh
Jan 24 at 17:02
$begingroup$
Ohh ok. Thanks. The problem has the "each" term missing here
$endgroup$
– Abhishek Ghosh
Jan 24 at 17:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The question given can be corrected as shown:
There are 20 people. Find the probability that among the 12 months in the year there are 4 months each containing exactly 2 birthdays and 4 months each containing exactly 3 birthdays?
So 4*2 + 4*3= 20
Now the total possible outcomes are(n) = 12^20
No of out comes favourable(m) =
12C4×8C4×20C8 × (8!/(2!)^4) × (12!/(3!)^4)
Required probability = m/n = 1.060420099×10^-3
answered Jan 25 at 15:42
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The question given can be corrected as shown:
There are 20 people. Find the probability that among the 12 months in the year there are 4 months each containing exactly 2 birthdays and 4 months each containing exactly 3 birthdays?
So 4*2 + 4*3= 20
Now the total possible outcomes are(n) = 12^20
No of out comes favourable(m) =
12C4×8C4×20C8 × (8!/(2!)^4) × (12!/(3!)^4)
Required probability = m/n = 1.060420099×10^-3
answered Jan 25 at 15:42
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
add a comment |
$begingroup$
The question given can be corrected as shown:
There are 20 people. Find the probability that among the 12 months in the year there are 4 months each containing exactly 2 birthdays and 4 months each containing exactly 3 birthdays?
So 4*2 + 4*3= 20
Now the total possible outcomes are(n) = 12^20
No of out comes favourable(m) =
12C4×8C4×20C8 × (8!/(2!)^4) × (12!/(3!)^4)
Required probability = m/n = 1.060420099×10^-3
answered Jan 25 at 15:42
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
add a comment |
$begingroup$
The question given can be corrected as shown:
There are 20 people. Find the probability that among the 12 months in the year there are 4 months each containing exactly 2 birthdays and 4 months each containing exactly 3 birthdays?
So 4*2 + 4*3= 20
Now the total possible outcomes are(n) = 12^20
No of out comes favourable(m) =
12C4×8C4×20C8 × (8!/(2!)^4) × (12!/(3!)^4)
Required probability = m/n = 1.060420099×10^-3
answered Jan 25 at 15:42
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
The question given can be corrected as shown:
There are 20 people. Find the probability that among the 12 months in the year there are 4 months each containing exactly 2 birthdays and 4 months each containing exactly 3 birthdays?
So 4*2 + 4*3= 20
Now the total possible outcomes are(n) = 12^20
No of out comes favourable(m) =
12C4×8C4×20C8 × (8!/(2!)^4) × (12!/(3!)^4)
Required probability = m/n = 1.060420099×10^-3
answered Jan 25 at 15:42
Abhishek GhoshAbhishek Ghosh
878
answered Jan 25 at 15:42
Abhishek GhoshAbhishek Ghosh
878
answered Jan 25 at 15:42
Abhishek GhoshAbhishek Ghosh
878
answered Jan 25 at 15:42
Abhishek GhoshAbhishek Ghosh
878
878
add a comment |
add a comment |
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$begingroup$
There are $4$ month that each contain exactly $3$ birthdays, and $4$ other months that each contain exactly $2$ birthdays. That accounts for all $20$ birthdays; the other months contain none.
$endgroup$
– saulspatz
Jan 24 at 17:00
$begingroup$
Ohh ok. Thanks. The problem has the "each" term missing here
$endgroup$
– Abhishek Ghosh
Jan 24 at 17:02