Probability problem; adding elements to a list then picking from that list
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It seems like a very simple problem but I couldn't figure out the solution myself (not homework)
Simplified it's the following:
10% chance to add 1 to a list
20% chance to add 2 to a list
30% chance to add 3 to a list
40% chance to add 4 to a list
and from that list I pick 1 number.
How much chance is there to get a specific number?
Running a simulation in matlab gave me:
29.3% to get 4;
20.9% to get 3;
13.5% to get 2;
6.2% to get 1;
30.1% to get 0 numbers in a list so no number could be picked.
probability random
asked Dec 23 '18 at 17:50
Gerben BanaanGerben Banaan
61
$endgroup$
add a comment |
$begingroup$
It seems like a very simple problem but I couldn't figure out the solution myself (not homework)
Simplified it's the following:
10% chance to add 1 to a list
20% chance to add 2 to a list
30% chance to add 3 to a list
40% chance to add 4 to a list
and from that list I pick 1 number.
How much chance is there to get a specific number?
Running a simulation in matlab gave me:
29.3% to get 4;
20.9% to get 3;
13.5% to get 2;
6.2% to get 1;
30.1% to get 0 numbers in a list so no number could be picked.
probability random
asked Dec 23 '18 at 17:50
Gerben BanaanGerben Banaan
61
$endgroup$
$begingroup$
The % to get 0 numbers is (1-0.1) * (1-0.2) * (1-0.3) * (1-0.4) = 30.24%
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:00
add a comment |
$begingroup$
It seems like a very simple problem but I couldn't figure out the solution myself (not homework)
Simplified it's the following:
10% chance to add 1 to a list
20% chance to add 2 to a list
30% chance to add 3 to a list
40% chance to add 4 to a list
and from that list I pick 1 number.
How much chance is there to get a specific number?
Running a simulation in matlab gave me:
29.3% to get 4;
20.9% to get 3;
13.5% to get 2;
6.2% to get 1;
30.1% to get 0 numbers in a list so no number could be picked.
probability random
asked Dec 23 '18 at 17:50
Gerben BanaanGerben Banaan
61
$endgroup$
It seems like a very simple problem but I couldn't figure out the solution myself (not homework)
Simplified it's the following:
10% chance to add 1 to a list
20% chance to add 2 to a list
30% chance to add 3 to a list
40% chance to add 4 to a list
and from that list I pick 1 number.
How much chance is there to get a specific number?
Running a simulation in matlab gave me:
29.3% to get 4;
20.9% to get 3;
13.5% to get 2;
6.2% to get 1;
30.1% to get 0 numbers in a list so no number could be picked.
probability random
probability random
asked Dec 23 '18 at 17:50
Gerben BanaanGerben Banaan
61
asked Dec 23 '18 at 17:50
Gerben BanaanGerben Banaan
61
asked Dec 23 '18 at 17:50
Gerben BanaanGerben Banaan
61
asked Dec 23 '18 at 17:50
Gerben BanaanGerben Banaan
61
asked Dec 23 '18 at 17:50
Gerben BanaanGerben Banaan
61
61
$begingroup$
The % to get 0 numbers is (1-0.1) * (1-0.2) * (1-0.3) * (1-0.4) = 30.24%
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:00
add a comment |
$begingroup$
The % to get 0 numbers is (1-0.1) * (1-0.2) * (1-0.3) * (1-0.4) = 30.24%
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:00
$begingroup$
The % to get 0 numbers is (1-0.1) * (1-0.2) * (1-0.3) * (1-0.4) = 30.24%
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:00
$begingroup$
The % to get 0 numbers is (1-0.1) * (1-0.2) * (1-0.3) * (1-0.4) = 30.24%
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't think there is an easy way. The chance you get $4$ is
$0.4$ (that you added $4$ at all) times $Big[(1-0.1)(1-0.2)(1-0.3)$ (that it is the only number on the list) plus $frac 12big(0.1(1-0.2)(1-0.3)+(1-0.1)0.2(1-0.3)+(1-0.1)(1-0.2)0.3big)$ (because one other number was added) plus $ldotsBig]$ and the dots have the cases where two or three other numbers were added.
answered Dec 23 '18 at 18:19
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Yeah I got as far but that is not an efficient solution; especially not when working with 20 cases; 0.4*0.7*0.8*0.9 + 0.5*0.4*0.3*0.8*0.9 + 0.5*0.4*0.7*0.2*0.9 + 0.5*0.4*0.7*0.8*0.1 + 0.3333*0.4*0.3*0.2*0.9 + 0.3333 * 0.4* 0.7 * 0.2 * 0.1 + 0.3333 * 0.4 * 0.3 * 0.8 * 0.1 + 0.25 * 0.4 * 0.3 * 0.2 * 0.1 (Case where you only can pick 4) + (Cases where you can pick 4 + 1 other number) (Cases where you can pick 4 + 2 other numbers) (Cases where you can pick 4 + 3 other numbers) I was thinking that I must be missing something
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:24
$begingroup$
I don’t think you were missing anything. Solving problems like this with a neat formula depends on the problem having symmetry. If the chance of adding each number were the same you could just find the chance you don’t get any number, subtract from $1$, and divide by $4$ because of the symmetry. Here you don’t have that, so you have to go through the cases.
$endgroup$
– Ross Millikan
Dec 23 '18 at 19:20
add a comment |
$begingroup$
Let $p(n) (nin{1,2,3,4})$ be the probability that number $n$ is added to the list (e.g., your $0.1, 0.2, 0.3, 0.4$). Then a formula for the probability that the outcome is the specific number $x$ is
$$1(Abcd)+{1over 2}(ABcd+AbCd+AbcD)+{1over 3}(ABCd+ABcD+AbCD)+{1over 4}(ABCD)
$$
where
$$begin{align}A&=p(x)\
B&=p(1+(xbmod 4))\
C&=p(1+((x+1)bmod 4))\
D&=p(1+((x+2)bmod 4))\
end{align}$$
and $a=1-A, b=1-B, c=1-C, d=1-D.$
Note that $B,C,D$ are simply the add-to-the-list probabilities for the three other numbers besides $x$, and they may be taken in any order. (The above formula just uses $bmod 4$ to express the choice of taking them in cyclic order.)
In Sage:
def p(x): return [1/10, 2/10, 3/10, 4/10][x-1]
def Pout(x):
A = p(x)
B = p(1 + (x%4))
C = p(1 + ((x+1)%4))
D = p(1 + ((x+2)%4))
a = 1 - A
b = 1 - B
c = 1 - C
d = 1 - D
return (A*b*c*d)+(A*B*c*d+A*b*C*d+A*b*c*D)/2+(A*B*C*d+A*B*c*D+A*b*C*D)/3+(A*B*C*D)/4
print [Pout(x) for x in [1..4]], 1-sum([Pout(x) for x in [1..4]])
print [Pout(x).n() for x in [1..4]], 1-sum([Pout(x).n() for x in [1..4]])
[473/7500, 1981/15000, 521/2500, 4411/15000] 189/625
[0.0630666666666667, 0.132066666666667, 0.208400000000000, 0.294066666666667] 0.302400000000000
answered Dec 26 '18 at 15:19
r.e.s.r.e.s.
7,66311953
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't think there is an easy way. The chance you get $4$ is
$0.4$ (that you added $4$ at all) times $Big[(1-0.1)(1-0.2)(1-0.3)$ (that it is the only number on the list) plus $frac 12big(0.1(1-0.2)(1-0.3)+(1-0.1)0.2(1-0.3)+(1-0.1)(1-0.2)0.3big)$ (because one other number was added) plus $ldotsBig]$ and the dots have the cases where two or three other numbers were added.
answered Dec 23 '18 at 18:19
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Yeah I got as far but that is not an efficient solution; especially not when working with 20 cases; 0.4*0.7*0.8*0.9 + 0.5*0.4*0.3*0.8*0.9 + 0.5*0.4*0.7*0.2*0.9 + 0.5*0.4*0.7*0.8*0.1 + 0.3333*0.4*0.3*0.2*0.9 + 0.3333 * 0.4* 0.7 * 0.2 * 0.1 + 0.3333 * 0.4 * 0.3 * 0.8 * 0.1 + 0.25 * 0.4 * 0.3 * 0.2 * 0.1 (Case where you only can pick 4) + (Cases where you can pick 4 + 1 other number) (Cases where you can pick 4 + 2 other numbers) (Cases where you can pick 4 + 3 other numbers) I was thinking that I must be missing something
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:24
$begingroup$
I don’t think you were missing anything. Solving problems like this with a neat formula depends on the problem having symmetry. If the chance of adding each number were the same you could just find the chance you don’t get any number, subtract from $1$, and divide by $4$ because of the symmetry. Here you don’t have that, so you have to go through the cases.
$endgroup$
– Ross Millikan
Dec 23 '18 at 19:20
add a comment |
$begingroup$
I don't think there is an easy way. The chance you get $4$ is
$0.4$ (that you added $4$ at all) times $Big[(1-0.1)(1-0.2)(1-0.3)$ (that it is the only number on the list) plus $frac 12big(0.1(1-0.2)(1-0.3)+(1-0.1)0.2(1-0.3)+(1-0.1)(1-0.2)0.3big)$ (because one other number was added) plus $ldotsBig]$ and the dots have the cases where two or three other numbers were added.
answered Dec 23 '18 at 18:19
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Yeah I got as far but that is not an efficient solution; especially not when working with 20 cases; 0.4*0.7*0.8*0.9 + 0.5*0.4*0.3*0.8*0.9 + 0.5*0.4*0.7*0.2*0.9 + 0.5*0.4*0.7*0.8*0.1 + 0.3333*0.4*0.3*0.2*0.9 + 0.3333 * 0.4* 0.7 * 0.2 * 0.1 + 0.3333 * 0.4 * 0.3 * 0.8 * 0.1 + 0.25 * 0.4 * 0.3 * 0.2 * 0.1 (Case where you only can pick 4) + (Cases where you can pick 4 + 1 other number) (Cases where you can pick 4 + 2 other numbers) (Cases where you can pick 4 + 3 other numbers) I was thinking that I must be missing something
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:24
$begingroup$
I don’t think you were missing anything. Solving problems like this with a neat formula depends on the problem having symmetry. If the chance of adding each number were the same you could just find the chance you don’t get any number, subtract from $1$, and divide by $4$ because of the symmetry. Here you don’t have that, so you have to go through the cases.
$endgroup$
– Ross Millikan
Dec 23 '18 at 19:20
add a comment |
$begingroup$
I don't think there is an easy way. The chance you get $4$ is
$0.4$ (that you added $4$ at all) times $Big[(1-0.1)(1-0.2)(1-0.3)$ (that it is the only number on the list) plus $frac 12big(0.1(1-0.2)(1-0.3)+(1-0.1)0.2(1-0.3)+(1-0.1)(1-0.2)0.3big)$ (because one other number was added) plus $ldotsBig]$ and the dots have the cases where two or three other numbers were added.
answered Dec 23 '18 at 18:19
Ross MillikanRoss Millikan
301k24200375
$endgroup$
I don't think there is an easy way. The chance you get $4$ is
$0.4$ (that you added $4$ at all) times $Big[(1-0.1)(1-0.2)(1-0.3)$ (that it is the only number on the list) plus $frac 12big(0.1(1-0.2)(1-0.3)+(1-0.1)0.2(1-0.3)+(1-0.1)(1-0.2)0.3big)$ (because one other number was added) plus $ldotsBig]$ and the dots have the cases where two or three other numbers were added.
answered Dec 23 '18 at 18:19
Ross MillikanRoss Millikan
301k24200375
answered Dec 23 '18 at 18:19
Ross MillikanRoss Millikan
301k24200375
answered Dec 23 '18 at 18:19
Ross MillikanRoss Millikan
301k24200375
answered Dec 23 '18 at 18:19
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
Yeah I got as far but that is not an efficient solution; especially not when working with 20 cases; 0.4*0.7*0.8*0.9 + 0.5*0.4*0.3*0.8*0.9 + 0.5*0.4*0.7*0.2*0.9 + 0.5*0.4*0.7*0.8*0.1 + 0.3333*0.4*0.3*0.2*0.9 + 0.3333 * 0.4* 0.7 * 0.2 * 0.1 + 0.3333 * 0.4 * 0.3 * 0.8 * 0.1 + 0.25 * 0.4 * 0.3 * 0.2 * 0.1 (Case where you only can pick 4) + (Cases where you can pick 4 + 1 other number) (Cases where you can pick 4 + 2 other numbers) (Cases where you can pick 4 + 3 other numbers) I was thinking that I must be missing something
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:24
$begingroup$
I don’t think you were missing anything. Solving problems like this with a neat formula depends on the problem having symmetry. If the chance of adding each number were the same you could just find the chance you don’t get any number, subtract from $1$, and divide by $4$ because of the symmetry. Here you don’t have that, so you have to go through the cases.
$endgroup$
– Ross Millikan
Dec 23 '18 at 19:20
add a comment |
$begingroup$
Yeah I got as far but that is not an efficient solution; especially not when working with 20 cases; 0.4*0.7*0.8*0.9 + 0.5*0.4*0.3*0.8*0.9 + 0.5*0.4*0.7*0.2*0.9 + 0.5*0.4*0.7*0.8*0.1 + 0.3333*0.4*0.3*0.2*0.9 + 0.3333 * 0.4* 0.7 * 0.2 * 0.1 + 0.3333 * 0.4 * 0.3 * 0.8 * 0.1 + 0.25 * 0.4 * 0.3 * 0.2 * 0.1 (Case where you only can pick 4) + (Cases where you can pick 4 + 1 other number) (Cases where you can pick 4 + 2 other numbers) (Cases where you can pick 4 + 3 other numbers) I was thinking that I must be missing something
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:24
$begingroup$
I don’t think you were missing anything. Solving problems like this with a neat formula depends on the problem having symmetry. If the chance of adding each number were the same you could just find the chance you don’t get any number, subtract from $1$, and divide by $4$ because of the symmetry. Here you don’t have that, so you have to go through the cases.
$endgroup$
– Ross Millikan
Dec 23 '18 at 19:20
$begingroup$
Yeah I got as far but that is not an efficient solution; especially not when working with 20 cases; 0.4*0.7*0.8*0.9 + 0.5*0.4*0.3*0.8*0.9 + 0.5*0.4*0.7*0.2*0.9 + 0.5*0.4*0.7*0.8*0.1 + 0.3333*0.4*0.3*0.2*0.9 + 0.3333 * 0.4* 0.7 * 0.2 * 0.1 + 0.3333 * 0.4 * 0.3 * 0.8 * 0.1 + 0.25 * 0.4 * 0.3 * 0.2 * 0.1 (Case where you only can pick 4) + (Cases where you can pick 4 + 1 other number) (Cases where you can pick 4 + 2 other numbers) (Cases where you can pick 4 + 3 other numbers) I was thinking that I must be missing something
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:24
$begingroup$
Yeah I got as far but that is not an efficient solution; especially not when working with 20 cases; 0.4*0.7*0.8*0.9 + 0.5*0.4*0.3*0.8*0.9 + 0.5*0.4*0.7*0.2*0.9 + 0.5*0.4*0.7*0.8*0.1 + 0.3333*0.4*0.3*0.2*0.9 + 0.3333 * 0.4* 0.7 * 0.2 * 0.1 + 0.3333 * 0.4 * 0.3 * 0.8 * 0.1 + 0.25 * 0.4 * 0.3 * 0.2 * 0.1 (Case where you only can pick 4) + (Cases where you can pick 4 + 1 other number) (Cases where you can pick 4 + 2 other numbers) (Cases where you can pick 4 + 3 other numbers) I was thinking that I must be missing something
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:24
$begingroup$
I don’t think you were missing anything. Solving problems like this with a neat formula depends on the problem having symmetry. If the chance of adding each number were the same you could just find the chance you don’t get any number, subtract from $1$, and divide by $4$ because of the symmetry. Here you don’t have that, so you have to go through the cases.
$endgroup$
– Ross Millikan
Dec 23 '18 at 19:20
$begingroup$
I don’t think you were missing anything. Solving problems like this with a neat formula depends on the problem having symmetry. If the chance of adding each number were the same you could just find the chance you don’t get any number, subtract from $1$, and divide by $4$ because of the symmetry. Here you don’t have that, so you have to go through the cases.
$endgroup$
– Ross Millikan
Dec 23 '18 at 19:20
add a comment |
$begingroup$
Let $p(n) (nin{1,2,3,4})$ be the probability that number $n$ is added to the list (e.g., your $0.1, 0.2, 0.3, 0.4$). Then a formula for the probability that the outcome is the specific number $x$ is
$$1(Abcd)+{1over 2}(ABcd+AbCd+AbcD)+{1over 3}(ABCd+ABcD+AbCD)+{1over 4}(ABCD)
$$
where
$$begin{align}A&=p(x)\
B&=p(1+(xbmod 4))\
C&=p(1+((x+1)bmod 4))\
D&=p(1+((x+2)bmod 4))\
end{align}$$
and $a=1-A, b=1-B, c=1-C, d=1-D.$
Note that $B,C,D$ are simply the add-to-the-list probabilities for the three other numbers besides $x$, and they may be taken in any order. (The above formula just uses $bmod 4$ to express the choice of taking them in cyclic order.)
In Sage:
def p(x): return [1/10, 2/10, 3/10, 4/10][x-1]
def Pout(x):
A = p(x)
B = p(1 + (x%4))
C = p(1 + ((x+1)%4))
D = p(1 + ((x+2)%4))
a = 1 - A
b = 1 - B
c = 1 - C
d = 1 - D
return (A*b*c*d)+(A*B*c*d+A*b*C*d+A*b*c*D)/2+(A*B*C*d+A*B*c*D+A*b*C*D)/3+(A*B*C*D)/4
print [Pout(x) for x in [1..4]], 1-sum([Pout(x) for x in [1..4]])
print [Pout(x).n() for x in [1..4]], 1-sum([Pout(x).n() for x in [1..4]])
[473/7500, 1981/15000, 521/2500, 4411/15000] 189/625
[0.0630666666666667, 0.132066666666667, 0.208400000000000, 0.294066666666667] 0.302400000000000
answered Dec 26 '18 at 15:19
r.e.s.r.e.s.
7,66311953
$endgroup$
add a comment |
$begingroup$
Let $p(n) (nin{1,2,3,4})$ be the probability that number $n$ is added to the list (e.g., your $0.1, 0.2, 0.3, 0.4$). Then a formula for the probability that the outcome is the specific number $x$ is
$$1(Abcd)+{1over 2}(ABcd+AbCd+AbcD)+{1over 3}(ABCd+ABcD+AbCD)+{1over 4}(ABCD)
$$
where
$$begin{align}A&=p(x)\
B&=p(1+(xbmod 4))\
C&=p(1+((x+1)bmod 4))\
D&=p(1+((x+2)bmod 4))\
end{align}$$
and $a=1-A, b=1-B, c=1-C, d=1-D.$
Note that $B,C,D$ are simply the add-to-the-list probabilities for the three other numbers besides $x$, and they may be taken in any order. (The above formula just uses $bmod 4$ to express the choice of taking them in cyclic order.)
In Sage:
def p(x): return [1/10, 2/10, 3/10, 4/10][x-1]
def Pout(x):
A = p(x)
B = p(1 + (x%4))
C = p(1 + ((x+1)%4))
D = p(1 + ((x+2)%4))
a = 1 - A
b = 1 - B
c = 1 - C
d = 1 - D
return (A*b*c*d)+(A*B*c*d+A*b*C*d+A*b*c*D)/2+(A*B*C*d+A*B*c*D+A*b*C*D)/3+(A*B*C*D)/4
print [Pout(x) for x in [1..4]], 1-sum([Pout(x) for x in [1..4]])
print [Pout(x).n() for x in [1..4]], 1-sum([Pout(x).n() for x in [1..4]])
[473/7500, 1981/15000, 521/2500, 4411/15000] 189/625
[0.0630666666666667, 0.132066666666667, 0.208400000000000, 0.294066666666667] 0.302400000000000
answered Dec 26 '18 at 15:19
r.e.s.r.e.s.
7,66311953
$endgroup$
add a comment |
$begingroup$
Let $p(n) (nin{1,2,3,4})$ be the probability that number $n$ is added to the list (e.g., your $0.1, 0.2, 0.3, 0.4$). Then a formula for the probability that the outcome is the specific number $x$ is
$$1(Abcd)+{1over 2}(ABcd+AbCd+AbcD)+{1over 3}(ABCd+ABcD+AbCD)+{1over 4}(ABCD)
$$
where
$$begin{align}A&=p(x)\
B&=p(1+(xbmod 4))\
C&=p(1+((x+1)bmod 4))\
D&=p(1+((x+2)bmod 4))\
end{align}$$
and $a=1-A, b=1-B, c=1-C, d=1-D.$
Note that $B,C,D$ are simply the add-to-the-list probabilities for the three other numbers besides $x$, and they may be taken in any order. (The above formula just uses $bmod 4$ to express the choice of taking them in cyclic order.)
In Sage:
def p(x): return [1/10, 2/10, 3/10, 4/10][x-1]
def Pout(x):
A = p(x)
B = p(1 + (x%4))
C = p(1 + ((x+1)%4))
D = p(1 + ((x+2)%4))
a = 1 - A
b = 1 - B
c = 1 - C
d = 1 - D
return (A*b*c*d)+(A*B*c*d+A*b*C*d+A*b*c*D)/2+(A*B*C*d+A*B*c*D+A*b*C*D)/3+(A*B*C*D)/4
print [Pout(x) for x in [1..4]], 1-sum([Pout(x) for x in [1..4]])
print [Pout(x).n() for x in [1..4]], 1-sum([Pout(x).n() for x in [1..4]])
[473/7500, 1981/15000, 521/2500, 4411/15000] 189/625
[0.0630666666666667, 0.132066666666667, 0.208400000000000, 0.294066666666667] 0.302400000000000
answered Dec 26 '18 at 15:19
r.e.s.r.e.s.
7,66311953
$endgroup$
Let $p(n) (nin{1,2,3,4})$ be the probability that number $n$ is added to the list (e.g., your $0.1, 0.2, 0.3, 0.4$). Then a formula for the probability that the outcome is the specific number $x$ is
$$1(Abcd)+{1over 2}(ABcd+AbCd+AbcD)+{1over 3}(ABCd+ABcD+AbCD)+{1over 4}(ABCD)
$$
where
$$begin{align}A&=p(x)\
B&=p(1+(xbmod 4))\
C&=p(1+((x+1)bmod 4))\
D&=p(1+((x+2)bmod 4))\
end{align}$$
and $a=1-A, b=1-B, c=1-C, d=1-D.$
Note that $B,C,D$ are simply the add-to-the-list probabilities for the three other numbers besides $x$, and they may be taken in any order. (The above formula just uses $bmod 4$ to express the choice of taking them in cyclic order.)
In Sage:
def p(x): return [1/10, 2/10, 3/10, 4/10][x-1]
def Pout(x):
A = p(x)
B = p(1 + (x%4))
C = p(1 + ((x+1)%4))
D = p(1 + ((x+2)%4))
a = 1 - A
b = 1 - B
c = 1 - C
d = 1 - D
return (A*b*c*d)+(A*B*c*d+A*b*C*d+A*b*c*D)/2+(A*B*C*d+A*B*c*D+A*b*C*D)/3+(A*B*C*D)/4
print [Pout(x) for x in [1..4]], 1-sum([Pout(x) for x in [1..4]])
print [Pout(x).n() for x in [1..4]], 1-sum([Pout(x).n() for x in [1..4]])
[473/7500, 1981/15000, 521/2500, 4411/15000] 189/625
[0.0630666666666667, 0.132066666666667, 0.208400000000000, 0.294066666666667] 0.302400000000000
answered Dec 26 '18 at 15:19
r.e.s.r.e.s.
7,66311953
answered Dec 26 '18 at 15:19
r.e.s.r.e.s.
7,66311953
answered Dec 26 '18 at 15:19
r.e.s.r.e.s.
7,66311953
answered Dec 26 '18 at 15:19
r.e.s.r.e.s.
7,66311953
7,66311953
add a comment |
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$begingroup$
The % to get 0 numbers is (1-0.1) * (1-0.2) * (1-0.3) * (1-0.4) = 30.24%
$endgroup$
– Gerben Banaan
Dec 23 '18 at 18:00