A spider eats N flies a day. Until the fly fills his quota, a fly has 45% chance of survival if he attempts to pass a web.
Assume the M flies (M>N) have already made the attempt to pass. What is the probability that the (M+1)st fly will survive the attempt?
The probability of the complement, that is, that the spider will be eaten, is the sum from i = 0 to N-1 of the probability that exactly i flies have already been eaten, leaving the spider still hungry, multiplied by .55:
.55 * Sigma{i=0 to N-1} (.55^i * .45^(M-i) * C(M,i))
so the probability requested is
1 - .55 * Sigma{i=0 to N-1} (.55^i * .45^(M-i) * C(M,i))
Implementing this:
clc
fprintf(' ')
for m=1:15
fprintf(' %2d ',m)
end
disp(' ')
for n=1:20
fprintf('%2d',n)
for m=1:15
fprintf(' %5.3f',prob(m,n))
end
disp(' ')
end
function p=prob(M,N)
p=0;
if M<N
p=.45;
else
for i=0:N-1
p=p+(.55^i * .45^(M-i) * nchoosek(M,i));
end
p=1-p*.55;
end
end
It includes the 0.45 probability the fly escapes if M<N
Probability the fly escapes
\ M
N | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
|
1 0.752 0.889 0.950 0.977 0.990 0.995 0.998 0.999 1.000 1.000 1.000 1.000 1.000 1.000 1.000
2 0.450 0.616 0.766 0.867 0.928 0.962 0.980 0.990 0.995 0.998 0.999 0.999 1.000 1.000 1.000
3 0.450 0.450 0.542 0.665 0.776 0.860 0.916 0.951 0.973 0.985 0.992 0.996 0.998 0.999 0.999
4 0.450 0.450 0.450 0.500 0.591 0.693 0.785 0.857 0.909 0.944 0.966 0.980 0.989 0.994 0.997
5 0.450 0.450 0.450 0.450 0.478 0.540 0.624 0.712 0.792 0.856 0.904 0.939 0.962 0.977 0.986
6 0.450 0.450 0.450 0.450 0.450 0.465 0.506 0.571 0.649 0.727 0.798 0.857 0.902 0.935 0.958
7 0.450 0.450 0.450 0.450 0.450 0.450 0.458 0.485 0.532 0.596 0.668 0.740 0.804 0.858 0.900
8 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.455 0.471 0.505 0.555 0.617 0.685 0.750 0.809
9 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.453 0.463 0.486 0.524 0.575 0.636 0.699
10 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.451 0.458 0.473 0.501 0.542 0.593
11 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.451 0.455 0.465 0.485 0.516
12 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.453 0.459 0.473
13 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.452 0.456
14 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.451
15 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450
16 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450
17 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450
18 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450
19 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450
20 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450
>>
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Posted by Charlie
on 2024-10-31 10:02:16 |
proposed solution