Ask The Wizard #400
Farmer Brown lets his six of sheep graze a grass-covered fenced-in portion of his field. It takes them three days to clear the field of grass.
He then let's the grass regrow to the original height.
He then lets three of his sheep into the same field and it takes them seven days to clear the field.
He then let's the grass grow again to the original height and let's one sheep into the field. How long will it take that one sheep to clear it?
Assume that sheep eat grass at a constant rate and the grass also grows at another constant rate.
Let i = days it takes one sheep to eat the initial field of grass, assuming it didn't grow.
Let g = grass growth in one day.
We are given that it takes six sheep three days to consume the initial grass and three days worth of growth. We can express that as a formula as:
i + 3g = 3*6
We are also given that it takes three sheep seven days to consume the initial grass and three days worth of growth. We can express that as a formula as:
i + 7g = 7*3
We have two equations and two unknowns:
i + 3g = 18
i + 7g = 21
It is easy to solve for i and g as:
i = 63/4 = 15.75
g = 3/4 = 0.75
The question asks how long will it take one sheep to clear the field? Let's let x be that answer. We can express the equation as:
i + xg = x
(63/4) + (3/4)g = x
63/4 = x/4
x = 63.
So, it would take one sheep 63 days to clear the field.
A cuboid has dimensions x by y by z. It is composed of xyz individual cubes. Somebody paints all the exterior sides. What are the dimensions if the number of cubes that were pained equal the number that were not painted?
I come up with 20 distinct dimensions that work. Here they are.
- 5 X 13 X 132
- 5 X 14 X 72
- 5 X 15 X 52
- 5 X 16 X 42
- 5 X 17 X 36
- 5 X 18 X 32
- 5 X 20 X 27
- 5 X 22 X 24
- 6 X 9 X 56
- 6 X 10 X 32
- 6 X 11 X 24
- 6 X 12 X 20
- 6 X 14 X 16
- 7 X 7 X 100
- 7 X 8 X 30
- 7 X 9 X 20
- 7 X 10 X 16
- 8 X 8 X 18
- 8 X 9 X 14
- 8 X 10 X 12
You have mentioned, many times, the average trials needed for an event of probability p to occur is 1/p. My challenge to you is to prove it is true.
Let x = The expected number of trials for an event to happen.
x = 1*p + (1-p)*(1+x)
x = p + 1 + x - p - px
Subtracting x from both sides:
0 = p + 1 - p - px
Canceling p and -p:
0 = 1 - px
px = 1
x = 1/p
Let's define q = 1-p. In other words the probability that an event did not happen.
Let x = The expected number of trials for an event to happen.
x = 1 * pr(one trial needed) + 2 * pr(two trials needed) + 3 * pr(three trials needed) + ...
= 1p + 2pq + 3pq^2 + 4pq^3 + ...
x/p = 1 + 2q + 3q^2 + 4q^3 + ...
x/p - 1 = 2q + 3q^2 + 4q^3 + ...
x/p - 1 = q * (2 + 3q + 4q^2 + 5q^3 + ...)
x/p - 1 = q * (1 + 2q + 3q^2 + 4q^3 + ... + 1 + q + q^2 + q^3 + ...)
x/p - 1 = q * (x/p + 1 + q + q^2 + q^3 + ...)
Let y = 1 + q + q^2 + q^3 + ...
y-1 = q + q^2 + q^3 + ...
y-1 = q * (1 + q + q^2 + q^3 + ... )
(y-1)/q = 1 + q + q^2 + q^3 + ...
(y-1)/q = y
y/q - y = 1/q
y*(1/q - 1) = 1/q
y*(1/q - q/q) = 1/q
y*[(1-q)/q] = 1/q
y*(1-q) = 1
y = 1/(1-q)
x/p - 1 = q * (x/p + 1/(1-q))
x/p - 1 = q * (x/p + 1/p)
x/p - 1 = q * (1+x)/p
x/p - q * (1+x)/p = 1
x/p - qx/p = 1 + q/p
x*(1/p - q/p) = 1+q/p
x*(1-q)/p = 1+q/p
x*p/p = 1+q/p
x = 1+q/p
x = 1 + (1-p)/p
x = p/p + (1-p)/p
x = 1/p