Math probability question help?

Number of times Donald gets fined = parkings x probability = 100 x 0.05 = 5 times

Total fines = $0 + $2 + $5 + $5 + $5 = $17

Cost of paying meter each time = 100 x $0.10 = $10

Difference = $17 - $10 = $ 7

Donald actually will lose $7. So he better use the meter :-)

1st: I guess the exercise should read "the EXPECTED difference". There is no way to tell the difference exactly.

That being said, the answer is the expected cost of the fines, minus 10$

The expected cost of the fines is simple

It is the sum P[X=0]fine(0) + P[X=1]fine(1) + P[X=2]fine(2) + ... + P[X=100]fine(100)

Where X is the number of time Donald has been caught. And fine(n) is the cost of being caught n times.

fine(0) and fine(1) is 0. (Never caught=no fine. Once caught=no fine either). Then fine(2)=2$. fine(3)=2+5$, fine(4)=2+5+5$ etc.

So, obviously fine(n)=5n-8 if n≥2, 0 else.

P[X=n] is the probability to get caught n times.

It is a binomial law.

Two possibilities (depending on whether your lesson includes usages of tools like R) from there

1) Just compute the sum.

In R :

dbinom(2,100,0.05)*2 + dbinom(3, 100, 0.05)*7 + dbinom(4, 100, 0.05)*12 + ...

Of course you should do a loop to compute

Σ dbinom(k, 100, 0.05)*(5k-8) for k in [2,100]

But even if you do the sum by hand, you will see that it converge really fast. After a while, no need to continue the additions.

2) Try to do a formal calculation

P[X=n] = C(100, n)*pⁿ*(1-p)¹⁻ⁿ

where p=0.05

And C(100,n)=100! / ((100-n)!n!) is the binomial coefficient

(I've done nothing here: that is just the definition)

So, what you are trying to do is to compute

Σ(for k=2 to 100) C(100,k)*pᵏ*(1-p)¹⁻ᵏ * (5k-8)

Which is

5Σ(for k=2 to 100) C(100,k)*pᵏ*(1-p)¹⁻ᵏ * k

- 8Σ(for k=2 to 100) C(100,k)*pᵏ*(1-p)¹⁻ᵏ

The second term of the sum is very easy to compute

Just remember that Σ(k=0 to 100) C(100,k)*pᵏ*(1-p)¹⁻ᵏ is (p + (1-p))¹⁰⁰

(That is just the development of the binomial, you have seen years ago. Hence the name of the binomial law)

Since p+1-p=1,

then Σ(k=0 to 100) C(100,k)*pᵏ*(1-p)¹⁻ᵏ = 1

So, from there, the second term of the sum should be easy to compute. The only difference between this formula (whose value is 1) and the one you are looking for is that k starts from 0 instead of 2.

The first term (5Σ(for k=2 to 100) C(100,k)*pᵏ*(1-p)¹⁻ᵏ * k )

is a little bit trickier.

But not that tricky!

Just replace C(100,k) by its value : 100! / (100-k)!k!

If you multiply this by k (since this is what you have to do in this second term), you get 100! / (100-k)!(k-1)!

So you are trying to calculate

5Σ(for k=2 to 100) 100! / ((100-k)!(k-1)!) *pᵏ*(1-p)¹⁻ᵏ

That's look quite similar to the 2nd term (the one we've calculated earlier)

The only difference is (k-1)! instead of k!

So be it. Make a variable change. Define i=k-1

And you get

5Σ(for i=1 to 99) 100! / ((99-i)! i!) * pⁱ⁺¹ * (1-p)⁹⁹⁻ⁱ

Don't you think that is looking a lot like the binomial development of

(p + (1-p)) ⁹⁹ = Σ(for i=0 to 99) 99! / ((99-i)! i!) * pⁱ * (1-p)⁹⁹⁻ⁱ

?

Since the value of this development is 1, you should be able to find quite easily the value you are looking for.

So, just subtract the 2 values we've computed here, and then subtract 10$

That is the answer to the exercise.

(And by the way, the answer you get that way is one of the proposed answers. Which is not the case of 14$ :-))

(Note: I can validate or invalidate your answer if you are unsure. But you have to try to apply the method I gave first)

Cost of fines = 0+0.1+98*0.25 =

$24.60

Cost of paying = $10.00

Difference = $14.60