Question

I cannot crack this one, somebody please assist me

Answer 1

These
`N` outcomes make up the entire sample space, so

`\displaystyle \sum_(k=1)^N P(e_k) = P(e_1) + P(e_2) + P(e_3) + \cdots + P(e_N) = 1`

We're given that
`P(e_(j+1)) = 2 P(e_j)` for all
`j\in\{1,2,\ldots,N-1\}`, so

`P(e_1) + 2 P(e_1) + 2^2 P(e_1) + \cdots + 2^(N-1) P(e_1) = 1 \\\\ \implies P(e_1) = \frac1{1 + 2 + 2^2 + \cdots + 2^(N-1)} = \frac1{2^N - 1}`

Then we can solve the recurrence relation to get the probability of the
`j`-th outcome,

`P(e_(j+1)) = 2 P(e_j) = 2^2 P(e_(j-1)) = 2^3 P(e_(j-2)) = \cdots \\\\ \implies P(e_(j+1)) = 2^j P(e_1) \\\\ \implies P(e_j) = 2^(j-1) P(e_1) = (2^(j-1))/(2^N - 1)`

The probability of getting this sequence of
`k` outcomes is then

`\displaystyle P(E_k) = P(e_1) + P(e_2) + \cdots + P(e_k) = \sum_(j=1)^k (2^(j-1))/(2^N-1) = (2^k-1)/(2^N-1)`

as required.

Some preliminary results: If
`S` is the sum of the first
`n` terms of a geometric series with first term
`a` and common ratio
`r`, then

`S = a + ar + ar^2 + \cdots + ar^(n-1)`

`\implies rS = ar + ar^2 + ar^3 + \cdots + ar^n`

`\implies S - rS = a(1 - r^n)`

`\implies S = (a(1 - r^n))/(1 - r)`

which gives us, for instance,

`1 + 2 + 2^2 + \cdots + 2^(N-1) = (1 - 2^N)/(1 - 2) = 2^N-1`