Question

We have an infinite collection of biased coins, indexed by the positive integers. Coin i has probability 2−i of being selected. A flip of coin i results in Heads with probability 3−i. We select a coin and flip it. What is the probability that the result is Heads? The geometric sum formula may be useful here: ∑i=1∞αi=α1−α, when |α|<1.

The probability that the result is Heads is

Answer 1

1/5.
P(heads) = P(heads|coin) * P(coin)
which is the multiplication of the two geometric sums
solving for 2-i * 3-i = 6-i or (1/6)^i. plugging this into the geometric sum formula we get 1/5

Answer 2

I wonder how you arrived to 1/5?
Geometric sum formula looks like this when using α=6^(-i)

6^(-i)/(1-6^(-i))

Answer 3

Because the first term of the geometric sequence is 1 => 1 + (-6/5) = -1/5 take the absolute value = 1/5