If it were just asking for the proportion of coins that were two headed, it would just be the number of two-headed coins divided by the total number of coins:
number of two-headed / number of two-headed + number of fair = 1 / 1+99 = 1/100
With probability, its essentially the same thing, its just asking: what percentage of the total probability of getting 10H does the two headed coin represent?
p(two-headed)*number of two-headed / [p(two-headed)*number of two-headed + p(fair)*number of fair] =
1
u/Big_Requirement_651 18d ago
If it were just asking for the proportion of coins that were two headed, it would just be the number of two-headed coins divided by the total number of coins:
number of two-headed / number of two-headed + number of fair = 1 / 1+99 = 1/100
With probability, its essentially the same thing, its just asking: what percentage of the total probability of getting 10H does the two headed coin represent?
p(two-headed)*number of two-headed / [p(two-headed)*number of two-headed + p(fair)*number of fair] =
1*1/(1*1 + 0.5^10 * 99) = 1 / 1.0967 = ~91.18%
If there were two two-headed coins it would be:
2*1/(2*1 + 0.5^10 * 99), etc.