Answer:
a) (1/2)^3 / (x^3 + (1/2)^3)
b) (1/2) (1/2 + 1 - x)
Where x is the probability that a prince has not hemophilia given that the queen is not a carrier
Step-by-step explanation:
We can use Bayes' theorem to calculate this porbability,
let:
A = The event for the queen has hemophilia
B = The event for a prince to have hemophilia
We are looing for the porbability P(A | 3B)
Using Bayes' theorem:
P(A | 3B) = P(3B | A) ( P(A) / P(3B) )
We know that:
P(3B | A) = (1/2)^3
P(A) = 1/2
Now let's calculate P(3B), here we will assume that the porpability that a prince does not have hemophilia given that the queen is not a carrier is x
Therefore
P(3B) = (1/2)x^3 + (1/2) (1/2)^3
Replacing all the values:
P(A | 3B) = (1/2)^3 / (x^3 + (1/2)^3)
If x=1, that is, if it is 100% probable that the prince will not have hemphilia given that the queen is not a carrier:
P(A | 3B) = 0.111
If x=0.9, that is, if it is 90% probable that the prince will not have hemphilia given that the queen is not a carrier:
P(A | 3B) = 0.121
Now lets calculate the probabiity for a fourth prince to have hemophilia:
If the queen has hemophilia, the probability is
(1/2)
If the queen does not have hemophilia, the probability is:
1-x
Therefore, the total probabilty is :
(1/2) (1/2 + 1 - x)
If x = 1, the probability is 1/4
If x=0.9, the porbaility is 0.3
