There is a 50–50 chance that the queen carries the gene for hemophilia. if she is a carrier, then each prince has a 50–50 chance of having hemophilia. if the queen has had three princes without the disease, what is the probability that the queen is a carrier? if there is a fourth prince, what is the probability that he will have hemophilia?

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

m∠vru = 80°

m∠urw = 15°

step-by-step explanation:

4sqrt5

step-by-step explanation:

the base has dimensions of 8 x 4, so the diagonal is 64 + 16 square rooted or 4sqrt5.