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

umumumumum

im not sure

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation: