/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 There is a \(50-50\) chance that... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 70

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?

Short Answer

Expert verified
The probability that the queen is a carrier given that her three sons do not have hemophilia is approximately 11.11%. If there is a fourth prince, the probability that he will have hemophilia is around 5.56%.

Step by step solution

01

Understand the probabilities

We know that the queen has a 50% chance of being a carrier (C) and a 50% chance of not being a carrier (NC). If the queen is a carrier (C), each prince has a 50% chance of having hemophilia (H) and a 50% chance of not having hemophilia (NH). We are given that the first three princes do not have hemophilia (NH1, NH2, NH3).
02

Apply Bayes' theorem

Bayes' theorem states that P(A|B) = (P(B|A) * P(A)) / P(B), where A and B are events. In this case, we want to find the probability that the queen is a carrier (C) given that her three sons do not have hemophilia (NH1, NH2, NH3). Therefore, we need to find P(C|NH1, NH2, NH3). P(C|NH1, NH2, NH3) = (P(NH1, NH2, NH3|C) * P(C)) / P(NH1, NH2, NH3)
03

Calculate the probabilities

First, we find the probabilities: P(C) = 0.5 (given) P(NC) = 0.5 (given) P(NH1, NH2, NH3|C) = 0.5 * 0.5 * 0.5 = 0.125 (since the probabilities are all independent) P(NH1, NH2, NH3|NC) = 1 * 1 * 1 = 1 (since the probabilities are all 100% if the queen is not a carrier) Now, calculate P(NH1, NH2, NH3) using the law of total probability: P(NH1, NH2, NH3) = P(NH1, NH2, NH3|C) * P(C) + P(NH1, NH2, NH3|NC) * P(NC) P(NH1, NH2, NH3) = (0.125 * 0.5) + (1 * 0.5) = 0.0625 + 0.5 = 0.5625
04

Plug the probabilities into Bayes' theorem

Now, we plug these probabilities into Bayes' theorem: P(C|NH1, NH2, NH3) = (0.125 * 0.5) / 0.5625 = 0.0625 / 0.5625 鈮� 0.1111
05

Find the probability of the fourth prince having hemophilia

If the queen is a carrier (now with a probability of approximately 0.1111), there is a 50% chance that the fourth prince will have hemophilia (H4). So, the probability is: P(H4|C) = 0.5 We use the law of total probability again to find the probability that the fourth prince will have hemophilia (H4), given the new information that the queen is likely a carrier: P(H4) = P(H4|C) * P(C|NH1, NH2, NH3) + P(H4|NC) * P(NC|NH1, NH2, NH3) P(H4) = (0.5 * 0.1111) + (0 * (1 - 0.1111)) 鈮� 0.0556 So, the probability that the fourth prince will have hemophilia is around 5.56%.

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