Chapter 3: Q.3.70 (page 104)

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 the fourth prince, what is the probability that he will have hemophilia?

Short Answer

Expert verified

Events $E_{i}$ , that the $i$ -th son is a hemophiliac are conditionally independent given that the queen is carrier $(C)$

P (C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = \frac{1}{9} P (E_{4} ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = \frac{1}{18}

Step by step solution

Given Information

Events

$C$ - the Queen is a carrier

$E_{i}$ - the Queen's $i$ -th son is a hemophiliac

Probabilities:

$P (C) = \frac{1}{2}$

$P (E_{i} ∣ C) = \frac{1}{2}$

$P (E_{i} ∣ C^{c}) = 0$

Events $E_{i}$ are independent given $C$ .

Explanation

Compute: $P (C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}), P (E_{4} ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c})$ ,

Start with the definition of conditional probability:

$P (C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = \frac{P (C E_{1}^{c} E_{2}^{c} E_{3}^{c})}{P (E_{1}^{c} E_{2}^{c} E_{3}^{c})}$

Condition upon whether the queen is a carrier:

P (C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = \frac{P (E_{1}^{c} E_{2}^{c} E_{3}^{c} ∣ C) P (C)}{P (E_{1}^{c} E_{2}^{c} E_{3}^{c} ∣ C) P (C) + P (E_{1}^{c} E_{2}^{c} E_{3}^{c} ∣ C^{c}) P (C^{c})}

Explanation

Apply the conditional independence of $E_{i}$ 's - probability of intersection is the product of probabilities

$P (C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = \frac{P (E_{1}^{c} ∣ C) P (E_{2}^{c} ∣ C) P (E_{3}^{c} ∣ C) P (C)}{P (E_{1}^{c} ∣ C) P (E_{2}^{c} ∣ C) P (E_{3}^{c} ∣ C) P (C) + P (E_{1}^{c} ∣ C^{c}) P (E_{2}^{c} ∣ C^{c}) P (E_{3}^{c} ∣ C^{c}) P (C^{c})}$

Substitute $P (E_{i}^{c} ∣ C) = 1 / 2, P (C) = P (C^{c}) = 1 / 2, P (E_{i}^{c} ∣ C^{c}) = 1$ :

$P (C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = \frac{{(\frac{1}{2})}^{4}}{{(\frac{1}{2})}^{4} + 1^{3} \cdot \frac{1}{2}}$

$= \frac{1}{9}$

Explanation

$P (E_{4} ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c})$

Calculate the probability by conditioning upon $C$ . This is the Bayes formula with conditional probability $P (\cdot ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c})$

$P (E_{4} ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = P (E_{4} C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) + P (E_{4} C^{c} ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c})$

$= P (E_{4} ∣ C E_{1}^{c} E_{2}^{c} E_{3}^{c}) P (C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) + P (E_{4} ∣ C^{c} E_{1}^{c} E_{2}^{c} E_{3}^{c}) P (C^{c} ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c})$

Since events $E_{i}$ are conditionally independent given $C$ or $C^{c}$ , the conditioning upon more $E_{1}^{c}, E_{2}^{c}, E_{3}^{c}$ can be removed.

P (E_{4} ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = P (E_{4} ∣ C) P (C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) + P (E_{4} ∣ C^{c}) P (C^{c} ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c})

$= \frac{1}{2} \cdot \frac{1}{9} + 0 \cdot \frac{8}{9}$

$= \frac{1}{18}$

Final Answer

Events $E_{i}$ , that the $i$ -th son is a hemophiliac are conditionally independent given that the queen is carrier $(C)$ :

P (C ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = \frac{1}{9} P (E_{4} ∣ E_{1}^{c} E_{2}^{c} E_{3}^{c}) = \frac{1}{18}

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