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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 Ei, that the i-th son is a hemophiliac are conditionally independent given that the queen is carrier(C)

PC∣E1cE2cE3c=19PE4∣E1cE2cE3c=118.

Step by step solution

01

Given Information

Events

C - the Queen is a carrier

Ei - the Queen's i-th son is a hemophiliac

Probabilities:

P(C)=12

PEi∣C=12

PEi∣Cc=0

Events Eiare independent given C.

02

Explanation

Compute:PC∣E1cE2cE3c,PE4∣E1cE2cE3c,

Start with the definition of conditional probability:

PC∣E1cE2cE3c=PCE1cE2cE3cPE1cE2cE3c

Condition upon whether the queen is a carrier:

PC∣E1cE2cE3c=PE1cE2cE3c∣CP(C)PE1cE2cE3c∣CP(C)+PE1cE2cE3c∣CcPCc
03

Explanation

Apply the conditional independence of Ei's - probability of intersection is the product of probabilities

PC∣E1cE2cE3c=PE1c∣CPE2c∣CPE3c∣CP(C)PE1c∣CPE2c∣CPE3c∣CP(C)+PE1c∣CcPE2c∣CcPE3c∣CcPCc

Substitute PEic∣C=1/2,P(C)=PCc=1/2,PEic∣Cc=1:

PC∣E1cE2cE3c=124124+13·12

=19

04

Explanation

PE4∣E1cE2cE3c

Calculate the probability by conditioning upon C. This is the Bayes formula with conditional probability P·∣E1cE2cE3c

PE4∣E1cE2cE3c=PE4C∣E1cE2cE3c+PE4Cc∣E1cE2cE3c

=PE4∣CE1cE2cE3cPC∣E1cE2cE3c+PE4∣CcE1cE2cE3cPCc∣E1cE2cE3c

Since events Ei are conditionally independent given C or Cc, the conditioning upon more E1c,E2c,E3c can be removed.

PE4∣E1cE2cE3c=PE4∣CPC∣E1cE2cE3c+PE4∣CcPCc∣E1cE2cE3c

=12·19+0·89

=118

05

Final Answer

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

PC∣E1cE2cE3c=19PE4∣E1cE2cE3c=118

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