Sickle Cell Anemia Sickle cell anemia is a genetic blood disorder where red blood cells lose their flexibility and assume an abnormal, rigid, sickle shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a 25% chance of having the disease, 50% chance of being a carrier, and 25% chance of neither having the disease nor being a carrier. If two parents who are carriers of the disease have 6 children, what is the probability that: (a) 0 will have the disease? (b) 5 will have the disease? (c) at least 2 will neither have the disease nor be a carrier? (d) the first child with the disease will be child number 1?

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Chapter 3: Distributions of Random Variables
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I am fir
Sickle Cell Anemia
Sickle cell anemia is a genetic blood disorder where red blood cells lose their flexibility and assume an abnormal, rigid,
sickle shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a
25% chance of having the disease, 50% chance of being a carrier, and 25% chance of neither having the disease nor
being a carrier. If two parents who are carriers of the disease have 6 children, what is the probability that:
(a) 0 will have the disease?
(b) 5 will have the disease?
(c) at least 2 will neither have the disease nor be a carrier?
(d) the first child with the disease will be child number 1?
Submit answer
M
US V 0 10
UCH
Transcribed Image Text:Chapter 3: Distributions of Random Variables OPEN I am fir Sickle Cell Anemia Sickle cell anemia is a genetic blood disorder where red blood cells lose their flexibility and assume an abnormal, rigid, sickle shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a 25% chance of having the disease, 50% chance of being a carrier, and 25% chance of neither having the disease nor being a carrier. If two parents who are carriers of the disease have 6 children, what is the probability that: (a) 0 will have the disease? (b) 5 will have the disease? (c) at least 2 will neither have the disease nor be a carrier? (d) the first child with the disease will be child number 1? Submit answer M US V 0 10 UCH
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Hello. Since your question has multiple sub-parts, we will solve first three sub-parts for you. If you want remaining sub-parts to be solved, then please resubmit the whole question and specify those sub-parts you want us to solve.

We have,

n = 6

p1 = probability that the child has disease = 25% = 0.25

p2 = probability that the child will be a carrier of the disease = 50% = 0.50

p3 = probability that the child neither will be a carrier of the disease nor will have the disease = 25% = 0.25

Part a:

Let X denote the number of children that will have the disease.

Thus, X~B(n = 6, p1 =0.25)

The probability that 0 will have the disease is computed as,

P(X=0)=C06(0.25)0(1-0.25)6-0                         =0.1780

Thus, the probability that 0 will have the disease is 0.1780.

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