Question

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 4 children, what is the probability that:

(a) 0 will have the disease?

(b) 4 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?

Answer 1

Given that,

p ( child has disease) = 25% = 0.25

n = number of child =4

The pmf of binomial distribution is,

a) P ( 0 will have disease)

= 0.3164 Here we use excel function =BINOM.DIST(0,4,0.25,FALSE)

b) P ( 4 will have disease)

= 0.0039   Here we use excel function =BINOM.DIST(4,4,0.25,FALSE)

c) P (neither having the disease nor being a carrier) =25% = 0.25

p( at least 2 will neither have the disease nor be a carrier)

= 0.2617 Here we use excel function for finding p( x=0) & p ( x=1)

d) to find p (the first child with the disease will be child number 1) we use

the pmf of geometric distribution is,

p( x= 1) = ( 1 - 0.25 )^{1 -1} * 0.25

= 0.25