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) 1 will have the disease?
(b) 4 will have the disease?
(c) at least 3 will neither have the disease nor be a carrier?
(d) the first child with the disease will be child number 4?

Answer 1

Let the random variable "X" : The number of children having the disease. "X" can take values 0,1,2,3 and 4 .The chance that a child is having the disease is 25% .

X ~ B (4, 0.25 )

a) Probability that 1 will have the disease P(X = 1 ) = ⁴C₁ * 0.25¹ * 0.75^4-1

= 0.4219

b)  Probability that 4 will have the disease P(X = 1 ) = ⁴C₄ * 0.25⁴ * 0.75^4-4

= 0.0039

c) Probability that at least 3 will neither have the disease nor be a carrier P(X > = 3)

= ⁴C₃ * 0.25³ * 0.75^4-+  ⁴C₄ *  0.25⁴ * 0.75^{4-4 = 0.05078}     

^d) Probability that the first child with the disease will be child number 4

= P (not 1^st) * P (not 2^nd) * P (not 3^rd​​​​​​​) * P (4^th​​​​​​​)  

= 0.75 * 0.75 * 0.75 * 0.25

= 0.1055