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 3 children, what is the probability that

(a) two will have the disease? (b) none will have the disease?

(c) at least one will neither have the disease nor be a carrier? (d) the first child with the disease will the be 3rd child?

Answer 1

Answer:

Step-by-step explanation:

Given that,

• p ( probability that the child has disease) = 25% = 0.25

• n = number of children = 3

The probability mass function of binomial distribution is,

• (P = X) = (nCx) X (p)^x X (1 - p)^n-x ; x = 0, 1, 2 ,3

• = 3Cx X (0.25)^x X (1 - 0.25)^3-x ; ( n = 3, p = 0.25

a) P ( two will have disease)

p ( X = 2) = 3C2 X (0.250^2 X (1 - 0.25) ^3-2

= 0.1406

b) P ( none will have disease)

p (X = 0) = 3C0 X (0.25)^0 X (1 - 0.25)^3-0

= 0.4219

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

The probability that at least one will neither having the disease nor being a carrier ;

P(X> or equals to) = 1 - P(X < 1)

= 1 - P( X = 0)

= 1 - 3C0 X (0.25)^0 X (1 - 0.25)^3-0

= 0.5781

d) p( the first child with the disease will the be 3rd child)

P(X = x) = (1-p)^x-1 X p

p( X= x) = ( 1 - 0.25 )^x -1 X 0.25

for third child = P(X = 3) = (1 - 0.25)^3-1 X (0.25)

= 0.1406