Question

1. In the US, 45% of the population has blood type O, 40% have blood type A, 11% have blood type B, and 4 percent have blood type AB. Two individuals are chosen independently.  

a. What is the probability they both have blood type AB?

b. What is the probability they both have blood type B?

c. In what situation must you consider conditional probabilities?

2. A large national sample of health care visits to the VA indicates that 9.1% of veterans over age 60 have chronic kidney disease. The rate is 14.1% among veterans in their 60s with diabetes and 6.4% among veterans in their 60s without diabetes.

a. P(chronic kidney disease) =

b. P(chronic kidney disease | diabetes) =

c. P(chronic kidney disease | no diabetes) =

Answer 1

1. P( type O) =0.45

P(type A) = 0.40

P(type B) = 0.11

P( type AB) = 0.04

a) Probability that both have blood type AB= P(type AB) *P(type AB) = 0.04*0.04 = 0.0016

b) Probability that both have blood type B= P(type B) *P(type B) = 0.11*0.11 = 0.0121

Note : As two individuals are selected indepedently ,probability of an individual having a blood group is independent of others , thus we multiply the individual probabilities to get the joint probabilities

Let B be the event that one has blood group B

P( both have blood B) = P(BB) =P(B)*P(B)

c) As two individuals are selected indepedently ,probability of an individual having a blood group is independent of others

But if the two individuals are selected in such a way that the two of them are next of kin ( siblings, parent child ) , then conditional probabilities are used.

2(a) P( chronic kidney disease ) = 0.091

Note : 9.1% have chronic kidney disease

P( chronic kidney disease I diabetes) = 0.141

Note : 14.1% with diabetes have chronic kidney disease

P( chronic kidney disease I no diabetes) = 0.064

Note : 6.4% without diabetes have chronic kidney disease