Question

Suppose that 39% of a town's population have type O blood, 15% are Rh-negative, and 7% have type o

blood and are Rh-negative. What is the probability that a randomly selected individual in the town will
neither have type O blood nor be Rh-negative?

Answer 1

39% of the population of your hypothetical town will have neither Rh-negative nor O type blood.

Answer 2

53 %

Explanation:

Let O represents the event of Type O blood and R represents the type Rh-negative blood,

According to the question,

P(O) = 39% = 0.39,

P(R) = 15% = 0.15,

P(O ∩ R) = 7% = 0.07,

We know that,

P(O ∪ R) = P(O) + P(R) - P(O ∩ R)

= 0.39 + 0.15 - 0.07

= 0.47

Hence, the probability of an individual who is neither have type O blood nor be Rh-negative = 1 - P(O ∪ R)

= 1 - 0.47

= 0.53

= 53 %