Answer:
(1) a. 0.0009
(2) d. 0.640
(3)
- a. P(A and B) = 0.06.
- b. P(A or B) = 0.70.
(4)Not disjoint
(5) a. nearly 0.
(6)b. 0.919
Step-by-Step Explanation:
(1)Probability of a baby being born with a birth defect =3%=0.03
The probability that both babies have birth defects=0.03 X 0.03= 0.0009.
(2)The probability of contracting the influenza virus each year = 20%=0.2
Therefore, the probability of not contracting the influenza virus =1-0.2=0.8
The probability that neither baby catches the flu in a given year:
=0.8 X 0.8
=0.64
(3)
P(A)=0.1
P(B)=0.6
P(A or B)=P(A)+P(B)=0.1 + 0.6 =0.7
P(A and B)=P(A)XP(B)=0.1 X 0.6 =0.06
(4)
P(A)=0.2
P(B)=0.9
Event A and B cannot be disjoint.
(5)
The probability of an American woman aged 20 to 24 having Chlamydia infection
The probability that three randomly selected women in this age group have the infection
(6)The probability of an American woman aged 20 to 24 not having Chlamydia infection
The probability that three randomly selected women in this age group do not have the infection
