Question

During an influenza epidemic, we note that in a household of mother, father and two children, the probability that mother gets influenza is 10%. Furthermore, we see that the probability that father get influenza is also 10% while the probability that both mother and father get influenza is 2%. For the children, we assume a probability of 20% that each child will get influenza and a probability of 10% that both children get the disease at the same time. (a) Are the events that mother and father get influenza independent of each other? (b) What is the probability that at least one child will get influenza? (c) What is the conditional probability that the father has influenza given the the mother has influenza? 5 (d) What is the conditional probability that one child has influenza given that the other child does not have influenza?

Answer 1

The independence of the parents' flu was assessed, and it was found that the events are not independent. The probability of at least one child having the flu is 36%, and various conditional probabilities concerning parental and sibling influenza were calculated.

Step-by-step explanation:

Probability Questions Answered

a) To check if the events that mother and father get influenza are independent, we compare the product of their individual probabilities to the joint probability. Since 10% (mother) * 10% (father) equals 1%, which is not equal to the given 2% joint probability, the events are not independent.

b) To find the probability that at least one child will get influenza, we calculate the complement of the probability that neither child gets sick. This is 1 - (80% * 80%) = 36%.

c) The conditional probability that the father has influenza given that the mother has influenza is given by the joint probability divided by the probability of the mother getting influenza, which is 2% / 10% = 20%.

d) To find the conditional probability that one child has influenza given that the other does not, we use the fact that there is a 10% probability both children get sick. So if one child isn't sick, the probability the other is sick is the individual probability minus the joint probability, which is 20% - 10% = 10%.

Answer 2

i could only work out (A)

Step-by-step explanation:

Given: P(MOTHER) = 0.10 P(FATHER) = 0.10 P(MOTHER AND FATHER)= 0.02 P(FIRST CHILD) = 0.20 P(SECOND CHILD) = 0.20 P(FIRST CHILD AND SECOND CHILD) = 0.10 So, we get: P(MOTHER) X P(FATHER) = 0.10X 0.10 = 0.01 Since P(MOTHER) X P(FATHER) = 0.01 P(MO