Probability of getting the flu Suppose that the probability of exposure to the flu during an epidemic is .6. Experience has
shown that a serum is 80% successful in preventing an inoculated person from acquiring the
flu, if exposed to it. A person not inoculated faces a probability of .90 of acquiring the flu if
exposed to it. Two persons, one inoculated and one not, perform a highly specialized task in a
business. Assume that they are not at the same location, are not in contact with the same people,
and cannot expose each other to the flu. What is the probability that at least one will get the flu?
The correct answer is 0.5952, I can't find my mistake. A: vaccinated person catching the flu
B: unvaccinated person catching the flu
F: exposed to flu
thus, I must find P(A ∪ B)
P(F)=0.60
P(A | F)=0.20
P(B | F)=0.90
A, B ⊆ F
$P(A│F)=(P( A∩F))/(P(F))→P(A│F)=P( A)/(P(F))→ 0.20=P( A)/0.60→P(A)=0.12$
$P(B│F)=(P( B∩F))/(P(F))→P(B│F)=P( B)/(P(F))→ 0.90=P( B)/0.60→P(B)=0.54$
$P(A ∪ B)=P(A)+ P(B) -P(A∩B)=0.12+0.54-0=0.66$
Ans: 0.66
A: The two events of $A$ and $B$ catching the flu are independent, meaning that $P(A\cap B)=P(A)P(B)$, not zero. (In the context of the question, you can't rule out the possibility that both of them catching it just because they don't come into contact)
Hence, $$P(A\cup B)=0.12+0.54-(0.12)(0.54)=0.5952$$ as required.
A: Response already given addressing OP's specific question - therefore open season.
Alternative approach:
Chance of inoculated person getting flu is
$$p_1 = .6 \times .2 = 0.12.$$
Therefore, chance of inoculated person not getting flu is
$$q_1 = 1 - p_1 = 0.88.$$
Chance of non-inoculated person getting flu is
$$p_2 = .6 \times .9 = 0.54.$$
Therefore, chance of non-inoculated person not getting flu is
$$q_2 = 1 - p_2 = 0.46.$$
Chance of both not getting flu is
$$q_1 \times q_2 = (0.88) \times (0.46) = 0.4048.$$
Therefore, chance of at least one getting flue is
$$1 - (q_1 \times q_2) = 0.5952.$$
