Question: You are currently quarantining in a house with 2 other people. All three of you decide to try and experimental vaccine, that is either effective(70% chance of preventing transmission) or ineffective(30% chance). A fourth friend who is infectious comes to stay with you. if all three of you subsequently become infected, what is the probability the vaccine is ineffective?
A. 70% B: 92.7% C: 86.4% D: 89.8%
Opinion: I found this question quite disburbing because even though this is meant as bayesian probability question, the events are definitely not clear enough.
If I write what the question wants as the conditional probility, it would be:
P(vaccine is ineffective|all three people are tested positive)
I don't know how to apply the law of total probability when solving this one, because I'm quite confused about the logical expression of the probability.
The only possibility that I understand would be,
P(all 3 people infected|vaccine is ineffective)
-> which means with 0.3 chance that the antibody didn't form, there the conditional probability that 3 people got infected given none of them formed an antibody, should be $1^3$.
P(vaccine is ineffective)
-> since we are talking about 3 people, we should get $0.3^3$
P(all 3 people infected|vaccine is effective)
-> This one gets me so confused. I interpreted as, although the antibody had formed for all 3 people, the vaccine turned out to be ineffective(which means $0.3^3$)
P(vaccine is effective)
-> if we follow the above expression, than it would be $0.7^3$
So by calculation, the probability should be $0.027/(0.027+(0.027 * 0.343)) = 0.7446$
but this is not in the options.
Could anyone give a different interpretation of this question, and possibly come up with the proper solution for this?
The problem makes the implicit assumption that the prior probability of effectiveness is $50\%$ (the so-called uniform noninformative prior):
$$P(\text{effective})=P(\text{ineffective})=0.5$$
Then with Bayes:
$$P(\text{ineffective }|\ 3 \text{ infected})=\\\frac{P(3 \text{ infected }| \text{ ineffective})P(\text{ineffective})}{P(3 \text{ infected }| \text{ ineffective})P(\text{ineffective})+P(3 \text{ infected }| \text{ effective})P(\text{effective})} = \frac{0.7^3\cdot 0.5}{0.7^3\cdot 0.5+0.3^3\cdot 0.5}\approx92.7\%$$
So the answer is B.