This one tripped me up for a second because the answer feels like it should be zero (lazy reviewers never go negative), and it basically is.
Recognize this as a Bayes' theorem problem where you need to find P(Lazy | Negative). First compute the overall probability of a negative review using the law of total probability, then apply Bayes' theorem to find the conditional probability.
Pro tip: State your assumptions clearly (e.g., independence of reviews) and walk through the calculation step-by-step to show your reasoning. Also, mention that the result is counterintuitive because despite lazy reviewers being only 20% of the population, they contribute a significant portion of negative reviews due to their zero false negative rate.
Let L be the event that a reviewer is lazy, C for careful, and N for a negative review. Given: P(L)=0.2, P(C)=0.8, P(Positive|L)=1, so P(Negative|L)=0; P(Positive|C)=0.6, so P(Negative|C)=0.4.
Use the law of total probability: P(N) = P(N|L)P(L) + P(N|C)P(C) = 0*0.2 + 0.4*0.8 = 0.32.
Compute P(L|N) = P(N|L)P(L) / P(N) = 0*0.2 / 0.32 = 0.
The probability is 0, meaning that given a negative review, it is impossible that it came from a lazy reviewer. This is because lazy reviewers never leave negative feedback.
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