In a clinic, 70% of patients are vaccinated against the flu. Among vaccinated patients, 90% do not contract the flu, while among unvaccinated patients, only 40% do not contract the flu.
If a patient is known to have contracted the flu, what is the probability that they were unvaccinated?
The Correct Answer and Explanation is:
To find the probability that a patient who contracted the flu was unvaccinated, we can use Bayes’ Theorem.
Step 1: Define the Events
Let’s define:
- VV: Patient is vaccinated
- UU: Patient is unvaccinated
- FF: Patient contracted the flu
- FcF^c: Patient did not contract the flu
Step 2: Given Information
- P(V)=0.70⇒P(U)=1−0.70=0.30P(V) = 0.70 \Rightarrow P(U) = 1 – 0.70 = 0.30
- P(Fc∣V)=0.90⇒P(F∣V)=1−0.90=0.10P(F^c|V) = 0.90 \Rightarrow P(F|V) = 1 – 0.90 = 0.10
- P(Fc∣U)=0.40⇒P(F∣U)=1−0.40=0.60P(F^c|U) = 0.40 \Rightarrow P(F|U) = 1 – 0.40 = 0.60
Step 3: Use the Law of Total Probability to Find P(F)P(F)
P(F)=P(F∣V)⋅P(V)+P(F∣U)⋅P(U)P(F) = P(F|V) \cdot P(V) + P(F|U) \cdot P(U) P(F)=(0.10⋅0.70)+(0.60⋅0.30)=0.07+0.18=0.25P(F) = (0.10 \cdot 0.70) + (0.60 \cdot 0.30) = 0.07 + 0.18 = 0.25
Step 4: Use Bayes’ Theorem to Find P(U∣F)P(U|F)
P(U∣F)=P(F∣U)⋅P(U)P(F)=0.60⋅0.300.25=0.180.25=0.72P(U|F) = \frac{P(F|U) \cdot P(U)}{P(F)} = \frac{0.60 \cdot 0.30}{0.25} = \frac{0.18}{0.25} = 0.72
✅ Final Answer: 0.72\boxed{0.72} or 72%
Explanation (300+ Words)
This problem is a classic application of Bayes’ Theorem, which allows us to reverse conditional probabilities — in this case, determining the likelihood that a patient was unvaccinated, given that they contracted the flu.
We begin by breaking the population into two groups: vaccinated (70%) and unvaccinated (30%). Among those vaccinated, only 10% contract the flu, thanks to the vaccine’s effectiveness. Conversely, 60% of unvaccinated individuals get the flu, highlighting the higher risk in this group.
To find the overall chance that any patient gets the flu, we use the law of total probability. This combines the probabilities of flu in both groups, weighted by how common each group is. We calculate: P(F)=0.07(vaccinatedflucases)+0.18(unvaccinatedflucases)=0.25P(F) = 0.07 (vaccinated flu cases) + 0.18 (unvaccinated flu cases) = 0.25
This means 25% of all patients contract the flu.
Now, to find the likelihood that a flu patient was unvaccinated, we apply Bayes’ Theorem: P(U∣F)=P(F∣U)⋅P(U)P(F)=0.180.25=0.72P(U|F) = \frac{P(F|U) \cdot P(U)}{P(F)} = \frac{0.18}{0.25} = 0.72
So, there is a 72% chance that a patient who contracted the flu was unvaccinated.
This result makes intuitive sense. Even though unvaccinated patients are the minority (30%), they account for a disproportionate number of flu cases (60% chance of infection), so they form the majority (72%) of those who end up with the flu. This highlights the protective effect of vaccination and supports the importance of increasing vaccine coverage to reduce overall flu incidence.