Question

In a certain medical survey, 45 percent of the people surveyed had the type A antigen in their blood and 3 percent had both the type A antigen and the type B antigen. Which of the following is closest to the percent of those with the type A antigen who also had the type B antigen? [Official GMAT-2018]

Hint:

To find the percent of people with both conditions in conditional probability, divide the probability of both conditions by the probability of the given condition.

Answer 1

Step 1: Define the given percentages.
Let:
- \( P(A) = 45\% \) be the percent of people with the type A antigen.
- \( P(A \cap B) = 3\% \) be the percent of people with both the type A antigen and the type B antigen.
Step 2: Calculate the percent of people with the type A antigen who also had the type B antigen.
We need to find \( P(B \mid A) \), the conditional probability that a person with the type A antigen also has the type B antigen. The formula for conditional probability is:
\[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \]
Substitute the values:
\[ P(B \mid A) = \frac{3\%}{45\%} = \frac{3}{45} = \frac{1}{15} \approx 0.0667 = 6.67\% \] Step 3: Conclusion.
Thus, the percent of people with the type A antigen who also had the type B antigen is approximately 6.67%.