If 15% of adults had neither a flu shot nor an annual exam, what percent had at least one of the two? Then think about how this relates to the formula for $P(A\text{ or }B)$.

Use the relationship $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$ with the values you know to find the percent who had both a flu shot and an exam.

Once you know the percent who had both, divide by the percent who had an exam to get the conditional probability. Then convert that to a percent and round to the nearest tenth.

In Desmos, type the expression (0.6 + 0.7 - 0.85) / 0.7 * 100. This uses $0.6$ and $0.7$ for the two individual percents, $0.85$ for the percent with at least one (since $1 - 0.15 = 0.85$), divides by $0.7$ (the exam group), and multiplies by $100$ to convert to a percent. Read off the numerical result shown by Desmos and round it to the nearest tenth of a percent.

Translate the percentages into probabilities (or think of them as percents directly):

• Flu shot: $60\%$
• Annual exam: $70\%$
• Neither: $15\%$

From 15% with neither, it follows that $100\%-15\%=85\%$ of adults had at least one of the two (flu shot or exam).

Let $F$ be "had a flu shot" and $E$ be "had an annual exam."

We know:

• $P(F)=60\%$
• $P(E)=70\%$
• $P(F\text{ or }E)=85\%$ (because they had at least one)

Use the union formula:

Plug in the known values and solve for $P(F\text{ and }E)$ (the percent who had both).

From the formula:

So

Rearrange to get

So 45% of adults had both a flu shot and an annual exam.

The question asks: "what percent of adults who had an annual physical exam also received a flu shot?" That is the conditional probability $P(F\mid E)$, which is

Substitute the values:

Compute this fraction as a decimal, then multiply by 100 to express it as a percent and round to the nearest tenth. The result is 64.3%, which matches choice C.