To solve this problem, we will use the concept of conditional probability. Specifically, we need to find the probability that a person will buy popcorn given that they have already bought soda.
Let:
- [tex]\( P(A) \)[/tex] be the probability that a person buys a soda.
- [tex]\( P(B \cap A) \)[/tex] be the probability that a person buys both popcorn and soda.
We are looking to find [tex]\( P(B|A) \)[/tex], the conditional probability that a person buys popcorn given that they have bought soda. The formula for conditional probability is:
[tex]\[
P(B|A) = \frac{P(B \cap A)}{P(A)}
\][/tex]
For this problem:
- [tex]\( P(A) = 0.60 \)[/tex] (since 60% of the moviegoers bought a soda)
- [tex]\( P(B \cap A) = 0.45 \)[/tex] (since 45% of the moviegoers bought both popcorn and soda)
Plugging these values into the formula:
[tex]\[
P(B|A) = \frac{0.45}{0.60}
\][/tex]
To simplify this fraction, we can divide the numerator and the denominator by their greatest common divisor, which is 0.15:
[tex]\[
P(B|A) = \frac{0.45 \div 0.15}{0.60 \div 0.15} = \frac{3}{4}
\][/tex]
Hence, the probability that a person will buy popcorn given that they have already decided to buy soda is:
[tex]\[
P(B|A) = \frac{3}{4}
\][/tex]
So, the correct answer is:
[tex]\[
\boxed{\frac{3}{4}}
\][/tex]
