Questions: An internet research company surveyed 90 online shoppers, each of whom made one purchase today. The company recorded the type of purchase each shopper made. Here is a summary. Type of purchase Number of shoppers ------ Electronics 12 Beauty supplies 19 Food 34 Toys 25 Three shoppers from the survey are selected at random, one at a time without replacement. What is the probability that the first shopper purchased beauty supplies and the other two did not?

Solution Steps

We need to find the probability that the first shopper purchased beauty supplies and the other two shoppers did not. This scenario can be modeled using the hypergeometric distribution.

The parameters for the hypergeometric distribution are defined as follows:

• \( N = 90 \): Total number of shoppers.
• \( K = 19 \): Total number of shoppers who purchased beauty supplies.
• \( n = 3 \): Total number of shoppers selected.
• \( k = 1 \): Number of shoppers who purchased beauty supplies in our selection.

The probability can be calculated using the formula for the hypergeometric distribution:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

Substituting the values, we have:

\[ P(X = 1) = \frac{\binom{19}{1} \binom{71}{2}}{\binom{90}{3}} \]

Calculating this gives:

\[ P(X = 1) = 0.4019 \]

Final Answer