A one-week study revealed that $$ 60 \% $$ of a warehouse store's customers are women and t $$ 30 \% $$ of women customers spend at least $$ \$ 250 $$ on a single visit to the store. Find the probability that a randomly chosen customer will be a woman who spends at least $$ \$ 250 $$.

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The probability is $$ 0.18 $$ or $$ 18\% $$. To find the probability that a randomly chosen customer is a woman who spends at least \$250, we can use the information provided: 1. The probability that a customer is a woman, $$ P(W) $$, is $$ 60\% $$ or $$ 0.6 $$. 2. The probability that a woman customer spends at least \$250, $$ P(S | W) $$, is $$ 30\% $$ or $$ 0.3 $$. We want to find the joint probability that a customer is a woman and spends at least \$250, which can be calculated using the formula for conditional probability: $$ P(W \cap S) = P(W) \times P(S | W) $$ Substituting the values we have: $$ P(W \cap S) = 0.6 \times 0.3 $$ Calculating this gives: $$ P(W \cap S) = 0.18 $$ Thus, the probability that a randomly chosen customer is a woman who spends at least \$250 is $$ \boxed{0.18} $$ or $$ 18\% $$.

A one-week study revealed that \( 60 \% \) of a warehouse store's customers are women and t \( 30 \% \) of women customers spend at least \( \$ 250 \) on a single visit to the store. Find the probability that a randomly chosen customer will be a woman who spends at least \( \$ 250 \).

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A one-week study revealed that \( 60 \% \) of a warehouse store's customers are women and t \( 30 \% \) of women customers spend at least \( \$ 250 \) on a single visit to the store. Find the probability that a randomly chosen customer will be a woman who spends at least \( \$ 250 \).

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