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1 August, 01:54

A leather store performs an observational survey of women walking through a mall. There were 30 women that walked by in an hour. Of those women, 18 were carrying purses, 12 were wearing belts, and 6 were both carrying purses and wearing belts.

What is the probability that a woman was wearing a belt, given that the woman was also carrying a purse?

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  1. 1 August, 04:44
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    P (B/C) = 0.3333

    Step-by-step explanation:

    let's call B the event that a woman was wearing a belt and C the event that a woman was carrying a purse.

    The probability P (B/C) that a woman was wearing a belt, given that the woman was also carrying a purse can be calculated as:

    P (B/C) = P (B∩C) / P (C)

    Where P (C) is the probability that a woman was carrying purse and P (B∩C) is the probability that the woman was both carrying purse and wearing belt.

    So, P (C) is calculated as:

    P (C) = 18 / 30 = 0.6

    Because there were 30 women that walked by in an hour and of those women, 18 were carrying purses.

    At the same way, P (B∩C) is equal to:

    P (B∩C) = 6 / 30 = 0.2

    Finally, P (B/C) is equal to:

    P (B/C) = 0.2/0.6 = 0.3333
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