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7 February, 00:30

When I visit a local cafe, there's an $80/%$ probability that I'll order a sandwich, and a $70/%$ probability that I'll order a soup. If there's a $5/%$ probability that I don't order the sandwich or soup, then what's the probability that I order both the sandwich and soup

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  1. 7 February, 00:54
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    There is 55% probability that I order both the sandwich and soup

    Step-by-step explanation:

    P (sandwich) = 0.8

    P (Soup) = 0.7

    P (neither sandwich nor soup) = 0.05

    P (sandwich or soup) = 1 - P (neither sandwich nor soup)

    P (sandwich or soup) = 1 - 0.05 = 0.95

    P (Sandwich & Soup) = x

    P (Sandwich only) = 0.8 - x

    P (Soup only) = 0.7 - x

    P (sandwich or soup) = P (Sandwich only) + P (Soup only) + P (Sandwich & Soup)

    Note that P (neither sandwich nor soup) has already been used to get the P (sandwich or soup) and should not be included in the above formula. Don't make that mistake!

    0.95 = 0.8 - x + 0.7 - x + x

    0.95 = 1.50 - x

    x = 1.50 - 0.95

    x = 0.55

    There is 55% probability that I order both the sandwich and soup
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