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Probability - White Balls in Box

This topic has 1 expert reply and 0 member replies
fourteenstix Senior | Next Rank: 100 Posts Default Avatar
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Test Date:
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Probability - White Balls in Box

Post Sat Sep 28, 2013 2:38 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])

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    Post Sat Sep 28, 2013 3:36 pm
    Quote:
    Each of the 25 balls in a certain box is either red, blue, or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

    (1) The probability that the ball will both be white and have an even number painted on it is 0.
    (2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2.
    Target question: What is the value of P(white or even)?

    To solve this, we need P(A or B) = P(A) + P(B) - P(A & B)
    So, P(white or even) = P(white) + P(even) - P(white & even)

    Statement 1: P(white & even) = 0
    We can add this to our probability equation to get: P(white or even) = P(white) + P(even) - 0
    Still need P(white), and we need P(even)
    INSUFFICIENT

    Statement 2: P(white) - P(even)= 0.2
    We have no idea about the sum of P(white) and P(even), and we don't know the value of P(white & even)
    INSUFFICIENT

    Statements 1 and 2 combined:
    Given P(white) - P(even)= 0.2 does not tell us the individual values of P(white) and P(even), and it doesn't tell us the value of P(white) + P(even).

    So, since we can't determine the value of P(white) + P(even) - P(white & even), the statements combined are not sufficient.

    Answer: E

    Cheers,
    Brent

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