Probability Question?

Q1) Two balls are randomly selected without replacement from a box containing three black balls numbered 1, 2, 3 and two white balls numbered 4 and 5. Assuming that all outcomes are equally likely. 

Find out the probabilities of following events.

a) Probability that the color of second ball is white.

b) Probability that the color of second ball is black.

c) Probability that both balls are black.

d) Probability that both balls are white.

2 Answers

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  • 2 months ago

    a) P(First White & Second White) + P(First Black & Second White) =  

        (2/5)(1/4) + (3/5)(2/4) = 2/20 + 6/20 = 8/20 = 2/5

    b) P(First Black & Second Black) + P(First White & Second Black) = 

        (3/5)(2/4) + (2/5)(3/4) = 6/20 + 6/20 =12/20 = 3/5

    c) P(First Black & Second Black) = (3/5)(2/4) = 6/20 = 3/10

    d) P(First White & Second White) = (2/5)(1/4) = 2/20 = 1/10

  • 2 months ago

    a) There are two possibilities

    P(White then White) or P(Black then White)

    i.e. (2/5) x (1/4) or (3/5) x (2/4)

    so, 2/20 + 6/20 = 8/20 = 2/5

    b) P(White then Black) or P(Black then Black)

    i.e. (2/5) x (3/4) or (3/5) x (2/4)

    so, 6/20 + 6/20 = 12/20 = 3/5

    c) P(Black then Black) = 3/5 x 2/4 => 3/10

    d) P(White then White) = 2/5 x 1/4 => 1/10

    :)>

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