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A container has three white balls and two red balls. A first ball is drawn at random and not replaces. Then a second ball is drawn. give the following conditions. What is the probability that the second ball was red? The first ball was white? The first ball was red?

  • math -

    1. First ball was red
    No. of red balls: 2
    Total number of balls:5
    P(R)=2/5
    2. First ball was white
    No. of white balls: 3
    Total number of balls: 5
    P(W)=3/5
    3. Second ball was red:
    The probability is given by
    P(R,R)+P(W,R)

    Now we'll calculate P(R,R)
    P(R)=2/5
    After the first red, there is one more red left, and four balls altogether, therefore
    P(RR)=(2/5)*(1/4)
    =1/10
    Similarly,
    P(WR)=(3/5)*(2/4)
    =3/10
    Probability of the second ball being red is
    (1/10)+(3/10)
    =2/5

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