Anonymous
Anonymous asked in 科學及數學數學 · 8 years ago

Probability

There are 4 black balls, 2 white balls and 1 blue ball in a bag. One ball is drawn at a time at random from the bag without replacement. Find the probability that it takes at least 3 draws to get a black ball.

3 Answers

Rating
  • freda
    Lv 5
    8 years ago
    Favorite Answer

    P( takes at least 3 draws)

    =P(takes more than or equal to 3 draws)

    =P(1- 1 draw -2 draws)

    =1- 4/7 -(3/7)*(4/6)

    =1-4/7 -2/7

    =1/7 //

  • 8 years ago

    Sol

    A:第一次不是黑球

    B:第二次不是黑球

    C:第三次不是黑球

    D:第三是黑球

    E:第四次是黑球

    P=P(ABCE)+P(ABD)

    P(ABCE)

    =P(E|ABC)*P(ABC)

    =P(E|ABC)*P(C|AB)*P(AB)

    =P(E|ABC)*P(C|AB)*P(B|A)*P(A)

    =1*(1/5)*(2/6)*(3/7)

    P(ABD)

    =P(D|AB)*P(AB)

    =P(D|AB)*P(B|A)*P(A)

    =(4/5)*(2/6)*(3/7)

    P=1*(2/6)*(3/7)=1/7

  • For it to take at least 3 draws to get a black ball, the first 2 must be non-black ball with probability:

    3/7 x 2/6 = 1/7

    (Since in the first draw, there are 7 balls in which there are 3 not black and in the 2nd draw, there are 6 balls in which there are 2 not black)

    Source(s): 原創答案
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