Question 1053378: A bag containing 5 white, and 7 black balls. All of the balls are drawn at random from the bag one by one without replacement. Find the probability of drawing white ball on 6th attempt and the previous 3 balls being black.
Found 2 solutions by ewatrrr, Edwin McCravy:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! 12 balls: 5 white, and 7 black balls
P(?, ? , B,B,B , W) =
P(W, W , B,B,B , W) +
P(B, B , B,B,B , W) +
P(B, W , B,B,B , W) +
P(W, B , B,B,B , W)
P = (5/12)(4/11)(7/10)(6/9)(6/8)(3/7) + (7/12)(6/11)(5/10)(4/9)(3/8)(5/7)
+ (7/12)(5/11)(6/10)(5/9)(4/8)(4/7) + (5/12)(7/11)(6/10)(5/9)(4/8)(4/7)
P =
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website! A bag containing 5 white, and 7 black balls. All of the balls are drawn at
random from the bag one by one without replacement. Find the probability of
drawing a white ball on 6th attempt and the previous 3 balls being black.
We want to know the probability of drawing them in this order,
X,X,B,B,B,W,X,X,X,X,X,X
where the X's can be either color.
Since it doesn't matter what is drawn after the 6th ball, we only need
to consider the probability that the first 6 drawings are in this
order:
X,X,B,B,B,W
So there are four cases for the first 6 balls drawn:
Case 1: W,W,B,B,B,W
Probability = (5/12)(4/11)(7/10)(6/9)(5/8)(3/7) = 5/264
Case 2: W,B,B,B,B,W
Probability = (5/12)(7/11)(6/10)(5/9)(4/8)(4/7) = 5/198
Case 3: B,W,B,B,B,W
Probability = (7/12)(5/11)(6/10)(5/9)(4/8)(4/7) = 5/198
Case 4: B,B,B,B,B,W
Probability = (7/12)(6/11)(5/10)(4/9)(3/8)(5/7) = 5/264
The probability that one of those cases will occur equals
the sum of those probabilities, which is:
5/264 + 5/198 + 5/198 + 5/264 = 35/396
Edwin
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