Question 1197861: Box 1 contains one white and two black balls, box 2 contains one black and two white balls, and box 3 contains three black and three white balls. A die is rolled. If 1,2, or 3 shows up, box 1 is selected; if 4 shows up, box 2 is selected; and if 5 or 6 shows up, box 3 is selected. A ball is then drawn at random from the selected box. Let W be the event the the ball drawn is white. Let T, U, V be the events that the box selected is 1, 2, 3, respectively. Find P(U|W).
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
There are two methods (possibly more). I'll start with the approach that doesn't involve probability formulas.
We'll have a 6 by 6 table.
The 6 rows represent the sides of the die being rolled.
The first three rows are box 1, the fourth row is box 2, the last two rows represent box 3.
W = white
B = black
This row configuration is based on the fact that rolling 1 through 3 leads to box 1, rolling a 4 leads to box 2, and rolling 4 or 5 leads to box 3.
If we simply had one box per row, then that would imply each box has equal probability of being chosen (when that isn't the case).
In other words, box 1 has three times as many rows as box 2 for instance, because box 1 is three times as likely compared to box 2's chances.
The first three rows have fewer than 6 items in them. Extend the pattern out so that each row has 6 items exactly.
This is to get everything to be uniform.
The pattern WBB becomes WBBWBB
The pattern WWB becomes WWBWWB
This extension is valid because we haven't changed the relative frequency of either color for any of the boxes.
It's like changing the fraction 1/3 into 2/6.
In that extended table, there are 16 Ws
Of which, four of them come from box 2.
So 4/16 = 1/4 is the probability of selecting box 2 if we know 100% that the selected ball was white.
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Here's the other method using conditional probability formulas and the law of total probability. Bayes' theorem will show up as well.
P(U | W) = P(U given W)
P(U given W) = P(U and W)/P(W)
P(T) = 3/6 = 1/2
P(U) = 1/6
P(V) = 2/6 = 1/3
P(W given T) = (1 white)/(1 white+2 black) = 1/3
P(W given U) = (2 white)/(2 white+1 black) = 2/3
P(W given V) = (3 white)/(3 white+3 black) = 3/6 = 1/2
Apply the law of total probability
P(W) = P(W and T)+P(W and U)+P(W and V)
P(W) = P(W given T)*P(T)+P(W given U)*P(U)+P(W given V)*P(V)
P(W) = (1/3)*(1/2) + (2/3)*(1/6) + (1/2)*(1/3)
P(W) = 1/6 + 2/18 + 1/6
P(W) = 2/6 + 2/18
P(W) = 6/18 + 2/18
P(W) = 8/18
P(W) = 4/9
Then,
P(U given W) = P(U and W)/P(W)
P(U given W) = P(W and U)/P(W)
P(U given W) = P(W given U)*P(U)/P(W) ....... bayes' theorem
P(U given W) = (2/3)*(1/6)/(4/9)
P(U given W) = (2/3)*(1/6)*(9/4)
P(U given W) = 18/72
P(U given W) = 1/4
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Answer: 1/4
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