Question 331751: For a lottery drawing, balls numbered with the digits 1-9 are placed in three
bins and one ball is selected from each bin in order. What is the probability
that all three digits drawn will be odd?
Found 2 solutions by Alan3354, jrfrunner:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! For a lottery drawing, balls numbered with the digits 1-9 are placed in three
bins and one ball is selected from each bin in order. What is the probability
that all three digits drawn will be odd?
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5 of the 9 balls are odd.
--> (5/9)^3 = 125/729
=~ 0.1715
or 17%
Answer by jrfrunner(365)  (Show Source):
You can put this solution on YOUR website! you have to consider all the possibilities of the three bins (say bin#1, bin#2 and bin#3).
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You need to find P(all three digits are odd/draw one from each bin)
lets look at that
first, acknowledge that there are a total of 5 odd numbers out of the 9
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P(all three selected are odd, by selecting one from each of 3 bins) =
P(all 3 selections are odd/Bin#1 has 0 odd balls) =0
in fact any bin having no odd numbers makes the prob of getting all 3 odd numbers 0, since the 3 numbers are obtained one from each bin
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P(all 3 numbers are odd / bin#1 has 1 odd,bin#2 has 0 odd 1,bin#3 has 3 odd)= this is not possible since there exists 5 odd numbers and this makes the total of 4 odd.
P(all 3 numbers are odd / bin#1 has 1 odd,bin#2 has 1 odd,bin#3 has 3 odd)=1/3*1/3*3/3=3/27
P(all 3 numbers are odd / bin#1 has 1,bin#2 has 2,bin#3 has 2)=1/3*2/3*2/3=4/27
P(all 3 numbers are odd / bin#1 has 1,bin#2 has 3,bin#3 has 1)=1/3*3/3*1/3= 3/27
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P(all 3 numbers are odd / bin#1 has 2 odd,bin#2 has 0 odd,bin#3 has 2 odd) = 2/3*0/3*2/3=0
P(all 3 numbers are odd/bin#1 has 2 odd,bin#2 has 1 odd,bin#3 has 2 odd)=2/3*1/3*2/3=4/27
P(all 3 numbers are odd/bin#1 has 2,bin#2 has 2,bin#3 has 1)=2/3*2/3*1/3=4/27
P(all 3 numbers are odd/bin#1 has 2,bin#2 has 3,bin#3 has 0) = 2/3*3/3*0= 0
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P(all 3 numbers are odd/bin#1 has 3 odd,bin#2 has 0 odd,bin#3 has 2 odd) = 3/3*0/3*2/3=0
P(all 3 numbers are odd/bin#1 has 3 odd,bin#2 has 1 odd,bin#3 has 1 odd)=3/3*1/3*1/3=3/27
P(all 3 numbers are odd/bin#1 has 3,bin#2 has 2,bin#3 has 0)= 3/3*2/3*0/3= 0
So, summing all these probabilities =3/27+4/27+3/27+4/27+4/27+3/27=21/27
P(all three selected are odd, by selecting one from each bin) =21/27 or 7/9
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which is much higher than if you just put all the balls in one bin and selected 3 balls and asked the prob( all 3 are odd ) which would be 5/9*4/8*3/7=60/540=0.119
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