SOLUTION: URN 1 contains 4 red balls and 3 black balls. URN 2 contains 5 red balls and 3 black balls. URN 3 contains 2 red and 3 black. If an urn is selected at random and a ball is drawn, f
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-> SOLUTION: URN 1 contains 4 red balls and 3 black balls. URN 2 contains 5 red balls and 3 black balls. URN 3 contains 2 red and 3 black. If an urn is selected at random and a ball is drawn, f
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Question 1081622: URN 1 contains 4 red balls and 3 black balls. URN 2 contains 5 red balls and 3 black balls. URN 3 contains 2 red and 3 black. If an urn is selected at random and a ball is drawn, find the probability that it will be red. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! P(urn 1)=(1/3) and probability urn 1 and red ball is (4/7)(1/3)=(4/21)
P(urn 2)=(1/3) and probability urn 2 and red ball is (5/8)(1/3)=(5/24)
P(urn 3)=(1/3) and probability urn 3 and red ball is (2/5)(1/3)=(2/15)
The sum of those probabilities has common denominator 21*15*24
numerator is 4*(15*24)+5(21*15)+2(21*24)/(21*15*24)
numerator is 1440+1575+1008=4023
denominator is 7560
probability is 0.5321 ANSWER
by adding 4/21(0.190) and 5/24 (0.208) and 2/15(0.133), one gets 0.531, but there is rounding error.
