SOLUTION: Urn 1 contains 3 red balls and 4 black balls. Urn 2 contains 4 red balls and 2 black balls. Urn 3 contains 6 red balls and 5 black balls. If an urn is selected at random and a ball

Algebra ->  Probability-and-statistics -> SOLUTION: Urn 1 contains 3 red balls and 4 black balls. Urn 2 contains 4 red balls and 2 black balls. Urn 3 contains 6 red balls and 5 black balls. If an urn is selected at random and a ball      Log On


   

Question 1082210: Urn 1 contains 3 red balls and 4 black balls. Urn 2 contains 4 red balls and 2 black balls. Urn 3 contains 6 red balls and 5 black balls. If an urn is selected at random and a ball is drawn, find the probability it will be red.
Answer by jim_thompson5910(35256) About Me  (Show Source):
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Urn 1 = 3 red balls and 4 black balls
Urn 2 = 4 red balls and 2 black balls
Urn 3 = 6 red balls and 5 black balls

There are 3+4+6 = 13 red balls.
There are a total of 3+4+4+2+6+5 = 24 balls

P(red) = Probability of picking red ball
P(red) = (number of red balls)/(number of balls total)
P(red) = 13/24
P(red) = 0.54167 ... approximate
P(red) = 54.167% ... approximate

The answer as a fraction is 13/24
The answer as a decimal is approximately 0.54167
The answer as a percent is approximately 54.167%