SOLUTION: Urn 1 contains 7 red balls and 4 black balls. Urn 2 contains 4 red balls and 1 black ball. Urn 3 contains 5 red balls and 3 black balls. If an urn is selected at random and a ball

Algebra ->  Probability-and-statistics -> SOLUTION: Urn 1 contains 7 red balls and 4 black balls. Urn 2 contains 4 red balls and 1 black ball. Urn 3 contains 5 red balls and 3 black balls. If an urn is selected at random and a ball       Log On


   



Question 1196711: Urn 1 contains 7 red balls and 4 black balls. Urn 2 contains 4 red balls and 1 black ball. Urn 3 contains 5 red balls and 3 black balls. If an urn is selected at random and a ball is drawn, find the probability that it will be red.
Found 2 solutions by math_helper, ikleyn:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!

Each urn is selected with probability %281%2F3%29 so we multiply that by the probability of selecting red from each urn, and add them:
+%281%2F3%29%284%2F11%29+%2B+%281%2F3%29%281%2F5%29+%2B+%281%2F3%29%283%2F8%29+
which works out to +907%2F1320+ or about 0.687 (68.7% chance to pick red)
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EDIT: Oops, thanks tutor ikleyn. To check my answer, I had calculated the probability of obtaining black (0.313) and accidently copied down that calculation.

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
Urn 1 contains 7 red balls and 4 black balls.
Urn 2 contains 4 red balls and 1 black ball.
Urn 3 contains 5 red balls and 3 black balls.
If an urn is selected at random and a ball is drawn, find the probability that it will be red.
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            I came to fix incorrect writing  (typos ?)  in the post by @math_helper.


    P(red) = %281%2F3%29%2A%287%2F11%29 + %281%2F3%29%2A%284%2F5%29 + %281%2F3%29%2A%285%2F8%29 = 


           = %281%2F3%29%2A%287%2F11+%2B+4%2F5+%2B+5%2F8%29 = %281%2F3%29%2A%28%287%2A5%2A8+%2B+4%2A11%2A8+%2B+5%2A11%2A5%29%2F%2811%2A5%2A8%29%29 = 


           = %281%2F3%29%2A%28907%2F440%29 = 907%2F1320.    ANSWER

Solved.