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Murljashka [212]
3 years ago
8

An urn contains nine red and one blue balls. A second urn contains one red and five blue balls. One ball is removed from each ur

n at random and without replacement, and all of the remaining balls are put into a third urn. What is the probability that a ball drawn randomly from the third urn is blue?
Mathematics
1 answer:
il63 [147K]3 years ago
8 0

Answer:

0.362

Step-by-step explanation:

When drawing randomly from the 1st and 2nd urn, 4 case scenarios may happen:

- Red ball is drawn from the 1st urn with a probability of 9/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this case to happen is (9/10)*(1/6) = 9/60 = 3/20 or 0.15. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 5 blue)/(8 red + 1 blue + 5 blue) = 6/14 = 3/7.

- Red ball is drawn from the 1st urn with a probability of 9/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (9/10)*(5/6) = 45/60 = 3/4 or 0.75. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 4 blue)/(8 red + 1 blue + 1 red + 4 blue) = 5/14

- Blue ball is drawn from the 1st urn with a probability of 1/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (1/10)*(5/6) = 5/60 = 1/12. The probability that a ball drawn randomly from the third urn is blue given this scenario is (4 blue)/(9 red + 1 red + 4 blue) = 4/14 = 2/7

- Blue ball is drawn from the 1st urn with a probability of 1/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this event to happen is (1/10)*(1/6) = 1/60. The probability that a ball drawn randomly from the third urn is blue given this scenario is (5 blue)/(9 red + 5 blue) = 5/14.

Overall, the total probability that a ball drawn randomly from the third urn is blue is the sum of product of each scenario to happen with their respective given probability

P = 0.15(3/7) + 0.75(5/14) + (1/12)*(2/7) + (1/60)*(5/14) = 0.362

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