Question

Urn 1 contains 7 red balls and 1 black ball. Urn 2 contains 1 red ball and 2 black balls. Urn 3 contains 5 red balls and 2 black balls.

Probability and combinatorics
ANSWERED
asked 2021-09-13
Urn 1 contains 7 red balls and 1 black ball. Urn 2 contains 1 red ball and 2 black balls. Urn 3 contains 5 red balls and 2 black balls. If an urn is selected at random and ball is drawn, find the probability that it will be red.

Expert Answers (1)

2021-09-14
Urn 1 = 7 red balls and 1 black balls
Urn 2 = 1 red balls and 2 black balls
Urn 3 = 5 red balls and 2 black balls
There are 7+1+5 = 13 red balls.
There are a total of 7+1+5+1+2+2 = 18 balls
P(red) = (number of red balls)/(number of balls total)
P(red) = \(\displaystyle\frac{{13}}{{18}}\)
P(red) = \(\displaystyle{0.722}\)
14
 
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