Find the probability it will be red

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To find the probability of drawing a red ball from one of three urns, each selected at random, the calculations involve determining the probability from each urn. The first urn has a probability of 5/8 for red, the second 3/4, and the third 4/6. Each urn is chosen with a probability of 1/3, leading to the combined probability of (1/3)(5/8) + (1/3)(3/4) + (1/3)(4/6), which simplifies to 49/72. The final probability of drawing a red ball from a randomly selected urn is therefore 49/72.
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I need a help with this question please:

Urn 1 contains 5 red balls & 3 black balls.
Urn 2 contains 3 red balls & 1 black ball.
Urn 3 contains 4 red balls & 2 black balls.
If an urn is selected at random and a ball is drawn, find the probability it will be red.

Thanks for helping.
 
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If the urn is selected at random, can't you just treated as if all the balls were in 1 huge urn? I think it's just 12/18 right?
 
no, probability of getting each urn is 1/3

so

(1/3)(5/8)+(1/3)(3/4)+(1/3)(4/6)=49/72

I am thinking this way, but am not sre.
 
mutnauq said:
no, probability of getting each urn is 1/3

so

(1/3)(5/8)+(1/3)(3/4)+(1/3)(4/6)=49/72

I am thinking this way, but am not sre.
sounds right.
 
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