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Re: There are two bags with red and green balls only, the first bag contai [#permalink]
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JourneyToTheTop wrote:
There are two bags with red and green balls only, the first bag contains 4 red and 6 green balls, second contains 3 red and 7 green balls. Andy picked 1 ball from the first bag and placed it in the second bag, then Mira picked a ball from the second bag. What is the probability that Mira picked a red ball?

A) 8/55
B) 17/55
C) 9/55
D) 1/3
E) 23/110

Show SpoilerWorking below
This was my work:

First bag P(Red) * second bag P(Red) = 4/10*4/11 = 16/110 = 8/55
First bag P(Green) * second bag P(Red) = 6/10*3/11 = 18/110 = 9/55

Answer = 8/55 + 9/55 = 17/55


The question may look complicated but it is not.
There are two events that take place in sequence. The probability of the second event depends on the result of the first event.

We just need to consider the two results of the first event separately. It is like a partial binomial tree.

Attachment:
BT.jpeg
BT.jpeg [ 37.93 KiB | Viewed 3619 times ]


Probability of picking a Red from second bag = (4/10)*(4/11) + (6/10)*(3/11) = 17/55
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Re: There are two bags with red and green balls only, the first bag contai [#permalink]
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Re: There are two bags with red and green balls only, the first bag contai [#permalink]
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