Suppose that is the event of drawing eight red and two blue eggs.
First we must calculate .
.
This problem is asking you to find .
.
Can you finish?
Bag #1
900 red eggs
100 blue eggs
Bag #2
100 red eggs
900 blue eggs
You are given one of these two bags (50/50 chance) randomly.
You perform a set of trials where you take out one egg randomly from your bag, note the color, then put it back in, and repeat.
You drew 8 red eggs and 2 blue eggs from your bag during this trial. What is the probabilty (%) that you have Bag #1?
I thought it was 50% because you could have drawn 8 red and 2 blue from bag #2 as well, since drawing eggs and putting them back does not change the bag's contents...
Hello, daigo!
This is a Conditional Probability problem:
You are given one of these two bags randomly (50/50 chance).
You perform a set of trials where you take out one egg randomly from your bag,
note the color, then put it back in, and repeat.
You drew 8 red eggs and 2 blue eggs from your bag during this trial.
What is the probabilty (%) that you have Bag #1?
. . Given that 8R, 2B are drawn,
. . what is the probability that they were drawn from Bag #1?
Bayes' Theorem: . .[1]
Numerator: .
. . From Bag #1: .
. . Then: .
. . Hence: . .[2]
Denominator: .
. . From Bag #2: .
. . Then: .
. . Hence: . .(A)
And we have: . .[3]
Substitute [2] and [3] into [1]:
. .
Therefore: .
If we draw 8 Reds and 2 Blues from a bag,
. . the probability that they came from Bag #1 is virtually 100%.
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The answer is reasonable when we consider (A).
So if we have Bag #2, it is virtually impossible to draw 8 Reds and 2 Blues.