I don't have enough time/energy to do this at present, but attached
is a contingency tree which shows the urn contents and their probabilities
after the messing about, which someone may find a help with this.
RonLI have a homework question
three urns have the distributions of ball shown below
Black White Green
U1 7 3 10
U2 3 5 2
U3 5 7 3
Suppose a ball is selected at random from U1. A fair die is then tossed and if a number less than three appears the ball is placed in U2, otherwise the ball is placed in U3. You walk into the room and select U1 or U2 at random and select a ball.
A) which color are you most likely to select?
B) If the ball you selected is black, what is the probability a green ball was selected from U1?
Now we can work out the probability of drawing each colour for each of the
leaves on the contingency tree given in the earlier post.
I will work the first in detail, and you can check the rest your self.
For the top leaf we have the first two urns contain (6,3,10) and (4,5,2)
balls of each colour (Black, White, Green). As we select our urn at random
from these two we get a probability of:
The remaing leaf probabilities are (if my sums are right - this is very
labourious and error prone):
(0.3079, 0.3289, 0.3632)
(0.3206, 0.3253, 0.3541)
(0.3342, 0.3026, 0.3632)
(0.3206, 0.3062, 0.3732)
(0.3342, 0.3289, 0.3368)
Now the final probabilities for each coulou are found by taking the weighted
sum of the probabilities for each colour weighted by the leaf probability, so:
P(Black)=0.175x0.3397 + 0.175x0.3079 + 0.075x0.3206 +
.............0.075x0.3342 + 0.25x0.3206 +0.25x0.3342 ~=0.3261
So the answer to (a) is Green is the most likley colour.
This is the ratio of the sum leaf probs times the prob of selecting a black ballB) If the ball you selected is black, what is the probability a green ball
was selected from U1?
for those leaves where the first ball was green to the sum leaf probs times the prob of selecting a black ball for all leaves: