SOLUTION: Urn 1 contains 1 red and 9 white balls. Urn 2 contains 8 red and 1 white balls. A ball is drawn from urn 1 and placed in urn 2. Then a ball is drawn from urn 2. If the ball from ur

Algebra ->  Probability-and-statistics -> SOLUTION: Urn 1 contains 1 red and 9 white balls. Urn 2 contains 8 red and 1 white balls. A ball is drawn from urn 1 and placed in urn 2. Then a ball is drawn from urn 2. If the ball from ur      Log On


   

Question 1058275: Urn 1 contains 1 red and 9 white balls. Urn 2 contains 8 red and 1 white balls. A ball is drawn from urn 1 and placed in urn 2. Then a ball is drawn from urn 2. If the ball from urn is red, what is the probability that the ball drawn from urn 1 is red?
I was able to complete the diagram reflecting the numbers in decimal as shown form the textbook.
However, I don't understand the last 2 parts. The numeric value of the probability of the product from branch probabilities leading from R through R? Do you multiply the top two sections leading form R to R? Which in this case would be 0.3*05?
Also, on the last part to obtain the 2nd number to divide by. The sum of all branch points leading to R?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
So look at it from the possible choices,
W1,W2- White ball chosen from Urn 1, White ball chosen from Urn 2
W1,R2
R1,W2
R1,R2
Look at the corresponding probabilities,
P%28W%5B1%5D%2CW%5B2%5D%29=%289%2F10%29%282%2F10%29=18%2F100
P%28W%5B1%5D%2CR%5B2%5D%29=%289%2F10%29%288%2F10%29=72%2F100
P%28R%5B1%5D%2CW%5B2%5D%29=%281%2F10%29%281%2F10%29=1%2F100
P%28R%5B1%5D%2CR%5B2%5D%29=%281%2F10%29%288%2F10%29=9%2F100
So there are two possibilities when the urn 2 ball is red, 72%2F100 times it occurs because the urn 1 ball was white and 9%2F100 times it occurs because the urn 2 ball was red. That makes a total of 81%2F100 times. %2872%2F100%29%2F%2881%2F100%29=72%2F81=8%2F9 occur because of urn 1 ball is white, %289%2F100%29%2F%2881%2F100%29=9%2F81=1%2F9 because the urn 1 ball is red.
P=1%2F9