Probability of choosing balls in an urn.

Hello,

Quote:

Originally Posted by digitalis77

Urn A contains 5 red balls and three yellow balls. Urn B contains 7 red balls and 2 yellow balls. A fair six-sided die is tossed. If a one or two is tossed urn A is selected, and two balls are selected one at a time and without replacement. If a 3, 4, 5 or 6 are tossed, urn B is selected and two balls are selected simultaneously, and the number of red balls is observed. Create a tree diagram for the above random experiment, and determine the probability associated with each branch of the tree.

Please help, this one is really tough!!(Headbang)

Here is a go :

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First part of the tree : the dice
The probability for the dice to yield 1 or 2 is 2/6. So the probability to choose urn A is 2/6.
The probability to choose urn B is ...

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Second part of the tree : first ball to be chosen
In urn A, there are 5 red balls over a total of 5+3=8 balls. Hence the probability of getting a red ball when choosing in urn A is 5/8. The prob of getting a yellow ball is ...
In urn B, there are 7 red balls for a total of 7+2=9 balls. The probability of getting a red ball when choosing in urn B is ...

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Third part of the tree : second ball to be chosen
Let's study the case in urn A.
In the first choice, you either got a red ball, either a yellow ball.

If you choose a red ball, there are 4 red balls remaining, for a total of 4+3=7 balls. The probability of getting another red ball is 4/7.
If you choose a yellow ball, there are 5 red balls and 2 yellow balls remaining. So the probability to get a red ball after getting a yellow ball is 5/7.

The same reasoning stands for urn B.

Does this help ?