please find attached file as below:

i have try by using tree diagram until the part which i mark with BOX.other than that i have blur how to configure it..any advise

tq(Headbang)

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- Dec 5th 2010, 05:21 PMnikkBasic Prob
please find attached file as below:

i have try by using tree diagram until the part which i mark with BOX.other than that i have blur how to configure it..any advise

tq(Headbang) - Dec 5th 2010, 06:58 PMrtblue
There are a total of 9 red discs and 7 black disks. The bags have been mixed thoroughly, so it is almost as if the contents have been dumped onto the ground and mixed. After mixing, there are 9 discs in the first bag. The probability of there being exactly three is:

$\displaystyle \frac{9}{16}*\frac{8}{16}*\frac{7}{16}*\frac{7}{16 }*\frac{6}{16}*\frac{5}{16}*\frac{4}{16}*\frac{3}{ 16}*\frac{2}{16}$

The first three terms account for there being 3 red discs, and the last six terms account for there being 6 black discs. - Dec 5th 2010, 09:38 PMnikk
Dear rtblue..

tq for the post.

attached is the ans: - Dec 6th 2010, 05:49 AMSoroban
Hello, nikk!

Quote:

$\displaystyle \text{Bag 1 contains 3 red and 5 black discs.}$

$\displaystyle \text{Bag 2 contains 6 red and 2 black discs.}$

$\displaystyle \text{A disc is chosen at random from Bag 1 and placed in Bag 2.}$

$\displaystyle \text{Then a disc is chosen at rancom from Bag 2 and placed in Bag 1.}$

$\displaystyle \text{Find the probability that Bag 1 has 3 red discs.}$

We want the probability that Bag 1 ends up with its original contents.

There are two ways this can happen:

. . [1] A red is drawn from Bag 1 and a red is returned to Bag 1.

. . [2] A black is drawn from Bag 1 and a black is returned to Bag 1.

[1] A red is drawn from Bag 1: .$\displaystyle P(R_1) = \frac{3}{8}$

. . .Bag 2 now contains:7 red and 2 black discs:

. . .A red is drawn from Bag 2: .$\displaystyle P(R_2) \:=\:\frac{7}{9}$

. . . . . $\displaystyle P(R_1R_2) \:=\:\frac{3}{8}\cdot \frac{7}{9}$

[2] A black is drawn from Bag 1: $\displaystyle P(B_1) = \frac{5}{8}$

. . .Bag 2 now contains: 6 red and 3 black discs.

. . .A black is drawn from Bag 2: .$\displaystyle P(B_2) = \frac{3}{9}$

. . . . . $\displaystyle P(B_1B_2) \:=\:\frac{5}{8}\cdot\frac{3}{9}$

Therefore: .$\displaystyle P(R_1R_2\text{ or }R_1B_2) \;=\;\frac{3}{8}\cdot\frac{7}{9} + \frac{5}{8}\cdot\frac{3}{9} \;=\;\frac{7}{24} + \frac{5}{24} \;=\;\frac{12}{24} \;=\;\dfrac{1}{2}$