1. ## Probability Problem

One jar contains 6 blue marbles, 5 red marbles, 8 white marbles, and 8 goldenrod marbles. A second jar contains 8 blue marbles, 7 red marbles, 8 white marbles, and 6 goldenrod marbles. A marble is picked at random from the first jar and put into the second jar. A marble is then selected at random from the second jar. Find the probability that the marble selected from the second jar is a goldenrod. Give answer as a fraction reduced to lowest terms.

I think this problem is easier than I think but I keep getting messed up. Help please!

2. $P(G_2 ) = P(B_1 G_2 ) + P(R_1 G_2 ) + P(W_1 G_2 ) + P(G_1 G_2 )$
$P(G_2 ) = P(G_2 |B_1 )P(B_1 ) + P(G_2 |R_1 )P(R_1 ) + P(G_2 |W_1 )P(W_1 ) + P(G_2 |G_1 )P(G_1 )$
I will do ${P(G_2 |R_1 )P(R_1 )}=\frac{6}{30}\frac{5}{27}$

3. Hello, sweeetcaroline!

I'll simplify the wording . . .

One jar contains 8 Gold marbles and 19 Others.
A second jar contains 6 Gold marbles and 23 Others.
A marble is picked at random from the first jar and put into the second jar.
A marble is then selected at random from the second jar.
Find the probability that the marble selected from the second jar is Gold.
There are two cases to consider . . .

[1] The first marble drawn is Gold: . $P(\text{1st Gold}) \:=\:\frac{8}{27}$
. . .Then the second jar contains: 7 Gold and 23 Others: . $P(\text{2nd Gold}) \:=\:\frac{7}{30}$
. . . $P(\text{1st Gold }\wedge\text{ 2nd Gold}) \:=\:\frac{8}{27}\cdot\frac{7}{30} \:=\:\frac{56}{810}$

[2] The first marble drawn is an Other: . $P(\text{1st Other}) \:=\:\frac{19}{27}$

. . .Then the second jar contains: 6 Gold and 24 Others: . $P(\text{2nd Gold}) \:=\:\frac{6}{30}$

. . . $P(\text{1st Other }\wedge\text{ 2nd Gold}) \:=\:\frac{19}{27}\cdot\frac{6}{30} \:=\:\frac{114}{810}$

Therefore: . $P(\text{2nd Gold}) \;=\;\frac{56}{810} + \frac{114}{810} \;=\;\frac{170}{810} \;=\;\boxed{\frac{17}{81}}$