Vase and marginal probability Suppose we have a vase with different balls as follows:


*

*$15$ blue (B)

*$12$ red (R)

*$10$ green (G)

*$5$ white (W)


The probability to see a blue ball in the 1st pick and a green ball in the 3rd pick would be the marginal (without replacement: once one color is picked, all balls of that color are out of the vase):
$
P(B*G*) = P(BRGW)+P(BWGR)=
\big ( \dfrac{15}{42} \times \dfrac{12}{27} \times \dfrac{10}{15} \times \dfrac{5}{5} \big )
+
\big ( \dfrac{15}{42} \times \dfrac{5}{27} \times \dfrac{10}{22} \times \dfrac{12}{12} \big )
$
I have two questions:


*

*Is there any other way to compute $P(B*G*)$ (I mean without computing the marginals)?

*Can we claim that $\dfrac{P(B*G*)}{P(G*B*)} > \dfrac{P(R*G*)}{P(G*R*)}?$ The intuition behind is that the number of blue balls compared to green balls is higher than that of red and green balls.


Thanks for any help or hint.
 A: The problem is about the colors picked in the first and in the third draw, so you don't have to bother about the fourth one. You need, however, to consider the case with the second ball being green and the case of second ball being non-green.
First, from all possible triplets we can separate those starting with blue.
The probability of picking a blue ball in the first attempt is
$$P(B) = \frac{15}{15+12+10+5}=\frac{15}{42}$$
In the second attempt we pick from $41$ remaining balls with $10$ green among them, so we can either pick a green ball with conditional probability
$$P(*G|B) = \frac{10}{41}$$
or a non-green ball with probability
$$P(*\overline G|B) = \frac{41-10}{41}=\frac{31}{41}$$
If the second ball picked was green, we're left with $9$ green balls in the vase, so the probability of picking a green ball as a third one from remaining $40$ is
$$P(**G|BG)=\frac 9{40}$$
and in the opposite case
$$P(**G|B\overline G)=\frac{10}{40}$$
Finally the probability of drawing a three-ball sequence with the first ball blue and the third one green is
$$P(B*G) = P(BGG)+P(B\overline GG)\\
= P(B)\cdot P(*G|B) \cdot P(**G|BG) + P(B)\cdot P(*\overline G|B) \cdot P(**G|B\overline G) \\
= \frac{15}{42} \cdot \frac{10}{41} \cdot \frac 9{40} + \frac{15}{42} \cdot \frac{31}{41} \cdot \frac{10}{40} \\
= \frac{6000}{68880} \\
= \frac 1{11.48} \approx 0.087108$$
