Expected value of pulling blue marble from sack on nth pull

Question

I am doing a refresher for a work course and I'm a little stuck the question asks the following

You have a sack with 7 marbles inside 4 Blue and 3 Green

You draw all the marbles (without replacing them) and check their color.

If you let X represent the value of n if the first blue marble appears on the nth draw.

Obviously P(b) is 4/7 and P(g) = 3/7

My assumption was that: 4/7*(1+2+3+4+5+6+7) = (4/7)*1 + (4/7)*2 + (4/7)*3 ... (4/7)*7 = 16

But that seems entirely nonsensical value for X to take given that you can only pull 7 marbles from the sack.

I assume that is because the probability is obviously not static at 4/7.

If I look at the probability of the draws being green before being blue I end up with: X=1 = 3/7 X=2 = 2/6 X=3 = 1/5

[(3/7)*1+ (2/6)*2 + (1/5)*3 ] = 1.67

But after this I'm a little stuck

Where am I going wrong?

Answer 1

A hint

$X=1$ has probability $4/7$. Then $X=2$ has probability $3/7\times 4/6$ and so on.

Can you see what to do now?

Answer 2

• $P(X=1)=\frac{4}{7}=\frac{20}{35}$

• $P(X=2)=\frac{3}{7}\cdot \frac{4}{6}=\frac{10}{35}$

• $P(X=3)=\frac{3}{7}\cdot \frac{2}{6} \cdot \frac{4}{5}=\frac{4}{35} $

• $P(X=4)=\frac{3}{7}\cdot \frac{2}{6} \cdot \frac{1}{5}\cdot \frac{4}{4}=\frac{1}{35} $

Observe that $\Sigma_{i=1}^4 P(X=i)=1$ as it has to be...

Now, using the definition,

$$\mathbb{E}[X]=1\times\frac{20}{35}+2\times\frac{10}{35}+3\times\frac{4}{35}+4\times\frac{1}{35}=\frac{56}{35}=\frac{8}{5}$$