4 bulbs in the box, 2 are serviceable There are 4 bulbs in the box, two are intact and 2 are not.  Tom takes out one light bulb, checks it for work, and does not return it to the box.  Then he takes the next bulb, etc.  He does this until he obtains two serviceable bulbs.  What is the mathematical expectation of the number of bulbs that Tom got out of the box?
Denote the working bulb by T, and not working by F. We have such variants: TTFF (2 bulbs out of box), TFTF, FTTF(3 out of box), FFTT, FTFT, TFFT (4 out of box)
Mathematical expectation: (2+2*3+3*4)/6=20/6=10/3
Is it correct?
Thank you
 A: Let $X_n=1$ if the $n^{\mathrm{th}}$ bulb drawn is intact, and $0$ if it is broken. Let $S_n = \sum_{i=1}^n X_i$ and $\tau = \inf\{n>0:S_n=2\}$. Then 
\begin{align}
\mathbb P(\tau = 2) &= \mathbb P(S_2=2)\\
&= \mathbb P(X_1=1,X_2=1)\\
&= \mathbb P(X_2=1\mid X_1=1)\mathbb P(X_1=1)\\
&= \frac13\cdot\frac12\\
&= \frac16,\\
\\
\mathbb P(\tau = 3) &= \mathbb P(S_3=2,S_2<2)\\
&= \mathbb P(X_1=1,X_2=0,X_3=1) + \mathbb P(X_1=0,X_2=1,X_3=1)\\
&= \mathbb P(X_3=1\mid X_2=0,X_1=1)\mathbb P(X_2=0\mid X_1=1)\mathbb P(X_1=1) + \mathbb P(X_3=1\mid X_2=1,X_1=0)\mathbb P(X_2=1\mid X_1=0)\mathbb P(X_1=0)\\
&= \frac12\cdot\frac23\cdot\frac12 + \frac12\cdot \frac23\cdot\frac12\\
&= \frac13,\\
\\
\mathbb P(\tau=4) &= \mathbb P(S_4=2,S_3<2,S_2<2)\\
&=\mathbb P(X_1=1,X_2=0,X_3=0,X_4=1) + \mathbb P(X_1=0,X_2=1,X_3=0,X_4=1) + \mathbb P(X_1=0,X_2=0,X_3=1,X_4=1)\\
&= \frac12\cdot\frac23\cdot\frac12 + \frac12\cdot\frac23\cdot\frac12 + \frac12\cdot\frac13\\
&= \frac12.
\end{align}
Hence,
$$
\mathbb E[\tau] = 2\cdot\frac16 + 3\cdot\frac13 + 4\cdot\frac12 = \frac{10}3.
$$
