14. First group (from 8th grade. 3 points). The sequence consists of 19 ones and 49 zeros, arranged in a random order. We will call a group the maximal subsequence of identical symbols. For example, in the sequence 110001001111 there are five groups: two ones, then three zeros, then one one, then two zeros, and finally four ones. Find the expected value of the length of the first group.
Solution. Let's separately number the ones from 1 to 19 and the zeros from 1 to 49. Let $I_{k}$ be the indicator of the event "there are no zeros before the $k$-th one" and $J_{m}$ be the indicator of the event "there are no ones before the $m$-th zero", where $1 \leq k \leq 19$ and $1 \leq m \leq 49$. If at least one of the indicators $I_{k}$ is equal to one, then the first group consists of ones, and therefore all indicators $J_{m}$ are equal to zero. Conversely, if at least one of the indicators $J_{m}$ is equal to one, then all $I_{k}$ are equal to zero. The desired quantity $X$ "the length of the first group" is equal to

$$
I_{1}+I_{2}+\ldots+I_{19}+J_{1}+J_{2}+\ldots+J_{49}
$$

The probability of the event " $I_{k}=1$ " is $1 / 50$, since this event occurs only when, in a randomly shuffled sequence consisting of the $k$-th one and all 49 zeros, the one is in the first position. Therefore,

$$
\mathrm{E} I_{k}=\mathrm{P}\left(I_{k}=1\right)=1 / 50=0.02
$$

Similarly, $\mathrm{E} J_{m}=\mathrm{P}\left(J_{m}=1\right)=1 / 20=0.05$. Therefore, $\mathrm{E} X=19 \cdot 0.02+49 \cdot 0.05=2.83$.

Answer: 2.83.