49. On the user's computer desktop, there are two folders with files. In the first folder, there are 16 files, and 4 of them are less than 500 kilobytes in size. In the second folder, there are 20 files, and 5 of them are less than 500 kilobytes in size. Without considering the file sizes, the user moves one file from the first folder to the second folder, and then opens a file from the second folder. Find the probability that the file opened will be less than 500 kilobytes in size.
S o l u t i o n. Let's consider the following assumptions about the user's options for transferring files from the first folder to the second (hypotheses):

$B_{1}$ - a file less than $500 \mathrm{Kb}$ in size is transferred from the first folder to the second;

$B_{2}$ - a file of at least $500 \mathrm{Kb}$ in size is transferred from the first folder to the second.

We will find the probabilities of the hypotheses by applying formula (1):

$$
P\left(B_{1}\right)=\frac{4}{16}=\frac{1}{4} ; \quad P\left(B_{2}\right)=\frac{12}{16}=\frac{3}{4}
$$

Let's denote the event:

$A$ - the file opened by the user from the second folder is less than 500 Kb in size.

Event $A$ will occur either together with event $B_{1}$ or with event $B_{2}$. In the case of event $B_{1}$ occurring, there will be 21 files in the second folder, 6 of which are less than 500 Kb in size. Therefore, the conditional probability $P_{B_{1}}(A)=\frac{6}{21}$. In the case of event $B_{2}$ occurring, there will be 21 files in the second folder, 5 of which are less than 500 Kb in size. Therefore, the conditional probability $P_{B_{2}}(A)=\frac{5}{21}$.

The probability of event $A$ will be found using the formula for total probability (16) for $n=2$:

$$
P(A)=P\left(B_{1}\right) P_{B_{1}}(A)+P\left(B_{2}\right) P_{B_{2}}(A)=\frac{1}{4} \cdot \frac{6}{21}+\frac{3}{4} \cdot \frac{5}{21}=\frac{1}{4}
$$