Problem 4. A basketball player throws the ball at the hoop. The probability of hitting the hoop with one throw is 0.6. Formulate the probability distribution law for the number of throws, if the basketball player stops throwing as soon as he hits the hoop.
Solution. The number of throws $X$ until the first hit in the ring is a random variable that can take possible values $1,2,3, \ldots$.

$X=1$ means that the basketball player hit on the first throw. The probability of this value is 0.6, i.e., $p(X=1)=0.6$. $X=2$ means that the basketball player made two throws, which means the first throw was a miss, and the second was a hit. Therefore, $p(X=2)=0.4 \cdot 0.6=0.24$. $X=3$ means that three throws were made, with the first two being misses and the third being a hit.
Thus, $p(X=3)=0.4^{2} \cdot 0.6=0.096$, and so on: $p(X=k)=0.4^{k-1} \cdot 0.6$, since this means $k$ throws, $(k-1)$ of which were misses, and the last, $k$-th, was a hit. The table, or distribution series, of the values of $X$ is as follows:

| $x$ | 1 | 2 | 3 | $\ldots$ | $k$ | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $p$ | 0.6 | 0.24 | 0.096 | .. | $0.4^{k-1} \cdot 0.6$ | .. |