1495. A randomly selected phone number consists of 5 digits. What is the probability that in it: 1) all digits are different; 2) all digits are odd?
Solution. 1) Since any of the digits: $0,1,2,3,4,5,6,7,8$, 9 can stand in each of the five positions in a five-digit number, there will be $10^{5}$ different five-digit numbers (all of which can be enumerated as follows: number 00000 - 1st, $00001-2$-nd, $00002-3$-rd, ..., $99998-99999$-th and, finally, 99999 - 100,000-th). Numbers in which all digits are different are permutations of 10 elements taken 5 at a time. Therefore, the number of favorable cases is $A_{10}^{5}=10 \cdot 9 \cdot 8 \cdot 7 \cdot 6$ and the desired probability is

$$
p=\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{10^{5}}=0.3024
$$

2) From the 5 odd digits $(1,3,5,7,9)$, $5^{5}$ different five-digit numbers can be formed. $5^{5}$ is the number of favorable cases. Since there are $10^{5}$ equally possible cases, the desired probability is

$$
p=\frac{5^{5}}{10^{5}}=\left(\frac{1}{2}\right)^{5}=\frac{1}{32}=0.03125.
$$