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412-79 Exam Dumps : EC-Council Certified Security Analyst (ECSA)

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EC-Council Certified Security Analyst (ECSA) Questions and Answers

Question 1

What is the advantage in encrypting the communication between the agent and the monitor in an Intrusion Detection System?

Options:

A.

Encryption of agent communications will conceal the presence of the agents

B.

Alerts are sent to the monitor when a potential intrusion is detected

C.

An intruder could intercept and delete data or alerts and the intrusion can go undetected

D.

The monitor will know if counterfeit messages are being generated because they will not be encrypted

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Question 2

You are assigned to work in the computer forensics lab of a state police agency. While working on a high profile criminal case, you have followed every applicable procedure, however your boss is still concerned that the defense attorney might question weather evidence has been changed while at the laB. What can you do to prove that the evidence is the same as it was when it first entered the lab?

Options:

A.

make an MD5 hash of the evidence and compare it with the original MD5 hash that was taken when the evidence first entered the lab

B.

make an MD5 hash of the evidence and compare it to the standard database developed by NIST

C.

there is no reason to worry about this possible claim because state labs are certified

D.

sign a statement attesting that the evidence is the same as it was when it entered the lab

Question 3

A honey pot deployed with the IP 172.16.1.108 was compromised by an attacker . Given below is an excerpt from a Snort binary capture of the attack. Decipher the activity carried out by the attacker by studying the log. Please note that you are required to infer only what is explicit in the excerpt. (Note: The student is being tested on concepts learnt during passive OS fingerprinting, basic TCP/IP connection concepts and the ability to read packet signatures from a sniff dump.) 03/15-20:21:24.107053 211.185.125.124:3500 -> 172.16.1.108:111 TCP TTL:43 TOS:0×0 ID:29726 IpLen:20 DgmLen:52 DF ***A**** Seq: 0x9B6338C5 Ack: 0x5820ADD0 Win: 0x7D78 TcpLen: 32 TCP Options (3) => NOP NOP TS: 23678634 2878772 =+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=

03/15-20:21:24.452051 211.185.125.124:789 -> 172.16.1.103:111 UDP TTL:43 TOS:0×0 ID:29733 IpLen:20 DgmLen:84 Len: 64

01 0A 8A 0A 00 00 00 00 00 00 00 02 00 01 86 A0 ……………. 00 00 00 02 00 00 00 03 00 00 00 00 00 00 00 00 ……………. 00 00 00 00 00 00 00 00 00 01 86 B8 00 00 00 01 …………….

00 00 00 11 00 00 00 00 ……..

=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=

03/15-20:21:24.730436 211.185.125.124:790 -> 172.16.1.103:32773 UDP TTL:43 TOS:0×0 ID:29781 IpLen:20 DgmLen:1104 Len: 1084 47 F7 9F 63 00 00 00 00 00 00 00 02 00 01 86 B8

Options:

A.

The attacker has conducted a network sweep on port 111

B.

The attacker has scanned and exploited the system using Buffer Overflow

C.

The attacker has used a Trojan on port 32773

D.

The attacker has installed a backdoor