[Free] New Updated (October) ISC SSCP Real Exam 661-670

Ensurepass

 

QUESTION 661

Which of the following assertions is NOT true about pattern matching and anomaly detection in intrusion detection?

 

A.

Anomaly detection tends to produce more data

B.

A pattern matching IDS can only identify known attacks

C.

Stateful matching scans for attack signatures by analyzing individual packets instead of traffic streams

D.

An anomaly-based engine develops baselines of normal traffic activity and throughput, and alerts on deviations from these baselines

Correct Answer: C

Explanation:

This is wrong which makes this the correct choice. This statement is not true as stateful matching scans for attack signatures by analyzing traffic streams rather than individual packets. Stateful matching intrusion detection takes pattern matching to the next level.

 

As networks become faster there is an emerging need for security analysis techniques that can keep up with the increased network throughput. Existing network-based intrusion detection sensors can barely keep up with bandwidths of a few hundred Mbps. Analysis tools that can deal with higher throughput are unable to maintain state between different steps of an attack or they are limited to the analysis of packet headers.

 

The following answers are all incorrect:

 

Anomaly detection tends to produce more data is true as an anomaly-based IDS produces a lot of data as any activity outside of expected behavior is recorded.

 

A pattern matching IDS can only identify known attacks is true as a pattern matching IDS works by comparing traffic streams against signatures. These signatures are created for known attacks.

 

An anomaly-based engine develops baselines of normal traffic activity and throughput, and alerts on deviations from these baselines is true as the assertion is a characteristic of a statistical anomaly-based IDS.

 

Reference:

Official guide to the CISSP CBK. Pages 198 to 201

http://cs.ucsb.edu/~vigna/publications/2003_vigna_robertson_kher_kemmerer_ACSAC03.pdf

 

 

QUESTION 662

What is NOT true with pre shared key authentication within IKE / IPsec protocol?

 

A.

Pre shared key authentication is normally based on simple passwords

B.

Needs a Public Key Infrastructure (PKI) to work

C.

IKE is used to setup Security Associations

D.

IKE builds upon the Oakley protocol and the ISAKMP protocol.

 

Correct Answer: B

Explanation:

Internet Key Exchange (IKE or IKEv2) is the protocol used to set up a security association (SA) in the IPsec protocol suite. IKE builds upon the Oakley protocol and ISAKMP. IKE uses X.509 certificates for authentication which are either pre-shared or distributed using DNS (preferably with DNSSEC) and a Diffieellman key exchange to set up a shared session secret from which cryptographic keys are derived.

 

Internet Key Exchange (IKE) Internet key exchange allows communicating partners to prove their identity to each other and establish a secure communication channel, and is applied as an authentication component of IPSec.

 

IKE uses two phases:

 

Phase 1: In this phase, the partners authenticate with each other, using one of the following:

Shared Secret: A key that is exchanged by humans via telephone, fax, encrypted e-mail, etc.

Public Key Encryption: Digital certificates are exchanged. Revised mode of Public Key Encryption: To reduce the overhead of public key encryption, a nonce (a Cryptographic function that refers to a number or bit string used only once, in security engineering) is encrypted with the communicating partner’s public key, and the peer’s identity is encrypted with symmetric encryption using the nonce as the key. Next, IKE establishes a temporary security association and secure tunnel to protect the rest of the key exchange. Phase 2: The peers’ security associations are established, using the secure tunnel and temporary SA created at the end of phase 1.

 

The following reference(s) were used for this question:

 

Hernandez CISSP, Steven (2012-12-21). Official (ISC)2 Guide to the CISSP CBK, Third Edition ((ISC)2 Press) (Kindle Locations 7032-7048). Auerbach Publications. Kindle Edition.

RFC 2409 at http://tools.ietf.org/html/rfc2409

http://en.wikipedia.org/wiki/Internet_Key_Exchange

 

 

QUESTION 663

Where parties do not have a shared secret and large quantities of sensitive information must be passed, the most efficient means of transferring information is to use Hybrid Encryption Methods. What does this mean?

 

A.

Use of public key encryption to secure a secret key, and message encryption using the secret key.

B.

Use of the recipient’s public key for encryption and decryption based on the recipient’s private key.

C.

Use of software encryption assisted by a hardware encryption accelerator.

D.

Use of elliptic curve encryption.

 

Correct Answer: A

Explanation:

A Public Key is also known as an asymmetric algorithm and the use of a secret key would be a symmetric algorithm.

 

The following answers are incorrect:

 

Use of the recipient’s public key for encryption and decryption based on the recipient’s private key. Is incorrect this would be known as an asymmetric algorithm.

Use of software encryption assisted by a hardware encryption accelerator. This is incorrect, it is a distractor.

Use of Elliptic Curve Encryption. Is incorrect this would use an asymmetric algorithm.

 

 

QUESTION 664

The RSA Algorithm uses which mathematical concept as the basis of its encryption?

 

A.

Geometry

B.

16-round ciphers

C.

PI (3.14159…)

D.

Two large prime numbers

 

Correct Answer: D

Explanation:

Source: TIPTON, et. al, Official (ISC)2 Guide to the CISSP CBK, 2007 edition, page 254.

 

And from the RSA web site, http://www.rsa.com/rsalabs/node.asp?id=2214 :

The RSA cryptosystem is a public-key cryptosystem that offers both encryption and digital signatures (authentication). Ronald Rivest, Adi Shamir, and Leonard Adleman developed the RSA system in 1977 [RSA78]; RSA stands for the first letter in each of its inventors’ last names.

 

The RSA algorithm works as follows: take two large primes, p and q, and compute their product n = pq; n is called the modulus. Choose a number, e, less than n and relatively prime to (p-1)(q-1), which means e and (p-1)(q-1) have no common factors except 1. Find another number d such that (ed – 1) is divisible by (p-1)(q-1). The values e and d are called the public and private exponents, respectively. The public key is the pair (n, e); the private key is (n, d). The factors p and q may be destroyed or kept with the private key.

 

It is currently difficult to obtain the private key d from the public key (n, e). However if one could factor n into p and q, then one could obtain the private key

D.Thus the security of the RSA system is based on the assumption that factoring is difficult. The discovery of an easy method of factoring would “break” RSA (see Question 3.1.3 and Question 2.3.3).

 

Here is how the RSA system can be used for encryption and digital signatures (in practice, the actual use is slightly different; see Questions 3.1.7 and 3.1.8):

Encryption

 

Suppose Alice wants to send a message m to Bob. Alice creates the ciphertext c by exponentiating: c = me mod n, where e and n are Bob’s public key. She sends c to Bob. To decrypt, Bob also exponentiates: m = cd mod n; the relationship between e and d ensures that Bob correctly recovers m. Since only Bob knows d, only Bob can decrypt this message.

 

Digital Signature

Suppose Alice wants to send a message m to Bob in such a way that Bob is assured the message is both authentic, has not been tampered with, and from Alice. Alice creates a digital signature s by exponentiating: s = md mod n, where d and n are Alice’s private key. She sends m and s to Bob. To verify the signature, Bob exponentiates and checks that the message m is recovered: m = se mod n, where e and n are Alice’s public key.

 

Thus encryption and authentication take place without any sharing of private keys: each person uses only another’s public key or their own private key. Anyone can send an encrypted message or verify a signed message, but only someone in possession of the correct private key can decrypt or sign a message.

 

 

QUESTION 665

Which of the following offers security to wireless communications?

 

A.

S-WAP

B.

WTLS

C.

WSP

D.

WDP

 

Correct Answer: B

Explanation:

Wireless Transport Layer Security (WTLS) is a communication protocol that allows wireless devices to send and receive encrypted information over the Internet. S- WAP is not defined. WSP (Wireless Session Protocol) and WDP (Wireless Datagram Protocol) are part of Wireless Access Protocol (WAP).

Source: KRUTZ, Ronald L.& VINES, Russel D., The CISSP Prep Guide: Mastering the Ten Domains of Computer Security, John Wiley & Sons, 2001, Chapter 4: Cryptography (page 173).

 

QUESTION 666

Which of the following would best define a digital envelope?

 

A.

A message that is encrypted and signed with a digital certificate.

B.

A message that is signed with a secret key and encrypted with the sender’s private key.

C.

A message encrypted with a secret key attached with the message. The secret key is encrypted with the public key of the receiver.

D.

A message that is encrypted with the recipient’s public key and signed with the sender’s private key.

 

Correct Answer: C

Explanation:

A digital envelope for a recipient is a combination of encrypted data and its encryption key in an encrypted form that has been prepared for use of the recipient.

It consists of a hybrid encryption scheme in sealing a message, by encrypting the data and sending both it and a protected form of the key to the intended recipient, so that one else can open the message.

In PKCS #7, it means first encrypting the data using a symmetric encryption algorithm and a
secret key, and then encrypting the secret key using an asymmetric encryption algorithm and the public key of the intended recipient.

Source: SHIREY, Robert W., RFC2828: Internet Security Glossary, may 2000.

 

 

QUESTION 667

Which type of encryption is considered to be unbreakable if the stream is truly random and is as large as the plaintext and never reused in whole or part?

 

A.

One Time Pad (OTP)

B.

One time Cryptopad (OTC)

C.

Cryptanalysis

D.

Pretty Good Privacy (PGP)

 

Correct Answer: A

Explanation:

OTP or One Time Pad is considered unbreakable if the key is truly random and is as large as the plaintext and never reused in whole or part AND kept secret.

 

In cryptography, a one-time pad is a system in which a key generated randomly is used only once to encrypt a message that is then decrypted by the receiver using the matching one-time pad and key. Messages encrypted with keys based on randomness have the advantage that there is theoretically no way to “break the code” by analyzing a succession of messages. Each encryption is unique and bears no relation to the next encryption so that some pattern can be detected.

 

With a one-time pad, however, the decrypting party must have access to the same key used to encrypt the message and this raises the problem of how to get the key to the decrypting party safely or how to keep both keys secure. One-time pads have sometimes been used when the both parties started out at the same physical location and then separated, each with knowledge of the keys in the one-time pad. The key used in a one- time pad is called a secret key because if it is revealed, the messages encrypted with it can easily be deciphered.

 

One-time pads figured prominently in secret message transmission and espionage before and during World War II and in the Cold War era. On the Internet, the difficulty of securely controlling secret keys led to the invention of public key cryptography.

 

The biggest challenge with OTP was to get the pad security to the person or entity you wanted to communicate with. It had to be done in person or using a trusted courrier or custodian. It certainly did not scale up very well and it would not be usable for large quantity of data that needs to be encrypted as we often time have today.

 

The following answers are incorrect:

 

One time Cryptopad: Almost but this isn’t correct. Cryptopad isn’t a valid term in cryptography.

 

Cryptanalysis: Sorry, incorrect. Cryptanalysis is the process of analyzing information in an effort to breach the cryptographic security systems.

 

PGP – Pretty Good Privacy: PGP, written by Phil Zimmermann is a data encryption and decryption program that provides cryptographic privacy and authentication for data. Still isn’t the right answer though. Read more here about PGP.

 

The following reference(s) was used to create this question:

 

To get more info on this QUESTION NO: s or any QUESTION NO: s of Security+, subscribe to the CCCure Holistic Security+ CBT available at:

http://www.cccure.tv

http://users.telenet.be/d.rijmenants/en/otp.htm

http://en.wikipedia.org/wiki/One-time_pad

http://searchsecurity.techtarget.com/definition/one-time-pad

 

 

QUESTION 668

Which of the following answers is described as a random value used in cryptographic algorithms to ensure that patterns are not created during the encryption process?

 

A.

IV – Initialization Vector

B.

Stream Cipher

C.

OTP – One Time Pad

D.

Ciphertext

 

Correct Answer: A

Explanation:

The basic power in cryptography is randomness. This uncertainty is why encrypted data is unusable to someone without the key to decrypt.

 

Initialization Vectors are a used with encryption keys to add an extra layer of randomness to encrypted data. If no IV is used the attacker can possibly break the keyspace because of patterns resulting in the encryption process. Implementation such as DES in Code Book Mode (CBC) would allow frequency analysis attack to take place.

 

In cryptography, an initialization vector (IV) or starting variable (SV)is a fixed-size input to a cryptographic primitive that is typically required to be random or pseudorandom. Randomization is crucial for encryption schemes to achieve semantic security, a property whereby repeated usage of the scheme under the same key does not allow an attacker to infer relationships between segments of the encrypted message. For block ciphers, the use of an IV is described by so-called modes of operation. Randomization is also required for other primitives, such as universal hash functions and message authentication codes based thereon.

 

It is define by TechTarget as:

An initialization vector (IV) is an arbitrary number that can be used along with a secret key for data encryption. This number, also called a nonce, is employed only one time in any session.

The use of an IV prevents repetition in data encryption, making it more difficult for a hacker using a dictionary attack to find patterns and break a cipher. For example, a sequence might appear twice or more within the body of a message. If there are repeated sequences in encrypted data, an attacker could assume that the corresponding sequences in the message were also identical. The IV prevents the appearance of corresponding duplicate character sequences in the ciphertext.

 

The following answers are incorrect:

 

Stream Cipher: This isn’t correct. A stream cipher is a symmetric key cipher where plaintext digits are combined with pseudorandom key stream to product cipher text.

 

OTP – One Time Pad: This isn’t correct but OTP is made up of random values used as key material. (Encryption key) It is considered by most to be unbreakable but must be changed with a new key after it is used which makes it impractical for common use.

 

Ciphertext: Sorry, incorrect answer. Ciphertext is basically text that has been encrypted with key material (Encryption key)

 

The following reference(s) was used to create this question:

 

For more details on this TOPIC and other QUESTION NO: s of the Security+ CBK, subscribe to our Holistic Computer Based Tutorial (CBT) at

http://www.cccure.tv

whatis.techtarget.com/definition/initialization-vector-IV

en.wikipedia.org/wiki/Initialization_vector

 

 

QUESTION 669

Which of the following encryption algorithms does not deal with discrete logarithms?

 

A.

El Gamal

B.

Diffie-Hellman

C.

RSA

D.

Elliptic Curve

 

Correct Answer: C

Explanation:

The security of the RSA system is based on the assumption that factoring the product into two original large prime numbers is difficult

Source: KRUTZ, Ronald L.& VINES, Russel D., The CISSP Prep Guide: Mastering the Ten Domains of Computer Security, John Wiley & Sons, 2001, Chapter 4: Cryptography (page 159).

Shon Harris, CISSP All-in-One Examine Guide, Third Edition, McGraw-Hill Companies, August 2005, Chapter 8: Cryptography, Page 636 – 639

 

 

QUESTION 670

The Data Encryption Algorithm performs how many rounds of substitution and permutation?

 

A.

4

B.

16

C.

54

D.

64

 

Correct Answer: B

Explanation:

Source: TIPTON, Hal, (ISC)2, Introduction to the CISSP Exam presentation.

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