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Q:

What is multicast routing?

Answer

Sending a message to a group is called multicasting, and its routing algorithm is called multicast routing

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Subject: Networking
Job Role: Network Engineer

Q:

What is region?

Answer

When hierarchical routing is used, the routers are divided into what we call regions, with each router knowing all the details about how to route packets to destinations within its own region, but knowing nothing about the internal structure of other regions

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Subject: Networking
Job Role: Network Engineer

Q:

What is silly window syndrome?

Answer

It is a problem that can ruin TCP performance. This problem occurs when data are passed to the sending TCP entity in large blocks, but an interactive application on the receiving side reads 1 byte at a time.

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Subject: Networking
Job Role: Database Administration

Q:

What are Digrams and Trigrams?

Answer

The most common two letter combinations are called as digrams. e.g. th, in, er, re and an. The most common three letter combinations are called as trigrams. e.g. the, ing,and, and ion.

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Subject: Networking
Job Role: Network Engineer

Q:

What is wide-mouth frog?

Answer

Wide-mouth frog is the simplest known key distribution center (KDC) authentication protocol

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Subject: Networking
Job Role: Database Administration

Q:

What is Mail Gateway

Answer

It is a system that performs a protocol translation between different electronic mail delivery protocols

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Subject: Networking
Job Role: Network Engineer

Q:

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

A) 2/91 B) 1/22
C) 3/22 D) 2/77
 
Answer & Explanation Answer: A) 2/91

Explanation:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = 15C3  =15*14*133*2*1= 455.  

Let E = event of getting all the 3 red balls.

 n(E) = 5C3 = 5*42*1 = 10.

 => P(E) = n(E)/n(S) = 10/455 = 2/91.

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Filed Under: Probability

Q:

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

A) 3/20 B) 29/34
C) 47/100 D) 13/102
 
Answer & Explanation Answer: D) 13/102

Explanation:

Let S be the sample space.

 

Then, n(S) = 52C2=(52 x 51)/(2 x 1) = 1326.

 

Let E = event of getting 1 spade and 1 heart.

 

n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = 13C1*13C1 = 169.

 

P(E) = n(E)/n(S) = 169/1326 = 13/102.

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Filed Under: Probability