After some transmission on Ethernet, current frame of A's is collided 2 times and B's is collided 4 times. The probability that A wins the present back off race is____.

Number of times A has been involved in collision = 2

Number of times B has been involved in collision = 4

A can choose any number between (0 to 2^{2}-1 ) , i.e, (0,1,2,3)

B can choose any number between (0 to 2^{4}-1 ) , i.e, (0,1,2,3,......,15)

It is given that A wins the race, so choices for (A,B) will be:

(0,1) , (0,2) ,(0,3)................ , (0,15)

(1,2), (1,3),...........................,(1,15)

(2,3),(2,4),............................,(2,15)

(3,4),(3,5),.............................,(3,15)

Total choices = 15 + 14 + 13 + 12 = 54 out of 64 choices.

So probability = 54 / 64 = 0.84

Please explain the solution for the 3rd question of the quiz.

@techtud @pritam @sumitverma

pls explain how to solve this-

How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame?

@dashish @limitless This might help http://gateoverflow.in/3278/gate2008-it-18## Pages