GATE CSE 2015 Set 1 | Question: 29

Question

Consider a LAN with four nodes $S_1, S_2, S_3,$ and $S_4$. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmits in the same slot. The probabilities of generation of a frame in a time slot by $S_1, S_2, S_3,$ and $S_4$ are $0.1, 0.2, 0.3$ and $0.4$ respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is__________________.

Comments

Slotted Aloha + probability ka question hai

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Slotted Aloha + probability ka question hai,wrong tag lagaya hai

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if not given rounded off or upto 2 decimals then should we write whole answer?

Answer 1

Required probability=

=P(only one frame can transmit in a given time slot)

=P(only S1 sends)+ P(only S2 sends)+ P(only S3 sends)+ P(only S4 sends)

=(0.1*0.8*0.7*0.6)+(0.9*0.2*0.7*0.6)+(0.9*0.8*0.3*0.6)+(0.9*0.8*0.7*0.4)

So that which can not send than do minus from 1-(which cannot send on this slot of time expect which are sending in that slot of time we get  probability of sending )

=0.0336+ 0.0756 + 0.1296 + 0.2016

=0.4404

Answer 2

probablity=0.1*(1-0.2)*(1-0.3)*(1-0.4)+(1-0.1)*0.2*(1-0.3)*(1-0.4)+(1-0.1)*(1-0.2)*).3*(1-0.4)+(1-0.1)*(1-0.2)*(1-0.3)*0.4

=>0.0336+0.0756+0.1296+0.2016=.4404

Answer 3

Required probability

=P(only one frame can transmit in a given time slot)

=P(only S1 sends)+ P(only S2 sends)+ P(only S3 sends)+ P(only S4 sends)

=(0.1*0.8*0.7*0.6)+(0.9*0.2*0.7*0.6)+(0.9*0.8*0.3*0.6)+(0.9*0.8*0.7*0.4)

=0.0336+ 0.0756 + 0.1296 + 0.2016

=0.4404

Answer 4

it is a purely Probability Question, and it is flagged wrongly as question of Computer network
To avoid collision we must take care that while one station sending others should not send so for this. there will be as 

S1 sending and others not  OR
 S2 sending and others not OR 
S3 sending and others not OR 
S4 sending and others not 

now to make this statement mathematical we will add Probabilities of success and failure and also we will treat And as multiplication and OR as addition so equation will be :
 

   0.1 * (1 - 0.2) * (1 - 0.3) *(1 - 0.4) + 
  (1 -0.1) * 0.2 * (1 - 0.3) *(1 - 0.4) + 
  (1 -0.1) * (1 - 0.2) * 0.3 *(1 - 0.4) + 
  (1 -0.1) * (1 - 0.2) * (1 - 0.3) * 0.4

which gives result as  : 0.4404