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In a communication network, a packet of length $L$ bits takes link $L_1$ with a probability of $p_1$ or link $L_2$ with a probability of $p_2$. Link $L_1$ and $L_2$ have bit error probability of $b_1$ and $b_2$ respectively. The probability that the packet will be received without error via either $L_1$ or $L_2$ is

  1. $(1 - b_1)^Lp_1 + (1 - b_2)^Lp_2$
  2. $[1 - (b_1 + b_2)^L]p_1p_2$
  3. $(1 - b_1)^L (1 - b_2)^Lp_1p_2$
  4. $1 - (b_1^Lp_1 + b_2^Lp_2)$
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4 Answers

Best answer
91 votes
91 votes

Probability of choosing link $L_1=p_1$

Probability for no bit error (for any single bit)$=(1-b_1)$

Similarly for link $L_2$ 

Probability of no bit error $=(1-b_2)$

Packet can go either through link $L_1$ or $L_2$ they are mutually exclusive events (means one event happens other won't be happening and so we can simply add their respective probabilities for the favorable case).

Probability packet will be received without any error = Probability of $L_1$ being chosen and no error in any of the $L$ bits + Probability of $L_2$ being chosen and no error in any of the $L$ bits

$=(1-b_1)^Lp_1+(1-b_2)^Lp_2.$

Hence, answer is option A .

---------

Why option D is not correct choice here?

Option D here is giving the probability of a frame being arrived with at least one bit correctly - i.., all the bits are not errors. 

edited by
11 votes
11 votes

it should be A
Number of bits in a packet = L bits

Probability of link ‘l1′ = p1
Bit error probability for link ‘l1′ = b1
No bit error probability for link ‘l1′ = (1 – b1)

Probability of link ‘l2′ = p2
Bit error probability for link ‘l2′ = b2
No bit error probability for link ‘l2′ = (1 – b2)

Link ‘l1′ and ‘l2′ are mutually exclusive.
Thus, probability that the packet will be received without error = (1 – b1)Lp1 + (1 – b2)Lp2

 
Thus, option (A) is correct.
source-geeksforgeeks

4 votes
4 votes
for those having confusion between a & d

lets take L =2

b1=b2=p1=p2= 1/2

a gives 1/4 d gives 3/4

now lets checks which one of this is correct

 L is of 2 bits , probability that it takes L1 is 1/2 , and probability that it arrives correctly is 1/4(How?

if L is 00 . it can get corrupted to  10,01,11 or remains same hence 1/4 )

hence by multiplication theorem  1/2*1/4 =1/8

similarly for  if L takes L2 probability is 1/8

1/8 +1/8 = 1/4

hence a correct

as pooja mentioned d is

Option D here is giving the probability of a frame being arrived with at least one bit correctly & not all bits arrived correctly hence it can not be answer
Answer:

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