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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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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 .

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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
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Iam also getting option A, but. option D can also be correct..... which is.......1-collision probility.
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Answer should be 'd' because if  'a' was answer then  it should have been like:-

(1-b1)^L*p1*(1-p2)  +  (1-b2)^L*p2*(1-p1)

i.e (1-p1) and (1-p2) should also be included because for successful transmission packet should take link L1 with probability p1 and should not take L2 with probability p2 AND VICE-VERSA.
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D or A ??
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can't we think like this [ (1-b1)*p1  +  (1-b2)*p2 ]^L  means i am calculating for 1 bit then the same for L bits
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dhruvkc123  no u can't  becoz we have only 2 link for that probability is p1 and p2

but bit are in numbers of L on that case we take individual probability

@bikram sir check this pls

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still not clear..!!!!
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@dhruvkc123

No,  [ (1-b1)*p1  +  (1-b2)*p2 ]^L will be wrong. b1 and b2 are probabilities associated with bits whereas p1 and p2 are associated with the entire packet. The probability that a bit will not be erroneous is (1-b1). The probability that the entire packet of L bits will be error free is (1-b1)^L. And now, the probability that this packet takes link L1 is p1. Hence, it is (1-b1)^L*p1.

[ (1-b1)*p1  +  (1-b2)*p2 ]^L this is calculating the probability of individual bits taking different paths. Hence its wrong.
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@pooja plz elaborate option d.

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

+1 vote
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
–1 vote

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