GATE IT 2005 | Question: 74

Question

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

• A. $(1 - b_1)^Lp_1 + (1 - b_2)^Lp_2$
• B. $[1 - (b_1 + b_2)^L]p_1p_2$
• C. $(1 - b_1)^L (1 - b_2)^Lp_1p_2$
• D. $1 - (b_1^Lp_1 + b_2^Lp_2)$

Comments

For those who are confused in option, d

option d says 1 - P(all bits are error bits) = Not all bits are error bits.

If there are n bits then 1 bit is error free OR 2 bits are error-free OR 3 bits are error-free and so on which is same as saying at least 1 bit is error free.

But we want the entire packet to be error free. Hence option A.

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why not (p2)(1-p1){(1-b2)^l)}+p1(1-p2){(1-b1)^l}

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Why is Option D wrong?

$p1$ – Probability of taking the link $L1$

$p2$ – Probability of taking the link $L2$

$p1 + p2$ – Probability of taking the link $L1$ or $L2$

(this includes taking the link and ended up with a bit error $and$ taking the link and having no bit error) 

Now we need to subtract the probability of taking the link and having a bit error.

$p1 . b1$ – Probability of taking the link $L1$ and having a bit error. 

Similarly, $p2 . b2$ – Probability of taking the link $L2$ and having a bit error. 

So the final answer is $(p1 + p2) – (p1.b1 + p2.b2)  =  (1-b1).p1 + (1-b2).p2$

Option A is correct. 

Option D would be correct if $p1 + p2 = 1$

Answer 1

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. 

Comments

@Pooja Palod @Milicevic3306 

Please update this line "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."

Option D is giving the probability of frame being arrived with all error bits. And, we need to reject any frame with atleast one error bit. 

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why "L" is choosen as power?\

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@Salla shivateja L is the total number of bits that we need to transfer , In question we are given bit error probability which are b1 and b2 which means that it tells probability of error in only one bit whereas we need to calculate probability of error in L bits . So probability that L bits are error free are (1-b1)^L using L1 link and (1-b2)^2 using L2 link. 

Answer 2

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

Answer 3

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 4

 

$\therefore$ Probability that the packet will be received without any error = $\left ( 1-b1\right )^{L}p1 + \left ( 1-b2\right )^{L}p2$

Thus, option A is correct.