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During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of at most one bit in error in a block of n bits is given by
- a)pn
- b)1 – pn
- c)np(1 – p)n - 1 + (1 – p)n
- d)1 – (1 – p)n
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
During transmission over a certain binary communication channel, bit ...
Explanation:
To understand why option C is the correct answer, let's break down the problem step by step.
1. Probability of a bit error:
Given that bit errors occur independently with probability p, the probability of a bit being error-free is 1-p.
2. Probability of at most one bit in error:
To find the probability of at most one bit in error in a block of n bits, we need to consider two cases:
- Case 1: No bit is in error
- Case 2: Exactly one bit is in error
Case 1: No bit in error:
The probability of no bit being in error in a block of n bits is (1-p) for each bit. Since the bits are independent, the probability of all n bits being error-free is (1-p)^n.
Case 2: Exactly one bit in error:
The probability of exactly one bit being in error can occur in any one of the n positions in the block. So, we need to multiply the probability of a bit being in error (p) with the probability of all other bits being error-free (1-p)^(n-1). Therefore, the probability of exactly one bit being in error is n * p * (1-p)^(n-1).
Total probability:
To find the probability of at most one bit in error, we need to add the probabilities of Case 1 and Case 2:
P(at most one bit in error) = (1-p)^n + n * p * (1-p)^(n-1)
Simplifying this expression, we get:
P(at most one bit in error) = (1-p)^(n-1) * [(1-p) + n * p]
Final answer:
Comparing the simplified expression with the given options, we can see that option C matches the form of the expression:
np(1-p)^(n-1)
Therefore, option C is the correct answer.
To understand why option C is the correct answer, let's break down the problem step by step.
1. Probability of a bit error:
Given that bit errors occur independently with probability p, the probability of a bit being error-free is 1-p.
2. Probability of at most one bit in error:
To find the probability of at most one bit in error in a block of n bits, we need to consider two cases:
- Case 1: No bit is in error
- Case 2: Exactly one bit is in error
Case 1: No bit in error:
The probability of no bit being in error in a block of n bits is (1-p) for each bit. Since the bits are independent, the probability of all n bits being error-free is (1-p)^n.
Case 2: Exactly one bit in error:
The probability of exactly one bit being in error can occur in any one of the n positions in the block. So, we need to multiply the probability of a bit being in error (p) with the probability of all other bits being error-free (1-p)^(n-1). Therefore, the probability of exactly one bit being in error is n * p * (1-p)^(n-1).
Total probability:
To find the probability of at most one bit in error, we need to add the probabilities of Case 1 and Case 2:
P(at most one bit in error) = (1-p)^n + n * p * (1-p)^(n-1)
Simplifying this expression, we get:
P(at most one bit in error) = (1-p)^(n-1) * [(1-p) + n * p]
Final answer:
Comparing the simplified expression with the given options, we can see that option C matches the form of the expression:
np(1-p)^(n-1)
Therefore, option C is the correct answer.
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Community Answer
During transmission over a certain binary communication channel, bit ...
By Binomial distribution the probability of error is
Pe = nCr Pr (1- P) n - r
Probability of at most one error
= Probability of no error + Probability of one error
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During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of at most one bit in error in a block of n bits is given bya)pnb)1 – pnc)np(1 – p)n - 1 + (1 – p)nd)1 – (1 – p)nCorrect answer is option 'C'. Can you explain this answer?
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