Binary data are transmitted over a noisy communication channel. The probability that a received binary digit is in error due to channel noise is 0.05. Assume that such errors occur independently within the bit stream. (a) What is the probability that the 3rd error occurs on the 50th transmitted bit?
(b) On average, how many bits will be transmitted correctly before the first error?
(c) Consider a 32 bit transmissions over a noisy communication channel, What is the probability of exactly 2 errors in this word?

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Step-by-step explanation:

a.) P(x=3)=0.0166

When a random variable is defined as the kth success in the nth trial, the negative binomial distribution follows.

The probability mass function for the negative binomial distribution is defined as,

P(X=k)=\binom{n-1}{k-1}p^k(1-p)^{n-k}

where

k is the kth success = 3

n is the number of trials needed for kth success = 50

p = probability of success = 0.05

P(X=3)=\binom{50-1}{32-1}0.05^3(1-0.05)^{50-3}

P(X=3)=0.0166

b) On average, 20 bits will be transmitted before the first error.

When a random variable is defined as the number of trials needed until the first success occurs, the geometric distribution follows,

The expected value of the geometric random variable is obtained using the following formula,

E[X]=\frac{1}{p}=\frac{1}{0.05}=20

c) P(X = 2) = 0.0166

The probability mass function for the negative binomial distribution is defined as,

P(X=k)=\binom{n-1}{k-1}p^k(1-p)^{n-k}

Given

k = 2

n = 32

P(X=2)=\binom{32-1}{2-1}0.05^2(1-0.05)^{32-2}

P(X = 2) = 0.0166

3 1/3 x 2 3/5 = 8 2/3

step-by-step explanation:

just multiply the first number by the second number to get your answer.