Question

P [1r / 0s] = 0.03 and P [0r / 1s] = 0.02 conditional in an asymmetric binary communication channel

available. The probability of sending 0 in this channel is P [1s] = 0.56. 25 bits from this channel

It has been sent. Accordingly, find the possibility of fewer than 3 faulty bits in the 25 bits transmitted.

Note: 0s : 0 transmit, 1s : 1 transmit, 0r : 0 represent, 1r : 1 represent running events.

Answer 1

Answer:-

Given That:-

P [1r / 0s] = 0.03 and P [0r / 1s] = 0.02 conditional in an asymmetric binary communication channel available. The probability of sending 0 in this channel is P [1s] = 0.56. 25 bits from this channel It has been sent. Accordingly, find the possibility of fewer than 3 faulty bits in the 25 bits transmitted.

Note: 0s : 0 transmit, 1s : 1 transmit, 0r : 0 represent, 1r : 1 represent running events.

Given,

Probability of a single bit that is transmitted is faulty

= (Probability of transmitting a 0)*P[1r/0s] + (Probability of transmitting a 1)*P[0r/1s]

= 0.56*0.03 + (1-0.56)*0.02

= 0.0168+0.0088

= 0.0256.

Now,

let X denotes th number faulty of bits transmitted out of 25 transmitted bites.

Then X follows Binomial distribution with parameters, n = 25

and

p = 0.0256

Required probability = P(X<3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

   = 0.97466

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