Question 1159728: 4)Suppose the probability that a bit transmitted through a digital communication channel and received in error is 0.01. Assuming that the transmissions are independent events,
a) Find the probability that the first error occurs at the 8th bit.
b) Find the probability that the third error occurs at the 8th bit.
c) Find the probability that three or more errors occur among 100 transmitted bits.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! geometric function
probability is 0.99^7*0.01=0.0093
For the next part, we need the probability of 2 errors in 7 bits
there are 7C2 or 21 ways that can happen and the probability to get 2 errors is 21*0.99^5*0.01^2=0.00200 or 0.002
Then the eighth bit has a 0.01 probability of an error, so the joint probability is 0.0002 for the third error to occur on the 8th bit.
Find the probability of 0,1,2 errors among 100 bits
for 0, it is 0.99^100=0.3660
for 1 it is 100*0.99^99*0.01=0.3697
for 2 it is 100C2*0.99*98*0.01^2=0.1849
That sum is 0.9206
Also 2nd VARS to binomcdf (100,0.01,2)
BUT this is the probability we don't want. The complement, 3 or more errors, is 0.0794, and that is the answer.
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