Help..calculate the probability of finding 3 wrong bits, in a byte, if the BER is 0.5

Please help me...I'm a little confused.:|

Bit Error Rate (BER) = (No. of Error bits)/(Total number of bits transferred) in this case BER = 0.5

So number of error bits = 4 for a byte of data transferred..

Now, total number of error bits in a Byte will be 4. Hence probability of an error bit occurring in a byte of data = 4/8 = 1/2

Now, the probability of finding 3 error bits with probability of 1/2 = (1/2) * (1/8) * (3) = 3/16..

Hence probability of finding 3 wrong bits in a byte of data will be 3/16...

This is my calculation. Please update this thread if you get an answer different from this...

Hope to be helpful..

I think this formula applies to your problem: P = nCr x p^r x q^(n-r)
where:
n=no. of trial (8)
r=successful tries (3)
C=operation of 'combination'
p=probability of success (.5)
q=probabilty of failure (1-p=1-.5=.5)
P=probability

Therefore the probability of finding 3 errors in a transmission of byte with BER of .5 is P=0.21875=21.875%

The bit error rate or bit error ratio (BER) is the number of bit errors divided by the total number of transferred bits during a studied time interval. BER is a unitless performance measure, often expressed as a percentage.
The studied time interval is not specified in the problem. So I believe that one byte time interval is not necessary the studied time interval. We just have that BER=0.5.
When a tudied time interval tends to infinity, we have that BER=p=0.5, where p is bit error probability.
We will assume that p=0.5 in this problem.
The probability of finding 3 wrong bits in a byte is the probability of appearing 3 error bits and 5 true bits in a byte simultanuasly multiplied by the number of various ways in which 3 bit out of 8 may be in error.
So:
p=0.5 - error bit probability
(1-p) - true bit probability
p*p*p*(1-p)*(1-p)*(1-p)*(1-p)*(1-p) - the probability of appearing 3 error bits and 5 true bits in a byte simultanuasly
n!/(j!(n-j)!) - is the number of various ways in which j bit out of n may be in error.
And we have:
the probability of finding 3 wrong bits in a byte is:
P=p*p*p*(1-p)*(1-p)*(1-p)*(1-p)*(1-p)*n!/(j!(n-j)!)]=[0.5^8]*[8!/3!/5!]=0.00390625*40320/6/120=0.21875

And the formula given by Onigece is right

rsh456,

Are you telling about having exactly 3 wrong bits or to having at least 3 wrong bits?
The above formula holds for exactly.

Z