Recall that, an absolute value expression is a piece-wise in effect, thus we'd check the single-sided limits,
Answer:
(a) Probability that a triplet is decoded incorrectly by the receiving computer. = 0.010
(b)
(1 – p) = 0.010
(c)
E(x) = 25000 x 0.010
= 259.2
Explanation has given below.
Step-by-step explanation:
Solution:
(a) Probability that a triplet is decoded.
2 out of three
P = 0.94, n = 3
m= no of correct bits
m bit (3, 0.94)
At p(m≤1) = B (1; 3, 0.94)
= 0.010
(b) Using your answer to part (a),
(1 – p) = 0.010
Error for 1 bit transmission error.
(c) How does your answer to part (a) change if each bit is repeated five times (instead of three?
P( m ≤ 2 )
L = Bit (5, 0.94)
= B (2; 5, 0.94)
= 0.002
(d) Imagine a 25 kilobit message (i.e., one requiring 25,000 bits to send). What is the expected number of errors if there is no bit repetition implemented? If each bit is repeated three times?
Given:
h = 25000
Bits are switched during transmission between two computers = 6% = 0.06
m = Bit (25000, 0.06)
E(m) = np
= 25000 x 0.06
= 1500
m = Bit (25000, 0.01)
E(m) = 25000 x 0.010
= 259.2
2
8 - Or As a decimal it's 8.4
5
...........................
Step-by-step explanation:
42
=21+21
=42
OR
10+32
=42
