A computer chip is organized into bits, bytes, and cells. Each byte contains 8 bits, and each cell contains 112 bytes. The probability that any one bit will be bad (or corrupted) is 1.E-11 (i.e. 10-11). a) What is the probability that any one byte will contain a bad bit and so will be bad and give an error in a calculation? Note that you can neglect the probability that a byte will contain more than one bad bit. b) What is the probability there will be no bad bytes in a cell? c) What is the probability there will be exactly one bad byte in a cell? d) What is the probability there will be exactly two bad bytes (and so also exactly 110 good bytes) in a cell? e) What is the probability there will be exactly three bad bytes (and so also exactly 109 good bytes) in a cell? a) 8E-11 b) 0.999999991 d) 8.96E-09 3.97824E-17 1.16695E-25 f) What is the probability there will be two or more bad bytes in a cell? Calculate this in three ways: i) Use the results of some of parts (a) (b) (c). f) i) ii) Use the results of parts (d) and (e). iii) Use a cumulative probability. Do they give the same answer? If not, explain why not. ii) iii) 4.9903E-13