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## Here's the Solution to this Question

a)Each 'run' on Machine A is independent from other, so probability of two zeros in two 'runs' is multiplication of zero probability in one 'run'. So probability of line 0000011 is 0.5250.482, but order of zeros is irrelevant to as, so we need to multiply this probability on number of different ways to distribute 5 zeros in 7 'runs'. This number is 7!/5!(7-5)!=7*6/2=21. And the answer is approximately 18,4%

b)Firstly lets talk about Machine A. As I previously said each run is independent, and now order is important, so probability of '00110' is $0.52*0.52*0.48*0.48*0.52 \approx 0.032$ or 3.2%. Same way probability of '1001' is $0.48*0.52*0.52*0.48 \approx 0.062$ or 6.2%.

Macine B has three different states which affect probability 1 and 0. This states are empty memory slot(right before initialization), in memory slot and 1 in memory slot.

Empty memory state: First 0 has probability equal to 100%, 0 after 0 has probability of 39%, 0 after 1 has probability of 61%. Similarly after 0 has probability of 69% and after 1 has probability of 39%, but first has probability equal to 0%. So probability of '00110' is $1*0.39*0.61*0.39*0.61\approx 0.057$ or 5.7%. Probability of of '1001' is 0% since first to be generated is always 0.

0 in memory: The only change from previous case is probability of first and 0. They are equal to 61% and 39% respectively. So probability of '00110' is $0.39*0.39*0.61*0.39*0.61 \approx 0.022$ or 2.2% and probability of '1001' is $0.61*0.61*0.39*0.61 \approx 0.089$ or 8.9%.

in memory: This case is similar for case with 0 in memory, but probabilities were flipped and Machine B generate us 0 with 61% and with 39% on first 'run'. Following 'runs' probability remains unchanged, so probability of '00110' is $0.61*0.39*0.61*0.39*0.61 \approx 0.035$ or 3.5%, probability of '1001' is $0.39*0.61*0.39*0.61 \approx 0.057$ or 5.7%.