Question

You have two machines A and B that, each, generate binary digits. On each machine, when the “run” button is pressed, it will generate a single binary digit. Machine A generates a 0 - 52% of the times and a 1 - 48% of the times. Machine B, however, has a memory slot that stores the latest bit generated. Machine B always starts by generating a 0 and storing this in the memory slot. Each time, after this initialization, machine B generates the new bit by checking the bit in the memory slot, and generates the new bit by flipping the bit 61% of times and overwrites the memory with this new bit. When machine B is turned off, the memory slot clears itself.

(a) On using the ’run’ feature 7 times on Machine A, what is the probability that the outcome has exactly five 0’s?

(b) What is the probability of each machine generating ’00110’ and ’1001’?

Answer 1

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), 0 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 1 after 0 has probability of 69% and 1 after 1 has probability of 39%, but first 1 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 1 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%.

1 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 1 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%.