Vtyjkyuk

A system consisting of a micro controller and 3 peripheral devices. The system is said to be up when micro controller and at least 2 peripheral devices are in working condition. Determine the probability of system working when probability of failure of micro controller is 0.2 and probability of failure of each device is $0.3$ . All events are independent of each other.

Here is how I approach the problem:
Let the peripherals be A,B and C.

the probability of A,B peripherals not working = $0.3+0.3-(0.3times0.3) = 0.51$

Probability of A,B & B,C & A,C not working = $0.51 times 0.51 times 0.51 = 0.1326$

Probability of micro controller and no 2 combination of peripherals working = $0.2 + 0.1326 - (0.1326 times 0.2)
= 0.3061$

Probability of system working = $1-0.3061 = 0.693$

Does this approach lead to the answer correctly?

For the system to work, we certainly need the micro controller to be functioning, the probability of that would be $0.8$.

We also need at least two peripheral devices to work. $$binom32(0.7^2)(0.3)+(0.7)^3$$

We just have to multiply them up:

$$0.8(0.7)^2[ 0.9+0.7]=2(0.56^2)=0.6272$$

Note that the event that $(A,B)$ not working and $(B,C)$ not working are not independent.