When the system is executed, the system crashes with probability 0.05 if only Module A fails. The system crashes with probability 0.12 if only Module B fails. The system crashes with probability 0.60 if both Module A and Module B fail. The system crashes with probability 0.01 if neither fails. (a) When the system is executed, what is the probability the system will crash

P (system will crash) = 0.101528

P(A and B jails / System crash) = 0.5390

Step-by-step explanation:

The complete question is as stated below

"Suppose a system has two modules, A and B , that function independently. Module A fails with probability 0.24 and Module B fails with probability 0.38 ,when the system is executed.  When the system is executed, the system crashes with probability 0.05 if only Module A fails. The system crashes with probability 0.12 if only Module B fails. The system crashes with probability 0.60 if both Module A and Module B fail. The system crashes with probability 0.01 if neither fails.

(a) When the system is executed, what is the probability the system will crash?

(b) If the system crashes, what is the probability that both modules A and B crashed?"

Solution

P(A fails) = 0.24

P(B fails) = 0.38

P(A ∩ B) = P (A) * P (B) = 0.24 * 0.38 = 0.0912

P(only A fails) = P(A) - P(A ∩ B) =0.24 - 0.24*0.38 = 0.1488

P(only A fails) =P(B) - P(B ∩ A) = 0.38 - 0.24*0.38 = 0.2888

P(Both fails) = P(A) * P(B) = 0.24*0.38 = 0.0912

P(Neither fails) = P(A) * P(B) =1-(0.24+0.38-0.0912) = 0.4712

P(Add to 1)

a) P (system will crash) = 0.1488*0.05+0.2888*0.12+0.0912*0.6+0.4712*0.01 P (system will crash) = 0.101528

b) P(A and B jails / System crash) = 0.0912*0.6 / 0.101528

P(A and B jails / System crash) = 0.5390

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: