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

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)

answer: 2530

hope this