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Anton [14]
3 years ago
6

When the system is executed, the system crashes with probability 0.05 if only Module A fails. The system crashes with probabilit

y 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
Mathematics
1 answer:
Viefleur [7K]3 years ago
5 0

Answer:

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?"

<u>Solution</u>

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

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2\vec{v}-6\vec{u}=(-52,48) \\ \\ ||2\vec{v}-6\vec{u}||=75.28}

<h2>Step-by-step explanation:</h2>

In this problem we have two vectors:

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So we need to find two things:

2\vec{v}-6\vec{u}

and:

||2\vec{v}-6\vec{u}||

FIRST:

In this case we have the multiplication of vectors by scalars. A scalar is a simple number, so:

2\vec{v}-6\vec{u} \\ \\ Replace \ \vec{v} \ and \ \vec{u} \ by \ the \ given \ vectors: \\ \\ 2(-11,3)-6(5,-7) \\ \\ Multiply \ each \ component \ by \ the \ corresponding \ scalar:\\ \\ (2\times (-11),2\times 3)+(-6\times 5,-6\times (-7)) \\ \\ (-22,6)+(-30,42) \\ \\ Sum \ of \ vectors: \\ \\ (-22-30,6+42) \\ \\ \therefore \boxed{(-52,48)}

SECOND:

If we name:

\vec{w}=2\vec{v}-6\vec{u}

Then, ||2\vec{v}-6\vec{u}|| is the magnitude of the vector \vec{w}. Therefore:

||\vec{w}||=||2\vec{v}-6\vec{u}|| \\ \\ ||\vec{w}||=||(-52,48)|| \\ \\ ||\vec{w}||=\sqrt{(-58)^2+48^2} \\ \\ ||\vec{w}||=\sqrt{3364+2304} \\ \\ ||\vec{w}||=\sqrt{5668} \\ \\ \boxed{||\vec{w}||=75.28}

5 0
3 years ago
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