A machine consists of three components, which are components X, Y, and

Question

A machine consists of three components, which are components X, Y, and Z. These three components operate properly independently of one another. The machine operates properly only when component X operates properly and at least one of components Y and Z operates properly. The probability that component X operates properly is 0.8, the probability that component Y operates properly is 0.4, and the probability that component Z operates properly is 0.3. What is the probability that the machine operates properly?

A) 0.096
B) 0.252
C) 0.464
D) 0.583
E) 0.648

we can solve the problem:
P(X)*P(y)*nP(x)+P(x)*P(Z)*nP(y)+P(x)+P(y)+P(z)
= (0.8*0.4*0.7)+(0.8*0.3*0.6)+(0.8*0.4*0.3)
= 0.224+0.144+0.096
= 0.464, option C

A) 0.096 
B) 0.252 
C) 0.464 
D) 0.583 
E) 0.648

In all the Kaplan explanations for probability questions, it always says, "It is easier for us to find the probability of XYZ not occurring".

Can anyone explain why that is? Why can't we just find the probability of XYZ actually occurring?

For example, in the question above, they tell us outwardly the probability of Components Y and Z operating properly, 0.4 and 0.3 respectively. My inclination is to simply multiply the probability of Component X working (0.8) with the probabilities of Y or Z working, 0.8 * (0.4 + 0.3) = 0.56. While, I know that the incorrect answer, why doesn't this simple approach work?

The explanation states that you must first find the probability of Y and Z NOT operating properly, i.e.

1 - 0.4 = 0.6 
1 - 0.3 = 0.7

Then multiply them to get 0.42 and subtract it from 1 to find the probability of either Y or Z operating properly, which is 0.58. At that point you just multiply 0.58 by 0.8 and the answer is 0.464.

If anyone can shed light on this, I'd appreciate it. Thanks everyone!

pratik2310 · Answer

+   +  
=   +   +  
=0.224+0.144+0.096
=0.464
Ans -C

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shaarang · Answer

If I have understood this correctly, your approach assumes that the other (third) machine never works in each of your scenarios. This is incorrect. Let me explain.

0.8 * (0.4 + 0.3) = 0.8 * 0.4 + 0.8 * 0.3 = A working * B working + A working * C working

In these cases, you're not considering the probability of a third machine   working or not working, therefore, the implicit assumption in your approach is that C always doesn't work in Case 1 (P(C')=1) and B always doesn't work in Case 2 (P(B') = 1). But we know that's not the case because the question defines the possibilities for B and C working. Therefore, it is important to explicitly define the probability of the machine not working or else it defaults to x1 which is wrong.

Note that you're also ignoring the event when both B and C work along with A, so the approach is flawed overall.

Roma123 · Answer

Anyone please help me solve this question.

KarishmaB · Answer

We are given that Y and Z are independent components so they operate independently.

Machine operates if X operates AND (Y or Z operates)

P(Machine) = P(X) * P(Y or Z)

P (Y or Z) = P(Y)+P(Z) - P(Y and Z) = P(Y) + P(Z) - P(Y)*P(Z) = 0.4 + 0.3 - 0.4*0.3 = 0.58

P(MAchine) = 0.8*0.58 = 0.464

Answer (C)

KarishmaB · Answer

You are right. I wouldn't think in terms of "Not Y and Z" either because calculating P(Y or Z) is easy enough.

Note though that P(Y Or Z) is not P(Y) + P(Z). It is P(Y) + P(Z) - P(Y and Z) because P(Y and Z) is double counted (once in P(Y) and once in P(Z)). So you need to remove it out once.

Archit3110 · Answer

A machine consists of three components, which are components X, Y, and Z. These three components operate properly independently of one another. The machine operates properly only when component X operates properly and at least one of components Y and Z operates properly. The probability that component X operates properly is 0.8, the probability that component Y operates properly is 0.4, and the probability that component Z operates properly is 0.3. What is the probability that the machine operates properly?

A) 0.096
B) 0.252
C) 0.464
D) 0.583
E) 0.648

we can solve the problem:
P(X)*P(y)*nP(x)+P(x)*P(Z)*nP(y)+P(x)+P(y)+P(z)
= (0.8*0.4*0.7)+(0.8*0.3*0.6)+(0.8*0.4*0.3)
= 0.224+0.144+0.096
= 0.464, option C

A) 0.096 
B) 0.252 
C) 0.464 
D) 0.583 
E) 0.648

we can solve the problem:
P(X)*P(y)*nP(x)+P(x)*P(Z)*nP(y)+P(x)+P(y)+P(z)
= (0.8*0.4*0.7)+(0.8*0.3*0.6)+(0.8*0.4*0.3)
= 0.224+0.144+0.096
= 0.464, option C

Utkersh · Answer

In this question, it is mentioned that for the machine to work, Machine X and  atleast  one of Machine Y or Z needs to work. Does this not mean that the probability for the machine to work is

P(X) AND P(Y or Z) + P(X) AND P(Y) AND P(Z)

PSKhore · Answer

Find the probability where both the machines Y and Z fail = (1 - 0.4)*(1 - 0.3) = 0.6*0.7 = 0.42
Probability of at least one of Y or Z operating = (1 - 0.42) = 0.58

Probability that the machine operates = Z*(Y or Z) = 0.8*0.58 =  0.464

A machine consists of three components, which are components X, Y, and

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A machine consists of three components, which are components X, Y, and

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