Originally Posted by

**SadieKhan**
X Probability P(x)

0 (0.88)(0.88)(0.88) 0.681472

1 (0.12)(0.88)(0.88) 0.278784

2 (0.12)(0.12)(0.88) 0.038016

3 (0.12)(0.12)(0.12) 0.001728

You get the right probabilities but you're missing things. What that chart really should look like is:

Can you spot the difference?

b. Find the probability that the system does not work.

p(system doesn’t work)=P(A and C fail and B works)+P(B and C fail and A works)+P(A fails B and C fail)

=(0.12)(0.12)(0.88)+(0.12)(0.12(0.88)+(0.12)(0.12)(0.12)

=0.038016+0.038016+0.001728

=0.07776

And I mentioned you having the wrong things before because it seems to crop up here. Do your calculation again. Notice that

(.12)*(.12)*(.88) does not equal .038016.

c. How many parts are expected to fail?

x p(x) xp(x)

0 0.681472

0

1 0.278784

0.278784

2 0.038016 0.076032

3 0.001728 0.005184

Expectation of x=0.36

You should stop right here. I don't know what you're doing after this is and this is the correct answer.
How many parts are expected to fail?

=3*0.36=1.08.

Hence one part is expected to fail.

You should be getting 3*(.12)=0.36 as the expected number to fail. When I asked about the binomial distribution this is partly why. If you learn some basic distributions then these questions become quite easy. This process follows a binomial distribution. The expected value of a binomial is n*p so in this case 3*(.12) = .36.

d. If there are 10 of these machines, how many total parts are expected to fail?

=10*0.36=3.6

Once again just stop right here. When talking about the expected value NEVER round to the nearest integer. Nobody cares. Expected value is a mathematical property and it doesn't have to be an integer.
..or we may say almost 4 parts are expected to fail.