Question

Can some one explain how to solve this?

Answer 1

which parts?

Answer 2

i'm not sure on all of it

Answer 3

I'm unsure of a and b coz i didn't get that usual and unusual thing!!
I can explain b.
system succeeds when both the parts succeed or any one of them fail.
So, probability of system success = P(1st part success)xP(2nd part success) + P(1st part success)xP(2nd part failure) + P(1st part failure)xP(2nd part success)
So, probability of system success = (0.83)(0.83) + (0.83)(0.17) + (0.83)(0.17)
= 0.9711

Answer 4

ok what about c

Answer 5

I do not know any method for c..just tried with 3 parts but the probability came out 0.995087
So, I'm trying with 4 now!

Answer 6

with 4 parts, the probability is 0.99916479!!!

Answer 7

I think we can go with 5 parts

Answer 8

coz as we increase one part, the no.of 9's in the decimal places are increasing..take a look

Answer 9

? ok i'm confused

Answer 10

For two parts, P(success) > 0.9
For three, P(success) > 0.99
For four parts, P(success) > 0.999
So, obviously for five parts it would be P(success) > 0.9999 which is asked in the question

Answer 11

so c is 5?

Answer 12

solve for n\[0.17^{n} < 0.0001\]

Answer 13

In b, you can also use
P(system success) = 1 - P(system failure)
system fails only when all the parts fail
So, P(system success) = 1 - P(both the parts fail) = 1 - (0.17)(0.17) = 0.9711

Answer 14

So, now check it in this way for different number of parts and for six parts, you get probability of success greater than 0.9999
you get p(system success) = 0.9999759

Answer 15

ok so the answer is?

Answer 16

6

Answer 17

THX!!

Answer 18

welcome:)