Question

Suppose a machine has five independent components each of which has probability 0.01 of failing , (a) find the probability that none or all components will fail, (b) find the probability that at least two components will fail?

Answer 1

"Welcome to OpenStudy. I can guide regarding this useful site; ask your doubts from me, for it you can message me. Please use the chat for off topic questions. And remember to give the helper a medal, by clicking on "Best Answer". We follow a code of conduct, ( http://openstudy.com/code-of-conduct ). Please take a moment to read it."

Answer 2

all fail: \((0.01)^5\)

Answer 3

none fail
\((.99)^5\)

Answer 4

as the question is finding all fail or none fail, i should add (0.01)5 and (0.99)5

Answer 5

oh i thought that was two different numbers
yes, since they are obviously mutually exclusive, add them up

Answer 6

oh, thank you, could you help me in finding the next part in the question i.e., finding probability that at least two components will fail

Answer 7

sure

Answer 8

"at least two" translates as 2 or 3 or 4 or 5
which is a lot to compute
much easier to think of "at least 2" as
"not 0, not 1"
compute those, subtract from 1

Answer 9

you already got 0, what was \(.99^5\)

Answer 10

do you know how to get 1?

Answer 11

could you tell me how to get for 1?

Answer 12

sure
one fails \(.01\)
4 do not fail \(.99^4\) and there are 5 ways for this to happen,
first one fails others don't
second one fails, others don't, etc

Answer 13

so
\[5\times (0.1)\times (.99)^4\]

Answer 14

it is the binomial probability with \(p=.99,1-p=.01, n=5\) general form is
\[P(x=k)=\binom{n}{k}p^k(1-p)^{n-k}\]

Answer 15

oh Thank You very much:>

Answer 16

yw

Answer 17

hey hi, could you solve this question for me
A circuit consists of 8 resistors, 2 inductors and 3 capacitors. 3 of the circuit elements are damaged. Find the following probabilities (a) p(only 1 capacitor is damaged) (b) p(at least one inductor is damaged) (c) p(no resistor are damaged).