Question

two out of six computers in a lab have problems with hard drives .if three computers are selected at random for inspection,what is the probability that none of them has hard drive problem

Answer 1

\[\frac{2}{6}=\frac{1}{3}\] is the probability that one has a hard drive problem, so probability that a computer does not have one is \(1-\frac{1}{3}=\frac{2}{3}\)

Answer 2

then the probability all three selected do not have a problem is \((\frac{2}{3})^3=\frac{8}{27}\)

Answer 3

oh wow sorry that was way wrong
i did not read the "in the lab" part!!

Answer 4

i thought there were many many computers, not just six
it should be
\[\frac{4}{6}\times \frac{3}{5}\times \frac{2}{4}\] as your answer

Answer 5

ok but here is my solution plz tell me if m right.
2 are defective so 4 will be non defective
P(non defective)=4/6X3/5X2/4X1/3=0.06
1- P(defective among selected)=2/3 X1/2=1/3=0.6

now P(non defective)=0.06X.6

Answer 6

must be 4/6x3/5x2/4

Answer 7

how is that? 4 would be good right?? plz explain it in steps

Answer 8

first let's see the question properly ... "what is the probability that none of them has hard drive problem"
here the question asks that: 3 computers selected out of 6 has NOT hard drive problem.

now we have to select 3 computers ..ie 3 places
__X__X__
what is probability for 1st place that it does not have hard drive problem
similarly for 2nd place and 3 rd place.

Answer 9

ok

Answer 10

3/6X2/5X1/4

Answer 11

NO .. for first place you have 4 non-damaged pc out of 6

Answer 12

ok for the first place its 4/6
for 2nd place it wud be 3/5
for 3rd 2/4
ok got it