Question

among 18 computers in some store ,six have defects .five computers are randomly selected.compute the probability that all five computers are non defective?
my solution: 12/18X11/17X10/16X9/15X8/14 .is it right?

Answer 1

let me try something, just for practice :)

\[P(x)=\frac{\binom{Number.fail}{0fail}\binom{number.success}{f.success}}{\binom{Total}{number.choosen}}\]

Answer 2

6C5 / 18C5

Answer 3

this tiny little screen, i might have got my success and failures sawpped up

Answer 4

Total 18
6 are defective
5 chosen
5 not defective

Answer 5

\[P(x)=\frac{\binom{6}{0}\binom{12}{5}}{\binom{18}{5}}\]

Answer 6

i don understand ur solution :S :(

Answer 7

its the hyper geometric distribution. but lets see if we end up the same in the end

12.11.10.9.8
------------
18.17.16.15.14

Answer 8

yep, we are the same in the end

Answer 9

@spd two different methods to arrive at the same answer, your method in the question is correct.
first is not times second is not given first is not times ... this will work

amistre wrote a different method:
total number of ways to choose 5 from 18 is \(\binom{18}{5}\) and you have to select all 5 from the 12 that are not defective, number of ways to do that is \(\binom{12}{5}\)
so take the ratio
\[\frac{\dbinom{12}{5}}{\dbinom{18}{5}}\] you will get the same answer