Question

a box contains 2 defective lights inadvertenly mixed with 8 nondefective lights. If the lights are selected one at a time without replacement and tested until both defective lights are found, what is the probability that both defective lights will be found after exactly 3 trials

Answer 1

easy

Answer 2

so take a general case

(D)(D)(ND)

where D= defective , ND = not defective

Answer 3

the probability of this one event happening is

P(D, D, ND) = 2/10 x 1/9 ( we stop after they found the two bad ones )

Answer 4

P(D,ND,D) is the only other case to consider

Answer 5

a couple ways to do this but easiest might be this.
there are 3 possibilities for getting exactly 2 defective in 3 tries:
ddn
dnd
ndd where d is defective and n is not. each of these is given by the probability
\[\frac{2}{8}\times \frac{1}{7} \times\frac{6}{6}\] so answer is
\[3\times \frac{2}{8}\times \frac{1}{7}=\frac{3}{28}\]

Answer 6

doesnt quite work

Answer 7

that way

Answer 8

messed that all up didn't i?

Answer 9

I think theres something wrong, ohh might be right, probabilities is not my very best

Answer 10

right idea though. two out of eight.
\[\frac{2}{10}\times \frac{1}{9}\times \frac{8}{8}\times 3\]

Answer 11

unless "exactly three:" means on the third trial in which case it is
dnd
ndd but not
ddn

Answer 12

yeh thats was what I was thinking lol

Answer 13

it says they stop after they found duds

Answer 14

not clear is it?

Answer 15

so u cant include them

Answer 16

well "exactly 3" might mean "it takes three trials" or it might mean "in three trials we have two defective" not sure which

Answer 17

if it means "in three trials we have two defective" i think my answer is right

Answer 18

if it means "it takes exactly three trials before we know" we have only two possibilities
DND
NDD

Answer 19

second case would be
\[2\times \frac{2}{10}\times \frac{8}{9}\times \frac{1}{8} \]