Question

A shipment of 20 cameras includes 3 that are defective. What is the minimum number of cameras that must be selected if we require that P(at least 1 defective)>=.8.
So, y>=1, N=20, r=3, n=? ....?

Answer 1

18

Answer 2

providing me with an answer doesn't help me to learn and understand the problem. :/

Answer 3

The method I used produced n=8

Answer 4

sorry i will explain

Answer 5

in first draw we take two camera from 20 that can be defective or not defective .second draw also we take the same as from 20 that can be defective or non defective .at the eight draw of drawing two camera one must be defective from 20 camera.so at the eight draw we get defective camera

Answer 6

atleast one is defective=1-34/57=23/57

Answer 7

17c2/20c2=17*16/20*19=68/95
atleast one is defective is=1-68/95=27/95=0.2842

Answer 8

@kon i dont follow?
i get something different
\[P(X \ge 1) = 1- P(X=0)\]
so
\[P(X=0) \le 0.2\]
for there to be none defective, each of the selected cameras must be of the "17"
\[P(X=0) = \frac{17CN * 3C0}{20C N}\]
\[=\frac{17!}{N! (17-N)!}*\frac{N! (20-N)!}{20!}\]
\[= \frac{(20-N)(19-N)(18-N)}{20*19*18} \le 0.2\]
\[\rightarrow N^{3}-57N^{2}+1082N-5472 \ge 0\]
\[N \ge 7.87\]
min number of cameras is 8