The required probability is [tex]\frac{36}{55}[/tex]
Solution:
Given, a shipment of 11 printers contains 2 that are defective.
We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.
Now, we know that, [tex]\text { probability }=\frac{\text { favourable outcomes }}{\text { total outcomes }}[/tex]
Probability for first draw to be non-defective [tex]=\frac{11-2}{11}=\frac{9}{11}[/tex]
(total printers = 11; total defective printers = 2)
Probability for second draw to be non defective [tex]=\frac{10-2}{10}=\frac{8}{10}=\frac{4}{5}[/tex]
(printers after first slot = 10; total defective printers = 2)
Then, total probability [tex]=\frac{9}{11} \times \frac{4}{5}=\frac{36}{55}[/tex]
