The required probability is 
<u>Solution:</u>
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, 
Probability for first draw to be non-defective 
(total printers = 11; total defective printers = 2)
Probability for second draw to be non defective 
(printers after first slot = 10; total defective printers = 2)
Then, total probability 
Answer:
x = 30
Step-by-step explanation:
by solving ;
- 18 / x = 6 / 10
- 1 / x = 6 / ( 18 × 10 )
- 1 / x = 6 / 180
- 1 / x = 1 / 30
- x = 30
Answer:
1.2218 atm
Step-by-step explanation:
From ideal gas formula,
PV=nRT
Where n= number of moles
V= volume
T= temperature=25.0°C
= (273 + 25) = 298 K
R= gas constant = 0.082 l atm /K mol.
Making P subject of formula we have
P= [nRT] /V
Then substitute the values we have
P= [0.5 x 0.082 x 298] / 10
P= 1.2218 atm