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 
17. The letter is J
The letter appears 3 times
18. Lefthanded students = 4 students
<h3>How to determine the factors</h3>
17. We have that 1/4 of the months of the year start with the same letter
It is important to note that:
January, June and July all start with the same letter
The letter is J
The letter appears 3 times
18. Total number of students = 24
5/ 6 are right handed
Right handed students = 24 × 5/ 6 = 4 × 5 = 20 students
Lefthanded students = 24 - righthanded students
Lefthanded students = 24 - 20
Lefthanded students = 4 students
Thus, we can see that the letter that appear in 3 months of the year is J
Learn more about word problems here:
brainly.com/question/13818690
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Answer:
a= 78
Step-by-step explanation:
To solve subtract:
180-102= 78
Hope this helps!
