No anon!. Geno answer the other questions correctly but this question is different than them. You have to use Binomial distribution cumulative probability to answer this one.
It look like this
\(\Pr(X \le k) = \sum_{i=0}^k {n\choose i}p^i(1-p)^{n-i} \)
The sideways M like thing is a sumation and it mean you have to add all of these.
When you do these separately then it look like this
\({3\choose 0}0.05^0(1-.05)^{3} = 0.857375 \\ {3\choose 1}0.05^1(1-.05)^{2}=0.135375 \\ {3\choose 2}0.05^2(1-.05)^{1}=0.007125 \\ {3\choose 3}0.05^3(1-.05)^{0}=0.000125 \\\)
This is for
none defective
only 1 defective
only 2 defective
only 3 defective
Then you add up the probabilities depending on the question.
For the question of at least one defective you add the probabilities of 1 and 2 and 3. Because it is at least 1 but could be 2 or 3.
Add up
0.135375 + 0.007125 + 0.000125 = 0.142625
So 0.142625 is the correct probability not 0.0975
Notice if you use this formula for the first two questions it give the same answer as Geno’s. When it is all or nothing you can do it the way Geno did it. It is easier to do.
I am sure this is right but a mod should check it.
