5. A production process is known to produce a particular item in such a way that 5 percent of these are defective. If three items are randomly selected as they come off the production line,
(a). What is the probability that all three items will be defective?
(b). What is the probability that none of the items will be defective?
Thank you guys ! or who ever will answer to me ! :)
+1316 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.
+23252 Since the probability for each one is defective is 0.05,
the probability for all three to be defective: 0.05 x 0.05 x 0.05 = 0.000 125
Since the probability that each one is not defective is 0.95,
the probability for all three to not be defective: 0.95 x 0.95 x 0.95 = 0.857 375
I get it !
Thank you,
So the probability that at least one item will be defective would be :
1 - 0.95 x 0.95 = 0.0975 ?
Right ?
Thank you !
+1316 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.
