Probability Question (Simple, but I am confused)

[lovemake1]

Homework Statement

1. The question asks for the value of P(~A and ~B) - What equation to use?

The Attempt at a Solution

1. I am confused though, is P(~A and ~B) same as 1 - P(A and B)
or should I use, Pnot(A or B), and this is equivalent to 1 - P (A or B) ?Extra question...

There is also another concept that I didn't fully grasp.
Lets say there is a lot of 20 pens.
and 60% that the lot has no defective pen.
30% contains 1 defective pen, and only 10% chance that the lot will have 2 defective pens.

So if a person selected 2 pens at random, and noticed they were NOT defective,
how does that change the probability of the rest of 18 pens still in the lot?? is it still 0.6 / 0.3 / 0.1 ?


Appreciate any help in advance.
Thanks

 

[vela]

Homework Statement

1. The question asks for the value of P(~A and ~B) - What equation to use?

The Attempt at a Solution

1. I am confused though, is P(~A and ~B) same as 1 - P(A and B)
or should I use, Pnot(A or B), and this is equivalent to 1 - P (A or B) ?

Try drawing a Venn diagram. It'll become clear the latter is correct.

Extra question...

There is also another concept that I didn't fully grasp.
Let's say there is a lot of 20 pens.
and 60% that the lot has no defective pen.
30% contains 1 defective pen, and only 10% chance that the lot will have 2 defective pens.

So if a person selected 2 pens at random, and noticed they were NOT defective,
how does that change the probability of the rest of 18 pens still in the lot?? is it still 0.6 / 0.3 / 0.1 ?


Appreciate any help in advance.
Thanks

It often helps to look at extreme cases. Suppose you have the same set-up except that a lot contains only 3 pens. If you pull two pens and find they're not defective, you now know that the lot can not contain two defective pens. Clearly, the new information changes the probabilities.

You want to look into Bayes' theorem to see how to calculate the revised probabilities.

 

[lovemake1]

Thanks for your response.
I tried drawing the venn diagram, I guess I tried before but it didn't click for me
P(not A and not B) = 1 - P(A or B)

does not A mean B, and not B mean A?
Because technically venn diagram consists of only 2 group of data. ones in A and the other in B.

Edit: so for example, p(~A or B), does this represent P(B) ?And I'll get back on the 2nd question after reading up on Baye's theorum like you recommended.

 

[vela]

No, not A means everything other than A. B may overlap part of A and part of not A.

Take a look at http://en.wikipedia.org/wiki/Venn_diagram. There are diagrams on the righthand side in the overview. Not A, where A is the left circle, is shown in the absolute complement.

 

[lovemake1]

No, not A means everything other than A. B may overlap part of A and part of not A.

Take a look at http://en.wikipedia.org/wiki/Venn_diagram. There are diagrams on the righthand side in the overview. Not A, where A is the left circle, is shown in the absolute complement.

oh so the overlapping part was where I got confused.
http://en.wikipedia.org/wiki/Symmetric_difference

So this is exactly how the diagram should look. and hence for the reason of 1-P(A or B)

 

[vela]

Not quite. Look here: http://www.math.hawaii.edu/~hile/math100/setsc.htm. Look at the orange and white diagrams. It's exactly this case.

 

[lovemake1]

ok I think I am getting the hang of it.
Can you check just if this is correct. I've drawn the diagram and it looks right to me.

P(~A or B)
The equation for this is 1-P(A)

 

[vela]

I think you're making the same mistake that ~A means B.

~A would be everything outside of A. When you take the union of this region with B, you get everything except the region that's inside only A.

 

[lovemake1]

Yes that's what I thought too. Maybe my formula is not right?

1 - P(A) gives everything else other than P(A)?

Oh my mistake, I remember p(A) covers the whole circle and not just moon shape...
so it is. 1 - P(A) + P(A and B)

 

[vela]

Right.

 

[lovemake1]

I am now tackling the 2nd question. I've read up on Bayem's Rule and starting to get the hang of it.

So we need to find the probability that the 2 pen out from a lot of 20 is not defective right?

This will be related to the equation

P(No defective in the lot|2/20 not defective) = [P(No defect) * P(2/20 Not defective)] /
[P(No defect)*(2/20 not defective) + P(Defect)*(1 - P(2/20 Not defective)]

am I going on the right path?

 

[Ray Vickson]

Thanks for your response.
I tried drawing the venn diagram, I guess I tried before but it didn't click for me
P(not A and not B) = 1 - P(A or B)

does not A mean B, and not B mean A?
Because technically venn diagram consists of only 2 group of data. ones in A and the other in B.

Edit: so for example, p(~A or B), does this represent P(B) ?And I'll get back on the 2nd question after reading up on Baye's theorum like you recommended.

NO: ~A does not mean B and ~B does not mean A. You cannot assume that something is either in A or in B. For example, we could have A = {person is under 30 years old} and B = {person makes more than $50,000 per year}. There are certainly people older than 30 and who make less than $50,000.

So, how do we get P(~A or B)? It is a bit easier to recognize that ~A or B = ~(A and ~B), so P(~A or B) = 1-P(A and ~B), and P(A and ~B) = P(A) - P(A and B). Therefore
[tex]P[(\sim A) \cup B] = 1 - P(A) + P(A \cap B). [/tex]

RGV

 

[vela]

I am now tackling the 2nd question. I've read up on Bayem's Rule and starting to get the hang of it.

So we need to find the probability that the 2 pen out from a lot of 20 is not defective right?

This will be related to the equation

P(No defective in the lot|2/20 not defective) = [P(No defect) * P(2/20 Not defective)] /
[P(No defect)*(2/20 not defective) + P(Defect)*(1 - P(2/20 Not defective)]

am I going on the right path?

The denominator is incorrect. Bayes' rule says
$$P(A\vert B) = \frac{P(B\vert A)P(A)}{P(B)}.$$ So let A = "no defects in lot" and B="two pens were not defective". Then to calculate P(no defects in lot | two pens were not defective), you'll need
\begin{align*}
P(B|A) &= P(\text{two pens were not defective }|{\text{ no defects in lot}}) \\
P(A) &= P(\text{no defects in lot}) \\
P(B) &= P(\text{two pens were not defective}).
\end{align*} What are these three probabilities equal to?

 

[Ray Vickson]

Homework Statement

1. The question asks for the value of P(~A and ~B) - What equation to use?

The Attempt at a Solution

1. I am confused though, is P(~A and ~B) same as 1 - P(A and B)
or should I use, Pnot(A or B), and this is equivalent to 1 - P (A or B) ?


Extra question...

There is also another concept that I didn't fully grasp.
Lets say there is a lot of 20 pens.
and 60% that the lot has no defective pen.
30% contains 1 defective pen, and only 10% chance that the lot will have 2 defective pens.

So if a person selected 2 pens at random, and noticed they were NOT defective,
how does that change the probability of the rest of 18 pens still in the lot?? is it still 0.6 / 0.3 / 0.1 ?


Appreciate any help in advance.
Thanks

For your extra question: let D0 = {0 pens defective in lot}, D1 = {1 defective in lot} and D2 = {2 defective in lot}, and G = {both selected pens are good}. We are told P(D0) = 0.6, P(D1) = 0.3 and P(D2) = 0.1.

You want P(D0|G), P(D1|G) and P(D2|G). To compute P(D2|G), use Bayes:
[tex] P(D2|G) = \frac{P(D2 \text{ and } G)}{P(G)} = \frac{P(G|D2) P(D2)}{P(G)}, [/tex]
so you need P(G|D2) and P(G). Can you compute P(G|D2)? Can you see how to compute P(G)?

RGV

 

[lovemake1]

The denominator is incorrect. Bayes' rule says
$$P(A\vert B) = \frac{P(B\vert A)P(A)}{P(B)}.$$ So let A = "no defects in lot" and B="two pens were not defective". Then to calculate P(no defects in lot | two pens were not defective), you'll need
\begin{align*}
P(B|A) &= P(\text{two pens were not defective }|{\text{ no defects in lot}}) \\
P(A) &= P(\text{no defects in lot}) \\
P(B) &= P(\text{two pens were not defective}).
\end{align*}
What are these three probabilities equal to?

P(A: No defects in lot) = 0.6 as given
Also would P(B|A: Two pents not defective given no defects in lot) be 1 ?
If we are given that there are no defects in the lot, the probability of two pen being not defective is 100% = 1 I am confused how to calculate P(B: Two pens were not defective)
struggled on this part for long time =(

 

[Ray Vickson]

P(A: No defects in lot) = 0.6 as given
Also would P(B|A: Two pents not defective given no defects in lot) be 1 ?
If we are given that there are no defects in the lot, the probability of two pen being not defective is 100% = 1


I am confused how to calculate P(B: Two pens were not defective)
struggled on this part for long time =(

Of course P(both good|all 20 good) = 1. So, how would you compute P(both good|19 good, 1 bad)? How would you compute P(both good|18 good, 2 bad)? Just do all those computations and then put the results together---but first, you need to do those computations!

RGV

 

[lovemake1]

P(both good|20good = (20/20)(19/19) = 1
P(both good|19 good, 1 bad) = (19/20)(18/19) = 0.9
P(both good|18 good, 2 bad) = (18/20)(17/20) = 0.765Since P(Both good) is union of all 3 cases, we multiply them correct? P(Both good) = 1 x 0.9 x 0.765 = 0.6885

 

[Ray Vickson]

P(both good|20good = (20/20)(19/19) = 1
P(both good|19 good, 1 bad) = (19/20)(18/19) = 0.9
P(both good|18 good, 2 bad) = (18/20)(17/20) = 0.765


Since P(Both good) is union of all 3 cases, we multiply them correct? P(Both good) = 1 x 0.9 x 0.765 = 0.6885

No, why would you think that? Look up Bayes Theorem in your textbook or course notes or on Google.

RGV

 

[Macrowave]

Hi, I'm new here... Can anyone tell me how he got

P(both good|20good = (20/20)(19/19) = 1
P(both good|19 good, 1 bad) = (19/20)(18/19) = 0.9

I'm lost how he got those numbers. I tried using Bayes theorem but that led me in a circle... When I tried Using conditional probability I'm confused at how to get P(Both good AND 19 good/1 bad). Any hints or help..?

 

[vela]

P(both good|20good = (20/20)(19/19) = 1
P(both good|19 good, 1 bad) = (19/20)(18/19) = 0.9
P(both good|18 good, 2 bad) = (18/20)(17/20) = 0.765

Recheck the last one.

Since P(Both good) is union of all 3 cases, we multiply them correct? P(Both good) = 1 x 0.9 x 0.765 = 0.6885

No, you need to use

P(both good) = P(both good | 0 bad) P(0 bad) + P(both good | 1 bad) P(1 bad) + P(both good | 2 bad) P(2 bad)

 

[lovemake1]

Recheck the last one.No, you need to use

P(both good) = P(both good | 0 bad) P(0 bad) + P(both good | 1 bad) P(1 bad) + P(both good | 2 bad) P(2 bad)

oop nevermind, I think I am good now !
Thanks for all the help !