Probability Question (Simple, but I am confused)

P[(\sim A) \cap (\sim B)].The rest of your reasoning is fine - but this is not the end of the solution. You did not really ask a question, so I will not provide more details. But, you want to use the information about the initial two pens to revise the probabilities for the remaining pens.
  • #1
lovemake1
149
1

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
 
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  • #2
lovemake1 said:

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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
vela said:
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)
 
  • #7
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)
 
  • #8
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.
 
  • #9
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)
 
  • #10
Right.
 
  • #11
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?
 
  • #12
lovemake1 said:
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
 
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  • #13
lovemake1 said:
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?
 
  • #14
lovemake1 said:

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
 
  • #15
vela said:
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 =(
 
Last edited:
  • #16
lovemake1 said:
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
 
  • #17
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
 
Last edited:
  • #18
lovemake1 said:
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
 
  • #19
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..?
 
  • #20
lovemake1 said:
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)
 
  • #21
vela said:
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 !
 
Last edited:

1. What is probability?

Probability is the likelihood of an event occurring. It is represented as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

2. How do you calculate probability?

To calculate probability, you divide the number of favorable outcomes by the total number of possible outcomes. This gives you a decimal or fraction that can be converted to a percentage.

3. What is the difference between theoretical and experimental probability?

Theoretical probability is based on mathematical calculations and predictions, while experimental probability is based on actual data and observations from real-life experiments or events.

4. How does probability relate to statistics?

Probability is a fundamental concept in statistics. It is used to analyze and interpret data, make predictions, and draw conclusions about the likelihood of events occurring.

5. Can probability be used to predict the outcome of a single event?

No, probability cannot predict the outcome of a single event. It only provides a measure of likelihood based on past data or theoretical expectations. The outcome of a single event is ultimately determined by chance and cannot be predicted with certainty.

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