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# Thread: 2 discrete probability questions

1. ## 2 discrete probability questions

Rains occur after 2 successful cloud seedings. The probability of a cloud seeding being successful is .3. What is the probability of a rain occurring between the 9th and 11th seeding inclusive?

Inclusive means count the outside numbers, not just the middle, right?
Here is how I go with this: 9 choose 2 .3^2 .7^7 + 10 choose 2 .3^2...11 choose 2...

Where am I wrong? do I need a negative bin. here? I am assuming we stop seeding after a rain so I won't need 1- Prob(1 or O secessful seedings)

A strip club has two groups of strippers. 6 in group A. 4 in group B. The lilkelihood of any stripper being atleast a D-cup is .2. Exactly 3 strippers are atleast a D. What is the probability none are in group B?

1- prob all in A. Prob(all 3 in A)? 6 choose 3 .2^3 .8^3? I am missing something.

The following problems are from real life dilemmas I have recently encountered. Any similarities to past or present exam questions are entirely coincidental.

2. Why are you making up prob. problems in a strip club? Surely there's better things to do.

3. They aren't made up. I need a little help with some probability problems I have encountered in my daily life. Now, if you will excuse me, there are some Silver Iodide loaded nipple tassells that I need to get back to. They're for breastfeeding research.

4. ## reward

\$1.50 payable via paypal for information leading to a solution to the second problem. I think I can get the first intuitively. Looks simple, but with 85 views and no help, I find vindication.

5. You say

lilkelihood of any stripper being atleast a D-cup is .2. Exactly 3 strippers are atleast a D.
The first one to me says P(stripper at least D) = .2, the 2nd is saying P(stripper at least D) = 3/10

Those to me don't reconcile, unless I'm misinterpreting something.

6. Originally Posted by dliving
\$1.50 payable via paypal for information leading to a solution to the second problem. I think I can get the first intuitively. Looks simple, but with 85 views and no help, I find vindication.
In the second problem, it is the probability that 3 ladies in Group A are more alekhine4149's type (6 ch 3)(.2 ^ 3)(.8 ^ 3) times the probability that all of the ladies in Group B are more alekhine4149's type (4 ch 0)(.8 ^ 4). Then you divide by the probability that 3 ladies altogether are not alekhine4149's type (10 ch 3)(.2 ^ 3)(.8 ^ 7).

You can put the \$1.50 in your mouth, walk up to the stage, and make a deposit in her paypal account.

7. It occurs to me there's an easier solution to the strip club problem. The probability of a stripper being at least D doesn't matter because it's already been determined that there are 3 in this sample. So you are just looking for their distribution.

(10 ch 3) = 120 ways the D girls can be distributed in both groups.
(6 ch 3)(4 ch 0) = 20 ways the D girls are all in the A group.

So 20/120 = 1/6 which is the same answer but more easily arrived at.

8. Originally Posted by alekhine4149
The probability of a stripper being at least D doesn't matter because it's already been determined that there are 3 in this sample. So you are just looking for their distribution.
alleluia alleluia alleluia all-e-lu-ia

I'll up the reward money to 2 dollars, so I won't risk choking on coins. I see what you are saying nomoreexams. Look at it like this. We are all at a table and three ladies come up to us to offer a dance that are alekhine's type. We want to know the prob that all are from group a.

Let me make sure I follow. Saying alekhine's type or not makes it more apparent as a bernouli trial. For example, if the question were about 10employees in the two groups and each employee has the prob of .2 of having atleast one accident and exactly three experience more than one accident, and we want the probablity all are in A, the .2 would not matter and it is a hypergeometric distribution.

Food for thought... What if the strippers in group A and B had prob .3 and .1 respectively of being alekhine's type instead of .2 for all. Does this change the solution?

9. Originally Posted by dliving
alleluia alleluia alleluia all-e-lu-ia

I'll up the reward money to 2 dollars, so I won't risk choking on coins. I see what you are saying nomoreexams. Look at it like this. We are all at a table and three ladies come up to us to offer a dance that are alekhine's type. We want to know the prob that all are from group a.

Let me make sure I follow. Saying alekhine's type or not makes it more apparent as a bernouli trial. For example, if the question were about 10employees in the two groups and each employee has the prob of .2 of having atleast one accident and exactly three experience more than one accident, and we want the probablity all are in A, the .2 would not matter and it is a hypergeometric distribution.
I didn't read everything, but you seem to owe alekhine4149 \$1.5 already. Here (I hope) is another justification, which also addresses to Nomoreexams's concern.

The question is to be interpreted as a conditional probability question: the given probability 0.2, 0.8 are in reference to a bigger population (the real sample space), and the 10 persons are just a small sample (subset) of the sample space.

More precisely (but not in all details), let S = A union B, so |S| = 10,
E = the event that exactly 3 from S are >= D,
F = the event that all 3 in E are from A. Then the required probability is

P(F | E) = (6C3*0.2^3*0.8^7)/(10C3*0.2^3*0.8^7) = 1/6.

Food for thought... What if the strippers in group A and B had prob .3 and .1 respectively of being alekhine's type instead of .2 for all. Does this change the solution?
This is similar. The answer is

(6C3*0.3^3*0.7^3*0.9^4) / [6C3*0.3^3*0.7^3*0.9^4 +
(6C2)*0.3^2*0.7^4*(4C1)*0.1*0.9^3 +
(6C1)*0.3*0.7^5*(4C2)*0.1^2*0.9^2 +
0.7^6*(4C3)*0.1^3*0.9]

ctperng

10. Very nice. I'll tip a stripper for you too.