Probability question...STAT?

Question

If the probability of successfully introducing a new product (A for example) to the maket is 18% and the probability of successfull introducing two new products (A and B) is 5%, then:
1. What is the probabil. that anu new prod (B for ex.) will fail?
2. probability that @ least 1 of the 2 A or B will succeed?
3. What is the prob that A nor B will succeed?
4. probabil. that A will succeed if you know B did?
5. that A or B was success but not both?
6. Given that @ least of A or B is successful, probability that A was successful?
Here's another prob.
Assume that IQ are normally distributed. with a mean of 100 and a standar dev. of 15.
1. probability that randomly selected adult has an IQ less than 130
2. IQ greater than 131.5
3. IQ btw 90 and 110
4. IQ btw 110 and 120
5. Find P10, which is the IQ score separating the bottom 10? from the top 10%
6. Find P60, which is IQ sepaating the bottom 60% frm top %40
7. Find IQ score separating the top 35%
8. Find IQ separ. top 85%

Answer 1

Dear Billie,

I'll start with a suggestion. In the future you would do well to type questions exactly as they are written, without leaving out any words. Leaving out words in probability questions can significantly change their meaning, or make them confusing so that people have to guess at what was meant. People will be less likely to answer questions when the meaning is not clear.

Problem I.
Usually the easiest way to think about this type of problem is to make a two-by-two table labeled A and ~A on the columns and labeled B and ~B on the rows. Thus the squares of the table will correspond to the events A & B, ~A & B, A & ~B, and ~A & ~B, and you should then fill them in with the probabilities of those events. To the right of the rows, also put the probabilities for B and ~B, respectively, and do the same for A and ~A beneath the columns. The numbers in each row and column of the table should sum to the respective numbers to the right and below the table, as just described. Also all four numbers in the table should sum to 1, which in turn makes the sum of the row totals and column totals both equal to 1. (The symbol "~" means "not," and the symbol "&" means "and," in case they are unfamiliar to you.) The completed two-by-two table (without the labels) looks like

0.05 0.13
0.13 0.69

and with the row and column totals looks like

0.05 0.13 0.18
0.13 0.69 0.82
0.18 0.82

to which you can check to see both the rows and the columns sum to 1 (i.e., 0.18 + 0.82 = 1), and this can be put on the table if desired, to look like

0.05 0.13 0.18
0.13 0.69 0.82
0.18 0.82 1.00 .

Once the table is properly constructed, the questions should be easy to answer by referring the the table.

1. P(~B) = 1 - P(B) = 1 - 0.18 = 0.82 .
Notice that this is just the sum of the second row of the table. You have the same result for P(~A) in the second column.

2. P(A OR B) = 1 - P(~A & ~B) = 1 - 0.69 = 0.31 .
Another way to get the same result is to write
P(A OR B) = P(A) + P(B) - P(A & B) = 0.18 + 0.18 - 0.05 = 0.31 .

3. I'm guessing you meant neither A nor B succeed, in which case
P(~A & ~B) = 0.69 .
If you really meant A succeeds but not B, then you would have
P(A & ~B) = 0.13 .

4. P(A | B) = P(A & B) / P(B) = 0.05 / 0.18 = 5/18,
where the vertical bar "|" means "given."

5. P(A OR B) was found in question 2. We can use that here to find
P((A OR B) & ~(A & B)) = P(A OR B) - P(A & B) = 0.31 - 0.05 = 0.26 .
We can also think of this as
P((A & ~B) OR (~A & B)) = P(A & ~B) + P(~A & B) = 0.13 + 0.13 = 0.26 .

6. P(A | (A OR B)) = P(A) / P(A OR B) = 0.18 / 0.31 = 18/31.


Problem II.
These are basic questions about your familiarity with the normal distribution. For the most part you need to either look up numbers in tables in a book or use a program that can compute them for you. If you do not understand how to do this then you MUST read the book or ask the teacher. You don't have to do the computations yourself; there is no heavy math required, so all you need to do is look numbers up in a table and sometimes subtract one number from another. (If you are using tables, they are usually in standardized form, so you might have to rescale your numbers into standardized form, which only needs a subtraction and a division.)

1. This is the total area under the normal density curve as far as two standard deviations above the mean. It will help your intuition to have a diagram to look at, and to draw what the problem describes. Notice that since the mean is 100 and standard deviation is 15, that 130 differs from 100 by two standard deviations.
P(IQ < 130) = 0.97725 (to five decimal places, both here and below).

2. Did you mean to ask for the probability of a person having an IQ greater than 131.5? If so, you said nothing in the question about probability. As prefaced above, it is vital that you ask questions exactly as they are written so that they do not change or lose their meaning.
P(IQ > 131.5) = 0.01786 .

3. What do you mean? See the comment on the previous question.
P(90 < IQ < 110) = 0.49501 .

4. Again, what do you mean here? I'm guessing, but shouldn't have to.
P(110 < IQ < 120) = 0.16128 .

5. Do you mean for P10 to be the IQ score separating the bottom 10% from the top 90%? Again, you need to type questions exactly to avoid confusion.
P10 = 80.77673 .

6. P60 = 103.80021 .

7. P65 = 105.77981 .

8. P15 = 84.45350 .

Answer 2

I suspect my argument is not going to be well received. I say the probability is 50% Let a ∈ℝ, Let b ∈ℝ and randomly select the values for a and b. As already noted, for a ≤ 0, P( a < b²) = 1, this is trivial. Only slightly less trivial is the idea that P(a < 0 ) = 1/2 and thus P( a < b² | a ≤ 0) = 1 and P( a < b² ) ≥ 1/2 Now consider what happens when a > 0 For a > 0, while it is easy to show there is a non zero probability for a finite b, the limit, the probability is zero. a < b² is equivalent to saying 0 < a < b², remember we are only looking at a > 0. If this a finite interval on an infinite line. The probability that a is an element of this interval is zero. P( a < b² | a > 0) = 0 As such we have a total probability P( a < b² ) = P( a < b² | a ≤ 0) * P(a ≤ 0) + P( a < b² | a > 0) * P(a > 0) = 1 * 1/2 + 0 * 1/2 = 1/2 Remember, this is because of the infinite sets. No matter what type of interval you draw on paper or on a computer you will find a finite probability that appears to approach 1. But this is due to the finite random number generators on the computer and if we had this question asked with finite values there would be a a solution greater than 50%. I don't mean to be condescending, but please explain why using the Gaussian to approximate a uniform distribution is a good idea? Aren't infinite numbers fun. Cantor when mad working with them! :)