What is the Probability of Selecting 9 CDs with Saved Data from a Box of 15 CDs?

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Discussion Overview

The discussion revolves around calculating the probability of selecting exactly 9 CDs with saved data from a box of 15 CDs, where 10 contain data and 5 do not. Participants explore the application of probability concepts, including binomial distribution and combinatorial counting, in the context of this problem.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Exploratory

Main Points Raised

  • Some participants express confusion about how to approach the problem, considering conditional probability and the relevance of independence in probability calculations.
  • One participant suggests that the situation can be modeled using a binomial distribution, given that CDs are not replaced after selection.
  • There is a discussion about calculating the number of ways to choose CDs, with some participants proposing the use of combinations (e.g., 15C12) to determine the total selections.
  • Participants debate the correct number of CDs to select from each category, with some asserting the need to choose 9 from the 10 data CDs and others suggesting different combinations involving the blank CDs.
  • Confusion arises regarding the correct interpretation of the problem, with some participants questioning the assumptions made in earlier posts.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem. There are multiple competing views regarding the calculations and interpretations of the selection process, leading to ongoing confusion and debate.

Contextual Notes

Limitations include unclear assumptions about the selection process and the dependence on definitions of probability terms. The discussion reflects uncertainty in the application of combinatorial methods and binomial distribution concepts.

NewtonianAlch
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There are 15 CDs in a box. On 10 of the CDs there are saved data files,
and the other CDs have no data files saved on them.

i) Suppose that 12 CDs are randomly selected. Determine the probability that exactly 9 of these CDs selected have saved data files.

I'm not sure how to do this. First I thought some kind of conditional probability? But I'm confused about that.

P(A|B) = P(A intersection B)/P(B)?

Is that simply [P(A)*P(B)]/P(B) ? Wouldn't the P(B)'s cancel out?

If not, then if 12 are selected. The probability of any of them having data is (10/15)*(12/15) = 8/15

Now,

(9/12)*(8/15) = 40% [probability of 9 of those having any data from the 12 selected]
 
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NewtonianAlch said:
P(A|B) = P(A intersection B)/P(B)?

Is that simply [P(A)*P(B)]/P(B) ? Wouldn't the P(B)'s cancel out?
No, P(A intersection B) is the probability of A and B occurring together. It only reduces to P(A)*P(B) if they're independent. In another extreme, A might imply B, in which case it reduces to P(A).
How many equally likely ways are there of picking 12 of the 15? In how many of these do you get exactly 9 with data?
 
As long as the CDs aren't being placed back in the box after each selection, it should be a binomial distribution.
 
haruspex said:
No, P(A intersection B) is the probability of A and B occurring together. It only reduces to P(A)*P(B) if they're independent. In another extreme, A might imply B, in which case it reduces to P(A).
How many equally likely ways are there of picking 12 of the 15? In how many of these do you get exactly 9 with data?


Hmm...15C12 = 455?

Then 9/455?
 
NewtonianAlch said:
Hmm...15C12 = 455?

Then 9/455?

455 is right but 9 is wrong. Need to pick 9 of the ten and one of the 5.
 
I don't quite follow, do you mean 15C12 * (9/10)*(1/5) ?
 
NewtonianAlch said:
I don't quite follow, do you mean 15C12 * (9/10)*(1/5) ?
No. Want
number of ways of choosing 9 from the 10 and 1 from the 5
Since these are independent, that's
(number of ways of choosing 9 from the 10) * (number of ways of choosing 1 from the 5)
right?
 
haruspex said:
455 is right but 9 is wrong. Need to pick 9 of the ten and one of the 5.
9 is right, but "one" is wrong. You need to pick 3 of the 5 blank CDs.

(edit: btw, the problem is somewhat easier to grasp, if you replace it with drawing 3 CDs, and ask about the probability of exactly one having data)
 
Last edited:
Now I'm super confused.

What exactly is happening here? Is this some kind of binomial distribution like someone posted earlier.
 
  • #10
Norwegian said:
9 is right, but "one" is wrong. You need to pick 3 of the 5 blank CDs.
Sorry - got confused between the number of CDs with data and the number to be chosen.
So it's:
number of ways of choosing 9 from the 10 and 3 from the 5
Since these are independent, that's
(number of ways of choosing 9 from the 10) * (number of ways of choosing 3 from the 5)
 

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