Help with solving a probability question (1 Viewer)

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1. The problem statement, all variables and given/known data
From 27 pieces of luggage, an airline handler damages a random sample of 4.

The probability that exactly one of the damaged items is insured is twice the probability that none of the damaged pieces are insured.

Calculate the probability that exactly two of the four damaged pieces are insured.

2. Relevant equations
-

3. The attempt at a solution
From the second line,
"P(one of the damaged items is insured) = 2 x P(none of the damaged pieces are insured)"
Assuming that the probability of one of the 27 luggage being insured is r/27, I'm trying to work out the two probabilities in the second line... so that I can solve for r, and then use that to answer the actual question "Calculate the probability that exactly two of the four damaged pieces are insured"

I am confused about working out the two probabilities in the second line.
For P(one of the damaged items is insured)... is the answer that there are four possibilities where one of the four damaged items is insured,so:
4 (r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24) ?
 

scottdave

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I don't think you can assume the probability of one is x/27 from the info given. I think you will need to use Bayes Theorem (P of A given B) to solve this.
 

FactChecker

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4 (r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24) ?
That looks Ok to me. Set it equal to a similar expression for the other probability. You should be able to cancel several factors from both sides and see if you can solve for r. I hope r turns out to be an integer.
 
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I don't think you can assume the probability of one is x/27 from the info given. I think you will need to use Bayes Theorem (P of A given B) to solve this.
I found a solution online and it takes the r/27 approach so I think have to go with that

That looks Ok to me. Set it equal to a similar expression for the other probability. You should be able to cancel several factors from both sides and see if you can solve for r. I hope r turns out to be an integer.
P(one of the damaged items is insured) = 4 (r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24)
P(none of the damaged items are insured) = (27-r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24)

For the first probability I multiplied by 4 because 4 patterns of having one insured item among 4 items. Similarly am I supposed to be multiplying the second by something?.I am a little confused thinking about this.. should I be accounting for all of the possible non-insured items that could have ended up being in the random sample of 4 damaged items
 

FactChecker

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Use your original equation, P(one of the damaged items is insured) = 2 x P(none of the damaged pieces are insured), and see where that takes you.
 
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Use your original equation, P(one of the damaged items is insured) = 2 x P(none of the damaged pieces are insured), and see where that takes you.
With
P(one of the damaged items is insured) = 4 (r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24)
P(none of the damaged items are insured) = (27-r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24)

I end up with r=9
But according to the solution, I should be getting r=8
 

FactChecker

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I see one problem with 4 (r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24). For the first non-insured selection, there are 27-r non-insured, not 27-r-1.
 
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I see one problem with 4 (r/27)(27-r-1/26)(27-r-2/25)(27-r-3/24). For the first non-insured selection, there are 27-r non-insured, not 27-r-1.
you're right, that's the mistake. I am now getting r=8 as I should

I am also confused about the meaning of independent vs dependent. Does all it mean is whether you are replacing after each selection or not? So if its was independent, would it be:
4 (r/27)(27-r/27)(27-r/27)(27-r/27) ?
 

PeroK

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you're right, that's the mistake. I am now getting r=8 as I should

I am also confused about the meaning of independent vs dependent. Does all it mean is whether you are replacing after each selection or not? So if its was independent, would it be:
4 (r/27)(27-r/27)(27-r/27)(27-r/27) ?
Independence in this case means that you are treating the 27 pieces of luggage as independent, each with the same probability of being insured. The problem and your solution assume this.

If, instead, all 27 pieces of luggage belonged to the same party, then they you could say that either they are all insured or they are all uninsured. In this case, whether one piece of luggage is insured would depend on whether the other pieces were insured. Then the problem wouldn't make much sense.
 

FactChecker

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Once you start drawing, replacement makes it so that the next drawing does not depend on what happened in the first drawing. So you are correct. Without replacement, the following probabilities are modified as you calculated. That makes them dependent results.
 

PeroK

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Once you start drawing, replacement makes it so that the next drawing does not depend on what happened in the first drawing. So you are correct. Without replacement, the following probabilities are modified as you calculated. That makes them dependent results.
The problem says that 4 pieces of luggage are damaged. The handler selects each piece of luggage in turn, processes it, and in the process may or may not damage it. In this problem we know everything relevant about every piece of luggage: 23 are undamaged and 4 are damaged. Replacement or not doesn't enter into the problem.

A different problem might involve quality control of luggage to see how many are damaged. In that type of problem replacement may be a factor.
 

FactChecker

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The damage is merely the selection process, not the insured/uninsured probability. The probabilities are associated with the insured/uninsured within the 27. There is no replacement, so the probability of the second damaged luggage being insured is changed by whether the first was insured.
 

PeroK

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The damage is merely the selection process, not the insured/uninsured probability. The probabilities are associated with the insured/uninsured within the 27. There is no replacement, so the probability of the second damaged luggage being insured is changed by whether the first was insured.
Okay, so the problem should read:

From 27 pieces of lugguge, ##r## of which are insured ...

Rather than, how I interpreted it:

There is a probability of ##p## that a random piece of airline luggage is insured ...
 

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