Probability Questions?!

[JamieS]

Missed a bunch of class, missed the part I need to know to do this!!!!

1 - "A class is given a list of 20 study problems, from which 10 will be part of an upcoming exam. If a given student knows how to solve 15 of the problems, find the probability that the student will be able to answer all of the 10 questions correctly"

2 - Four letters and envelopes are addressed to 4 different people. If the letters are randomly inserted into the envelopes, what's the probability that at least one will be correct?"


And we have to use combinatorics (sp?) to figure it out.

 

[soroban]

Hello, JamieS!

1) A class is given a list of 20 study problems, from which 10 will be part of an upcoming exam.
If a given student knows how to solve 15 of the problems,
find the probability that the student will be able to answer all of the 10 questions correctly.

Ten of the 20 problems will be selected for the exam.
There are: \(\displaystyle \,\begin{pmatrix}20 \\ 10\end{pmatrix}\:=\:184,756\) possible exams.

If the ten problems are from the student's list pf 15 problems,
there are: \(\displaystyle \,\begin{pmatrix}15 \\ 10\end{pmatrix}\:=\:3003\) ways.

The probability is: \(\displaystyle \,\frac{3003}{184,756} \:=\:\frac{21}{1292}\:\approx\:1.6\%\)

2) Four letters and envelopes are addressed to 4 different people.
If the letters are randomly inserted into the envelopes,
what's the probability that at least one will be correct?

This is not a simple problem.

There are \(\displaystyle 4!\,=\,24\) possible ways to fill the envelopes.

Now we must count the number of ways that:
\(\displaystyle \;\;\)exactly one letter is in its correct envelope,
\(\displaystyle \;\;\) exactly two letters are in their correct envelopes,
\(\displaystyle \;\;\)and all four letters are in their correct envelopes. \(\displaystyle \;\)*

Or we can count the number of ways that no letters are in their correct envelopes
\(\displaystyle \;\;\)and subtract from 24.

Both methods are quite intricate and tricky.

The answer is: \(\displaystyle \,\frac{5}{8}\) . . . I'll let you work it out.

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* \(\displaystyle \;\) Note that we cannot have exactly 3 letters in their correct envelopes.
(If we put 3 letters in their correct envelopes, how can we mess up the fourth one?)

 

[JamieS]

Okay, for the first one I'd been getting it right, but the answer in the back of the book said 21/something and i was typing in 12 instead of 21 so that's why I got so confused lol...

For the second one...that's the problem though, I can't figure out how many ways none of them can be in the right envelopes...I keep getting 24...which makes 0 in the numerator which is not right...

 

[pka]

Soroban gave you the correct answer.
He also said that is was very complicated!

In the above graphic, I show the process.
I think that you will agree that he is right: it is complicated!
The derangement of 4 things is 9: In 9 of 24 ways no letter is correct.
So in 15 ways at least one is correct!

 

[soroban]

Hello, JamieS!

For the second one, I can't figure out how many ways none of them can be in the right envelopes.

You probably included some incorrect ones . . .

Let ABCD be the proper order for the four letters.
Then there are only 9 ways: BADC, BCDA, BDAC, CADB, CDAB, CDBA, DABC, DCAB, DCBA

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This is a classic problem, often called "The Mis-addressed Envelopes".

The arrangements where no item is in its proper position
\(\displaystyle \;\;\) is called a "disarrangement" or "complete permutation".

For \(\displaystyle n\,=\,5\), the number is 44. \(\displaystyle \;\)For \(\displaystyle n\,=\,6\), the number is 265.
\(\displaystyle \;\;\)You can spent the next millenium looking for the general formula.

As \(\displaystyle n\) gets larger, the probability of a disarrangment approaches \(\displaystyle \frac{1}{e}\,=\,0.367879441...\)


Edit: \(\displaystyle \;\)Ha! . . . pka beat me to it ... again!
.