# # of sophomore girls? Taylor expansion? ...

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• Dec 2nd 2006, 10:49 PM
Ruichan
# of sophomore girls? Taylor expansion? ...
1. In a class, there are 4 freshman boys, 6 freshman girls and 6 sophomore boys.
How many sophomore girls must be present if sex and class are to be independent when a student is selected at random?

I don't really understand this question. No sophomore girls are given in the information?

2. Let X be a Poisson random variable with parameter λ.
Show that P{ X is even } = 1/2 [1 + e^-2λ]. Use Taylor expansion of e^-λ + e^λ.

I set it up as
P{X is even} = e^λ [summation λ^2n / (2n)!]

how do i expand the summation part?

3. If X is a geometric random variable, show analytically that
P{ X = n + k | X > n } = P{X = k}.
Give a verbal argument using the intepretation of a geometric random variable as to why the above is true.

==>I simplified and got P{ X = n + k | X > n } = p(1-p)^k-1
What I don't understand is how do I give a verbal argument?

Thanks for the help in advance. Finals are coming up in 2 weeks. My prof decided to give us tons of extra discussion questions to do. These are the some of the questions I didn't understand. As for the rest, I'm still trying to do them.. please kindly help because I'm afraid that such questions will come up in the finals. If so, I'll be deadmeat coz I have no idea how to go about them.
• Dec 2nd 2006, 11:02 PM
CaptainBlack
Quote:

Originally Posted by Ruichan
1. In a class, there are 4 freshman boys, 6 freshman girls and 6 sophomore boys.
How many sophomore girls must be present if sex and class are to be independent when a student is selected at random?

I don't really understand this question. No sophomore girls are given in the information?

If gender and class are independent then:

p(m,f)=p(m)p(f)

the probability of picking a male freshman is the product of picking a male
and that of picking a freshman.

Suppose there are n sophomore girls, the above becomes:

p(m,f)=4/(16+n)=p(m)p(f)=10/(16+n) 10/(16+n),

so we require:

4=100/(16+n),

or:

16+n=25,

so:

n=9.

(this assumes that a solution is possible you will need to check that
p(g,c)=p(g)p(c) for the other gender g, and class c combinations)

RonL
• Dec 2nd 2006, 11:14 PM
Ruichan
Thanks for the help! I understand your solution but I don't really understand why is it so.
The given info is that a class has 4 freshman boys, 6 freshman girls and 6 sophomore boys.
Therefore even if sophomore girls aren't even mentioned in the question, we can actually find the answer, like you've done. I'm not quite understanding the question because aren't there supposed to be only 16 students in class?
If the answer is 9 sophomore girls, aren't there more than 16 students in class? Is it because I don't see the question containing 'there are n numbers of sophomore girls in class', that's why I'm having problems seeing the way to do it?
:confused: :confused: :confused:
• Dec 2nd 2006, 11:22 PM
CaptainBlack
Quote:

Originally Posted by Ruichan
Thanks for the help! I understand your solution but I don't really understand why is it so.
The given info is that a class has 4 freshman boys, 6 freshman girls and 6 sophomore boys.
Therefore even if sophomore girls aren't even mentioned in the question, we can actually find the answer, like you've done. I'm not quite understanding the question because aren't there supposed to be only 16 students in class?
If the answer is 9 sophomore girls, aren't there more than 16 students in class? Is it because I don't see the question containing 'there are n numbers of sophomore girls in class', that's why I'm having problems seeing the way to do it?
:confused: :confused: :confused:

The existing group has 16 students, the question effectivly asks how many
soph girls must be added to make gender and class independent.

RonL
• Dec 2nd 2006, 11:37 PM
CaptainBlack
Quote:

Originally Posted by Ruichan
2. Let X be a Poisson random variable with parameter λ.
Show that P{ X is even } = 1/2 [1 + e^-2λ]. Use Taylor expansion of e^-λ + e^λ.

I set it up as
P{X is even} = e^λ [summation λ^2n / (2n)!]

how do i expand the summation part?

First for the poisson distribution:

$\displaystyle P(n)=\frac{e^{-\lambda} \lambda^n}{n!}$

Then:

$\displaystyle P(\rm{X\ even})=e^{-\lambda} \sum_{n=0}^{\infty}\frac{ \lambda^{2n}}{(2n)!}$

Now expand the expression you have been given to show that it is equal to
the above. (By the way 1/2 [1 + e^-2λ] is ambiguous it could be either
1/{2 [1 + e^-2λ]} or (1/2) [1 + e^-2λ]).

RonL
• Dec 2nd 2006, 11:41 PM
Ruichan
Quote:

Originally Posted by CaptainBlack
The existing group has 16 students, the question effectivly asks how many
soph girls must be added to make gender and class independent.

RonL

Oh.....now that you phrase it this way, I understand what is the qxn actually asking.
I initially thought that there are already the sophomore girls present and we are asked to pick randomly from the whole class.
Erm...think I'm not making sense......sorry

It's (1/2) [1 + e^-2λ]. Sorry about it.
• Dec 2nd 2006, 11:53 PM
Ruichan
Quote:

Originally Posted by CaptainBlack
First for the poisson distribution:

Then:

$\displaystyle P(\rm{X\ even})=e^{-\lambda} \sum_{n=0}^{\infty}\frac{ \lambda^{2n}}{(2n)!}$

Now expand the expression you have been given to show that it is equal to
the above.

RonL

How do you expand the summation part using the Taylor expansion?
I do know that summation λ^n / n! = e^λ.
I'm very when it comes to expanding with factorials.
• Dec 2nd 2006, 11:58 PM
Ruichan
#3 Explanation
Is it correct if I were to explain that since according to definition,
P( X = n ) = p(1-p)^n-1,
and when P{ X = n + k | X > n } technically gets the same answer as the definition, ie. p(1-p)^k-1,
P{ X = n + k | X > n } = P{ X = k}.
If the number of first n trials fail, starting with the k trials is the same as beginning all over again.

Does this explanation make sense or did I totally got it wrong?
• Dec 3rd 2006, 09:05 PM
CaptainBlack
Quote:

Originally Posted by Ruichan
Oh.....now that you phrase it this way, I understand what is the qxn actually asking.
I initially thought that there are already the sophomore girls present and we are asked to pick randomly from the whole class.
Erm...think I'm not making sense......sorry

It's (1/2) [1 + e^-2λ]. Sorry about it.

OK here goes:

$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2) e^{-\lambda} [e^{\lambda}+e^{-\lambda}]$

Now expanding the exponentials inside the square brackets gives:

$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2)e^{-\lambda} \left[ \sum \frac{\lambda^n}{n!}+\sum \frac{(-\lambda)^n}{n!} \right]$

so:

$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2)e^{-\lambda} \sum \left[ \frac{\lambda^n}{n!}+ \frac{(-\lambda)^n}{n!} \right]$

Now when $\displaystyle n$ is odd the terms in the sum cancell out and when $\displaystyle n$ is even they add so we have:

$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2)e^{-\lambda} \sum 2\, \frac{\lambda^{2n}}{(2n)!} =e^{-\lambda} \sum \frac{\lambda^{2n}}{(2n)!}$,

which is equal to $\displaystyle P(\mbox{X even})$.

RonL
• Dec 3rd 2006, 10:38 PM
Ruichan
Quote:

Originally Posted by CaptainBlack
OK here goes:

$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2) e^{-\lambda} [e^{\lambda}+e^{-\lambda}]$

Now expanding the exponentials inside the square brackets gives:

$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2)e^{-\lambda} \left[ \sum \frac{\lambda^n}{n!}+\sum \frac{(-\lambda)^n}{n!} \right]$

so:

$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2)e^{-\lambda} \sum \left[ \frac{\lambda^n}{n!}+ \frac{(-\lambda)^n}{n!} \right]$

Now when $\displaystyle n$ is odd the terms in the sum cancell out and when $\displaystyle n$ is even they add so we have:

$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2)e^{-\lambda} \sum 2\, \frac{\lambda^{2n}}{(2n)!} =e^{-\lambda} \sum \frac{\lambda^{2n}}{(2n)!}$,

which is equal to $\displaystyle P(\mbox{X even})$.

RonL

So instead of expanding the summation part, I use
$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2) e^{-\lambda} [e^{\lambda}+e^{-\lambda}]$
Man....i am so bad at this. I didn't even realise that I'm not trying to do the expansion on the summation part.
$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2)e^{-\lambda} \sum 2\, \frac{\lambda^{2n}}{(2n)!} =e^{-\lambda} \sum \frac{\lambda^{2n}}{(2n)!}$
Where did you get the 2 from? I know the power 2n is there because only the even powers remain. But the 2? Is it because you get 2x the same answer when you add the even terms together?

Thank you very much.

Oh, if you can, please look at the explanation for question #3. Is it right for me to explain it the way I did or did I go about it wrong? Do you want me to show the steps as to how I got the answer?
Thanks.
• Dec 3rd 2006, 11:11 PM
Ruichan
Airplane Engines
4. Suppose that airplane engines will fail, when in flight, with probability 1-p independently from engine to engine. If an airplane needs a majority of its engines operative to make a successful flight, for what values of p is a 5-engine plane preferable to a 3-engine plane?

I tried using the geometric random variable.
P(X=5) = $\displaystyle p(1-p)^4$
P(X=3) = $\displaystyle p(1-p)^2$

I'm doing something wrong here, am I not? How do i continue from here?
• Dec 3rd 2006, 11:53 PM
CaptainBlack
Quote:

Originally Posted by Ruichan
So instead of expanding the summation part, I use
$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2) e^{-\lambda} [e^{\lambda}+e^{-\lambda}]$
Man....i am so bad at this. I didn't even realise that I'm not trying to do the expansion on the summation part.
$\displaystyle (1/2)[1+e^{-2\lambda}]=(1/2)e^{-\lambda} \sum 2\, \frac{\lambda^{2n}}{(2n)!} =e^{-\lambda} \sum \frac{\lambda^{2n}}{(2n)!}$
Where did you get the 2 from? I know the power 2n is there because only the even powers remain. But the 2? Is it because you get 2x the same answer when you add the even terms together?

The 2 comes from there being two terms $\displaystyle \frac{\lambda^{2n}}{(2n)!}$ one from the series for $\displaystyle e^{\lambda}$, and the other from the series for $\displaystyle e^{-\lambda}$.

RonL
• Dec 4th 2006, 12:31 AM
Ruichan
Thanks. I get it now.
It came from adding twice the same term.
• Dec 4th 2006, 05:08 AM
CaptainBlack
Quote:

Originally Posted by Ruichan
4. Suppose that airplane engines will fail, when in flight, with probability 1-p independently from engine to engine. If an airplane needs a majority of its engines operative to make a successful flight, for what values of p is a 5-engine plane preferable to a 3-engine plane?

I tried using the geometric random variable.
P(X=5) = $\displaystyle p(1-p)^4$
P(X=3) = $\displaystyle p(1-p)^2$

I'm doing something wrong here, am I not? How do i continue from here?

The number of engines that don't fail has a binomial distribution, so the
probability that exactly n engines don't fail for an N engine plane is:

B(n,N,p)=(N!)/((N-n)! n!) p^n (1-p)^(N-n)

(which is the same as the probability that exactly N-n engines do fail)

Now to complete a flight a 3-engine aircraft requires either 2 of 3 to not fail
so the probability of success is:

P(n>=2,3,p)=B(2,3,p)+B(3,3,p)

The 5-engine aircraft requires 3 or more engines to not fail so:

P(n>=3,5,p)=B(3,5,p)+B(4,5,p)+B(5,5,p).

So the question is asking for what values of p is:

P(n>=3,5,p)>P(n>=2,3,p)

RonL
• Dec 5th 2006, 05:07 PM
Ruichan
Expected # of cards
5. Cards from an ordinary deck are turned face up one at a time (w/o replacement). Compute the expected number of cards that need to be turned face up in order to obtain 2 aces.

Captainblack, this was the way you taught me the last time for a question: Draw 2 cards without replacement from a deck of playing cards. What is the expected number of aces?

Let random variable x1 = 1 if an ace turns up on the turn, 0 otherwise.
E(x1) = 1/13

Let random variable x2 = 1 if an ace turns up on the 2nd turn, 0 otherwise.

E(x2) = 1/13

E(x1 + x2) = 2/13

Now, instead of # of ace turning up, it's finding the expected number of cards that needed to be turned to obtain 2 aces. Can I use the same way?