# Probability distribution table help w/ Tree diag .Two values coming strangely.

• Mar 30th 2011, 01:15 AM
SolCon
Probability distribution table help w/ Tree diag .Two values coming strangely.
Hi again.

I'm having a problem with a tree diagram and the subsequent prob.dist.table. Here's the question:

Every day Eduardo tries to phone his friend. Every time he phones there is a 50% chance that his friend will answer. If his friend answers, Eduardo does not phone again on that day. If his friend does not answer, Eduardo tries again in a few minutes’ time. If his friend has not answered after 4 attempts, Eduardo does not try again on that day.

(i) Draw a tree diagram to illustrate this situation.

The figure:

http://usera.ImageCave.com/biosyn/ss.PNG

This is correct.

(ii) Let X be the number of unanswered phone calls made by Eduardo on a day. Copy and complete the table showing the probability distribution of X.

This is what I'm seeing seeing:

x 0 1 2 3 4
P(X=x) 1/2 1/2 1/4 1/8 1/16

But, this seems to be wrong. Once again, I have no clue as to why this is so.

Any help? (Speechless)
• Mar 30th 2011, 04:08 AM
Plato

Suppose that you flip a coin until a head appears.
However, there are no more than four flips allowed.
Let X equal the number of tails to appear.
Then is it not the case that $P(X=1)=\frac{1}{2^2}~?$
That is exactly one tail. That means the second flip yields a head.
What is the probability of $TH~?$
Now apply this model to your problem.
• Mar 30th 2011, 07:42 AM
SolCon

Still don't get it though. In your example, we keep flipping a coin until we get a head, after which we would stop flipping the coin. But only four flips are allowed. X is the number of tails. So, for the first flip, we can either get H or T. Both have equal probability. For the first flip, getting H would be the same as getting X=0 ('0' or 'no' Tail). But if we get T on the first flip, would that not mean X=1 ('1' Tail)? And seeing as their probabilities are the same (0.5), why would we multiply it by 0.5 a second time? That would mean a Tail and another Tail.

???
• Mar 30th 2011, 08:13 AM
Plato
Quote:

Originally Posted by SolCon
Still don't get it though. In your example, we keep flipping a coin until we get a head, after which we would stop flipping the coin. But only four flips are allowed. X is the number of tails. So, for the first flip, we can either get H or T. Both have equal probability. For the first flip, getting H would be the same as getting X=0 ('0' or 'no' Tail). But if we get T on the first flip, would that not mean X=1 ('1' Tail)? And seeing as their probabilities are the same (0.5), why would we multiply it by 0.5 a second time? That would mean a Tail and another Tail.

In a given day he makes at most four calls.
The only way to get exactly one unanswered call is to fail the fist time and the succeed the second time.
That gives $(0.5)(0.5)$
• Mar 30th 2011, 08:15 AM
Soroban
Hello, SolCon!

Quote:

Every day Eduardo tries to phone his friend.
Every time he phones there is a 50% chance that his friend will answer.
If his friend answers, Eduardo does not phone again on that day.
If his friend does not answer, Eduardo tries again in a few minutes’ time.
If his friend has not answered after 4 attempts, Eduardo does not try again on that day.

(i) Draw a tree diagram to illustrate this situation.

The figure:

http://usera.ImageCave.com/biosyn/ss.PNG

This is correct.

(ii) Let $\,x$ be the number of unanswered phone calls made by Eduardo on a day.
. . Copy and complete the table showing the probability distribution of $\,x.$

This is what I'm seeing:

. . $\begin{array}{c|ccccc}
x & 0 &1 & 2 & 3 & 4 \\ \hline \\[-4mm]
P(x) & \frac{1}{2} & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16}
\end{array}$
. This is incorrect.

But this seems to be wrong.
Once again, I have no clue as to why this is so.

Any help?

You must use the probabilities out to the $A$'s.

We can talk our way through the branches.

. . $P(0) \:=\:\frac{1}{2}$

. . $P(U \wedge A) \:=\:(\frac{1}{2})^2 \,=\,\frac{1}{4}$

. . $P(U \wedge U \wedge A) \:=\:(\frac{1}{2})^3 \:=\:\frac{1}{8}$

. . $P(U\wedge U\wedge U\wedge A) \:=\:(\frac{1}{2})^4 \:=\:\frac{1}{16}$

For 4 unanswered calls, Eduardo stops calling.
. . $P(U\wedge U\wedge U\wedge U) \:=\:(\frac{1}{2})^4 \:=\:\frac{1}{16}$

Your table should look like this:

. . $\begin{array}{c|ccccc}
x & 0 & 1 & 2 & 3 & 4 \\ \hline \\[-4mm]
P(x) & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{16}
\end{array}$

• Mar 30th 2011, 11:00 PM
SolCon
Ahh, thank you both for the helpful explanations. :)

Didn't realize that for the first unanswered call, the second time had to be discontinued, otherwise he'd be trying again.

I've got it now.

Thanks again. :)