Help me please guys!!

Y

Yoha

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Feb 25, 2012
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I'm studying probabilities and I came across the following problem

The problem is:
Anew digital scale accuracy may be acceptable (A) or unacceptable (U). A certain lab experiment required 2 scales will be independently selected and tested until 2 acceptable scales have been found. Suppose 80% of the new scales have an acceptable degree of accuracy. Let Y denote the number of scales that must be tested.
a. what is the random variable in the above question?
b. what is p(y=2)
c. what is p(3)? note there are 2 different outcomes the result Y=3
d. In order to have Y=5, what must be true of the 5th scale selected?
e. list the four outcomes for which Y=5 and then calculate p(5)?
My answers.
This question seems -ve binomial distribution.
so,
a- the random variable is Y
b- using this formula b(x; n, P) = nCx * Px * (1 - P)n - x, we can find p(2)..etc
so we have
n=2
p=.8
q=.2
p(2)=.64
p(3)=.204
but I couldn't figure out how I can find the outcomes!! what does it mean to have two different outcomes that result Y=3

if we find the sample space out come it will be like this {AA,UA,AU,UU}

Can anybody help!!!

Many thanks
 
Yoha said:
I'm studying probabilities and I came across the following problem

The problem is:
Anew digital scale accuracy may be acceptable (A) or unacceptable (U). A certain lab experiment required 2 scales will be independently selected and tested until 2 acceptable scales have been found. Suppose 80% of the new scales have an acceptable degree of accuracy. Let Y denote the number of scales that must be tested.
a. what is the random variable in the above question?
b. what is p(y=2)
c. what is p(3)? note there are 2 different outcomes the result Y=3
d. In order to have Y=5, what must be true of the 5th scale selected?
e. list the four outcomes for which Y=5 and then calculate p(5)?
My answers.
This question seems -ve binomial distribution.
so,
a- the random variable is Y
Click to expand...

Okay, good.

b- using this formula b(x; n, P) = nCx * Px * (1 - P)n - x, we can find p(2)..etc
Click to expand...
so we have
n=2
p=.8
q=.2
p(2)=.64
Click to expand...

Equivalently, to get two good scales in only two trials, both must be good: (.8)(.8)= .64 as you say.

p(3)=.204
but I couldn't figure out how I can find the outcomes!! what does it mean to have two different outcomes that result Y=3

if we find the sample space out come it will be like this {AA,UA,AU,UU}
Click to expand...

That would be the case if you were asking about Y= 2. To get two acceptable scales in three trials you must have UAA or AUA. Those are the "two different outcomes". Notice that you would NOT count "AAU" because if you got two acceptable scales in the first two trys, you would NOT go on to a third trial so that would not count as "Y= 3".
To find P(Y= 3) find the probability of those two (they will be the same) and add: (.2)(.8)(.8)+ (.8)(.2)(.8)= .128+ .128= .256, NOT ".204". That is NOT the same as \(\displaystyle \begin{pmatrix}3 \\ 2\end{pmatrix}(.2)^1(.8)^2= 3(.2)(.8)^2= .384\) because that would include "AAU".

Can anybody help!!!

Many thanks
Click to expand...
 
Hi HallsofIvy,

Thank you very much for your clarification and your constructive thoughts
so, you mean p(3) would be =.256
so, with regards to question d and e.
I could answer e but I'm not sure if I'm right or wrong.
The four outcomes would be like this AUAUA,UAUAU,,UUAUA,UAUUA
and the p(5)= .02048+.00512+.00512+.00512=.03584
Is that right?
what about question( d )... what does that question mean? Mathematically speaking?!!

Thanks very much for your explanation!
 
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