# Help with stats concepts. Help. Statistics overwhelm me.

• Jun 3rd 2006, 05:56 PM
bombo31
Help with stats concepts. Help. Statistics overwhelm me.
1. A corporation has four investment possibilities: A, B, and C with respective gains of $10,$20, and $50 million and investment D with a loss of$30 million. If one investment will be made and the probabilities of choosing A, B, C, or D are .1, .4, .2, and .3, respectively, find the expected gain for the corporation.

2.. A marketing research survey shows that approximately 80% of the car owners surveyed indicated that their next car purchase will be either a compact or economy car. Assume the 80% figure is correct and five prospective buyers are interviewed. a. Find the probability that all five indicate that their next car purchase will be either a compact or economy car. b. Find the probability that at most one indicates that her next purchase will be either a compact or economy car.
• Jun 3rd 2006, 06:31 PM
ThePerfectHacker
Quote:

Originally Posted by bombo31
1. A corporation has four investment possibilities: A, B, and C with respective gains of $10,$20, and $50 million and investment D with a loss of$30 million. If one investment will be made and the probabilities of choosing A, B, C, or D are .1, .4, .2, and .3, respectively, find the expected gain for the corporation.

My intution telle me you need to find,
$(.1)10+(.4)20+(.2)50-.3(30)=1+8+10-9=10\mbox{ million}$
Cuz since 10% probability means you are expected to get 10% of A which means 10% of 10 million, similarly the other 3. Then you need to add them up to see what they give, while remember to subtract D
• Jun 3rd 2006, 06:38 PM
ThePerfectHacker
Quote:

Originally Posted by bambi31
2.. A marketing research survey shows that approximately 80% of the car owners surveyed indicated that their next car purchase will be either a compact or economy car. Assume the 80% figure is correct and five prospective buyers are interviewed. a. Find the probability that all five indicate that their next car purchase will be either a compact or economy car.

The probability of the first person saying yes is .8
The probability of two people is $.8\times .8$
The probability of three $.8\times .8\times.8$
The that all five is $(.8)^5\approx .33$

Quote:

Originally Posted by bambi31
b. Find the probability that at most one indicates that her next purchase will be either a compact or economy car.

Find the probability that NONE will says yes. That is,
$(.2)^5$ because given that probability is .8 for saying yes then the probability of saying no is 1-.8=.2 since there are five people you rasie to the fifth az in the first question. But you need at most 1 which is the same as saying what is the probability of this (no one saying yes) will not happen which is,
$1-(.2)^5=.99968$
• Jun 3rd 2006, 07:00 PM
Quick
Question 2
Quote:

Find the probability that NONE will says yes. That is,
(.2)^5 because given that probability is .8 for saying yes then the probability of saying no is 1-.8=.2 since there are five people you rasie to the fifth az in the first question. But you need at most 1 which is the same as saying what is the probability of this (no one saying yes) will not happen which is,
1-(.2)^5=.99968
The question was not what are the chances that at least one will say yes but AT MOST one will say an yes. Heres what I get
$\frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}=\frac{5}{3125} = 0.0016 = 0.16$%
• Jun 4th 2006, 01:10 AM
CaptainBlack
Quote:

Originally Posted by Quick
The question was not what are the chances that at least one will say yes but AT MOST one will say an yes. Heres what I get
$\frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}=\frac{5}{3125} = 0.0016 = 0.16$%

You want:

$P((n_y=0) \lor (n_y=1))=P(n_y=0)+P(n_y=1)$

Now:

$P(n_y=0)=0.2^5\approx 0.00032$,

and:

$P(n_y=1)={5 \choose 1}\ 0.2^4\ 0.8=5\ 0.2^4\ 0.8\approx 0.0064$

Therefore:

$P((n_y=0) \lor (n_y=1))\approx 0.00672$,

or $\approx 0.67 \%$

RonL
• Jun 4th 2006, 03:23 AM
Quick
Why
Well, it appears I forgot to multiply by five. :o
• Jun 4th 2006, 05:10 AM
ThePerfectHacker
Quote:

Originally Posted by Quick
Well, it appears I forgot to multiply by five. :o

Because you only considered one case, namely,
YNNNN

But there are 4 more:
NYNNN
NNYNN
NNNYN
NNNNY

This is why you need to multiply by five.
----
Note, in may probability exacmple you need to use binomial coeificcients because of this reason.
• Jun 4th 2006, 03:37 PM
bombo31
Thank you
Sorry about the double posts. I did not realize, and thanks for your help.