Question

3 out of 46 was chosen to do a survey on recommending their attorney. 26 said yes 11 said no 9 said not sure. what would the probability of two out of the chosen 3 would not recommend their attorney and the 3rd one not being sure?

Answer 1

(11/46)(10/45)(8/44)

Answer 2

huh?

Answer 3

Okay, so first you pick two yes people: \[
\left(\frac{11}{46}\right)\left(\frac{10}{45}\right)
\]Then there are 44 left, and you pick a not sure person: \[
\left(\frac{9}{44}\right)
\]So to get this order of yes-yes-unsure, you have a probably of: \[
\left(\frac{11}{46}\right)\left(\frac{10}{45}\right)\left(\frac{9}{44}\right)
\]Then the question becomes... can we do this another way?

Answer 4

For example, we could have done unsure-yes-yes. that would give us:\[
\left(\frac{9}{46}\right)\left(\frac{11}{45}\right)\left(\frac{10}{44}\right)
\]Notice that no matter the order, the top is still \(11\times 10\times 9\), and the bottom is still \(46\times 45\times 44\).

Answer 5

So each ordering will still have the same probability. It's just a mater of finding the number of orderings.

Answer 6

There are three spots. One of them needs to be given 'unsure' and two need to be given yes. This can be counted using combinations. In this case \(\large {3\choose 2}{1\choose 1}\) if you want to do yes then no or \(\large {3\choose 1}{2\choose 2}\). They are both equivalent.

Answer 7

So the final answer is: \[
{3\choose2}{1\choose1}
\left(\frac{11}{46}\right)\left(\frac{10}{45}\right)\left(\frac{9}{44}\right)
\]
Here is wolfram computation:
http://www.wolframalpha.com/input/?i=%7B3%5Cchoose2%7D%7B1%5Cchoose1%7D+%5Cleft%28%5Cfrac%7B11%7D%7B46%7D%5Cright%29%5Cleft%28%5Cfrac%7B10%7D%7B45%7D%5Cright%29%5Cleft%28%5Cfrac%7B9%7D%7B44%7D%5Cright%29

Answer 8

Oh one minor issue... I said 'yes' when I should have been saying 'no' before.

Answer 9

so which one is the answer

Answer 10

The math is still correct though.

Answer 11

9/44?

Answer 12

Do you know what the actual answer is?

Answer 13

no

Answer 14

Wolfram calculates it to \(\frac 3{92}\)

Answer 15

you get the 92 from?

Answer 16

\[
{3\choose2}{1\choose1}
\left(\frac{11}{46}\right)\left(\frac{10}{45}\right)\left(\frac{9}{44}\right)
\]
Calculating this and simplifying...

Answer 17

\[
{3\choose 2} = 3, {1\choose 1} = 1
\]\[
\frac{\color{red}{\cancel{11}}\times 10\times \color{blue}{\cancel{9}}}{46\times \color{blue}{\cancel{45}}5\times \color{red}{\cancel{44}}4}=\frac{\color{red}{\cancel{10}}\color{blue}{\cancel{2}}}{46\times \color{red}{\cancel{5}}\times \color{blue}{\cancel{4}}2} = \frac{1}{92}
\]So \[
{3\choose 2}{1\choose 1} \frac{11\times 10\times 9}{46\times 45\times 44} = \frac {3}{92}
\]

Answer 18

k I need your help with two more...is that cool?

Answer 19

dunno, maybe