# Is this right?

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• Sep 12th 2012, 05:44 PM
Past45
Is this right?
The way I solved this was (1/5)^6, however that answer came out to be 6.4e-5. However the answer is suppose to be 0.0003200. Can I get a little help on what I did wrong? Tt
• Sep 12th 2012, 06:36 PM
MarkFL
Re: Is this right?
You found the probability that all six witnesses picked a particular person, not that they all picked the same person. So, let the first witness pick any of the five, then find the probability that the next five witnesses pick that same person.
• Sep 12th 2012, 06:40 PM
SworD
Re: Is this right?
After the first witnesses pick a person, each of the remaining 5 witnesses must pick that person, each with probability $\displaystyle \frac{1}{5}$. So $\displaystyle p = \left (\frac{1}{5}\right )^{5} = 0.00032$. There's no restriction on who the first person is, just the remaining five must pick the same one.

Edit - sorry, I misread the above post. Yeah what Mark said is correct.
• Sep 12th 2012, 06:42 PM
MarkFL
Re: Is this right?
Hmmm, since the OP stated $\displaystyle P(X)=\left(\frac{1}{5} \right)^6$ it appeared to me he did not take that into account...

edit: LOL, I should have waited 10 seconds before posting again! (Giggle)
• Sep 12th 2012, 06:43 PM
SworD
Re: Is this right?
Yea, I'm developing a habbit of reading posts too quick. I just opened the link, saw the selected choice, and "is this right". Sorry for that xD
• Sep 13th 2012, 05:31 PM
Past45
Re: Is this right?
Quote:

Originally Posted by MarkFL2
Hmmm, since the OP stated $\displaystyle P(X)=\left(\frac{1}{5} \right)^6$ it appeared to me he did not take that into account...

edit: LOL, I should have waited 10 seconds before posting again! (Giggle)

Ty, even when I do 1/5^5, it gives me 3.2e-4... Is that a calculator issue(Have TI-84 Silver edition)?
• Sep 13th 2012, 05:34 PM
SworD
Re: Is this right?
3.2e-4 = $\displaystyle 3.2 \cdot 10^{-4} = 0.00032$
• Sep 13th 2012, 07:17 PM
johnsomeone
Re: Is this right?
Another way to see this is that the probability of all 6 witnesses picking a particular man is $\displaystyle (1/5)^6$. But there are $\displaystyle \left( \begin{smallmatrix} 5\\ 1 \end{smallmatrix} \right) = 5$ ways to choose a particular man from all 5 men. Therefore the probability of all 6 witness choosing the same man is $\displaystyle (5)(1/5)^6 = (1/5)^5 = 0.00032$.
• Sep 14th 2012, 09:09 AM
johnsomeone
Re: Is this right?
Yet another completely equivalent way to see the answer is to think of the space of possibilities, which is a set of all the ordered choices of the 6 witnesses:

Let $\displaystyle X$ = set of possibilities = $\displaystyle A \times A \times A \times A \times A \times A = A^6$, where $\displaystyle A = \{ man_1, man_2, man_3, man_4, man_5 \}$.

Note $\displaystyle |X| = |A|^6 = 5^6$.

So, for instance, the element $\displaystyle (man_3, man_5, man_2, man_2, man_3, man_4) \in A$ would correspond to witness #1 choosing man #3,

witness #2 choosing man #5, witness #3 choosing man #2, and so forth. It's one possibility of the choices of the 6 witnesses.

Let $\displaystyle S$ be the set of elements of $\displaystyle X$ which represent "all witnesses chose the same man". Then

$\displaystyle S$ = $\displaystyle \{ (man_1, man_1, man_1, man_1, man_1, man_1), (man_2, man_2, man_2, man_2, man_2, man_2), ... (man_5, man_5, man_5, man_5, man_5, man_5) \}$.

Note $\displaystyle |S| = 5$.

By the definition of probability, the probability that "all chose the same man" = $\displaystyle |S| / |X| = (5) / (5^6) = 1/5^5 = 0.00032$.