More statistics.
Binomial Probability
\[P(r)=\frac{n!}{r!(n-r)!}p^rq^{n-r}=C_{n,r}p^rq^{n-r}\]
The problem states 10% of adults deliberately do a one time fling (purchase clothing, wear 'em to an event, and return 'em).
In a group of 7 adults what is the probability that anyone has done a one time fling?
I figured that:
n=7
r=0
I plugged those into the equation:
\[P(0)=\frac{7!}{0!(7-0)!}p^0q^{7-0}=C_{7,0}p^0q^{7-0}\]
What about p and q?
The probability of success (p) is 0.1 (1% of people do one time flings)?
and q would be 0.9? Would that be right?

- anonymous

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

@hartnn

- anonymous

@mathslover ?

- hartnn

are u sure about n=7, r= 0 ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- hartnn

out of 7, any'one' has done a one....
so do u feel r should be 1 ?

- hartnn

the probability of success = probability of flinging = p = 10% = 0.1 \(\checkmark\)
the probability of failure = probability of not flinging = q = 90% = 0.9 \(\checkmark\)

- hartnn

just clear your confusion about, n and r

- anonymous

oh sorry I was gone for while. Let's see here. Give me a second to read everything. Sorry :S

- anonymous

Ok so about n and r.
#number of trials
How do I relate that to this problem? I guessed that n=7 because it looked nice, but I don't have a good explanation for why I picked it

- anonymous

I meant n=# of trials

- hartnn

number of trials or events or no. of creatures/people on which experiment is performed = n =7 here, \(\checkmark\)

- hartnn

r is the number of people 'chosen'
here probability that anyONE has done fling implies, r=1
make sense?
or more explanation?

- anonymous

so when we say r=0...

- hartnn

almost never.....when we don't perform experiment.....maybe there are some special case for this

- hartnn

make sure u understand why n=7, r=1 here

- anonymous

When no one has done a fling wouldn't that be r=0?

- hartnn

umm, yes i think so....(0.9)^7 seems right

- anonymous

a) No one has done a one-time fling? r=0
b) At least one person r=1
c.) No more than two people \[r \ge 2\]
That's how I calculated it, whatcha think?

- hartnn

b)P( atleast 1 person) =1-P (no one )

- hartnn

no more = less
r<=2

- anonymous

oh yes. sorry.

- hartnn

exactly 1---->r=1
atleast 1 = 1- none

- hartnn

atmost 1 --> r<=1

- hartnn

make sense ?

- anonymous

Yes, I think so.

- anonymous

Can I show you what I did for another problem. I think my thought process is right. Feel free to point out any errors.

- hartnn

i will.
yes show

- anonymous

Trevor is interested in purchasing the local
hardware/sporting goods store in the small town of Dove Creek, Montana. After
examining accounting records for the past several years, he found that the store
has been grossing over $850 per day about 60% of the business days it is open.
Estimate the probability that the store will gross over $850
(a) at least 3 out of 5 business days.
(b) at least 6 out of 10 business days.
(c) fewer than 5 out of 10 business days.
p=.60
q=.40
a.)
n=5 because
"number of trials or events or no. of creatures/people on which experiment is performed "
=> In this case we have a total number of 5 days (events/trials)
at least 3 out of 5 would mean:
r>=3
To calculate this I would do P[r=3]+P[r=4]+P[r=5]
b.)
n=10
r>=6
To calculate.... P[r=6]+P[r=7]+P[r=8]+P[r=9]+P[r=10]

- hartnn

and did u get correct answer for the problem posted ?
n=7,r=1,p=0.1,q=0.9

- anonymous

and for (c)
n=10
r<5

- hartnn

all n, r selections \(\huge \checkmark\)

- anonymous

Yaaay!!!!!
For the first problem I got
a.) 0.478
b.) 0.522
c.) 0.974

- hartnn

sorry, i didn't do calculation part, i thought u had answers to verify, method is correct

- anonymous

I did have the answers =)

- anonymous

Thank you once again my friend!

- hartnn

so did they match?
welcome ^_^

- anonymous

yes sir!

- hartnn

ok :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.