A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
More statistics.
Binomial Probability
\[P(r)=\frac{n!}{r!(nr)!}p^rq^{nr}=C_{n,r}p^rq^{nr}\]
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:
\[P(0)=\frac{7!}{0!(70)!}p^0q^{70}=C_{7,0}p^0q^{70}\]
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
 3 years ago
More statistics. Binomial Probability \[P(r)=\frac{n!}{r!(nr)!}p^rq^{nr}=C_{n,r}p^rq^{nr}\] 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: \[P(0)=\frac{7!}{0!(70)!}p^0q^{70}=C_{7,0}p^0q^{70}\] 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?

This Question is Closed

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2are u sure about n=7, r= 0 ?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2out of 7, any'one' has done a one.... so do u feel r should be 1 ?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2the 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
 3 years ago
Best ResponseYou've already chosen the best response.2just clear your confusion about, n and r

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh sorry I was gone for while. Let's see here. Give me a second to read everything. Sorry :S

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok 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
 3 years ago
Best ResponseYou've already chosen the best response.0I meant n=# of trials

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2number of trials or events or no. of creatures/people on which experiment is performed = n =7 here, \(\checkmark\)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2r is the number of people 'chosen' here probability that anyONE has done fling implies, r=1 make sense? or more explanation?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so when we say r=0...

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2almost never.....when we don't perform experiment.....maybe there are some special case for this

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2make sure u understand why n=7, r=1 here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When no one has done a fling wouldn't that be r=0?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2umm, yes i think so....(0.9)^7 seems right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a) No one has done a onetime 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
 3 years ago
Best ResponseYou've already chosen the best response.2b)P( atleast 1 person) =1P (no one )

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2exactly 1>r=1 atleast 1 = 1 none

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can I show you what I did for another problem. I think my thought process is right. Feel free to point out any errors.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Trevor 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
 3 years ago
Best ResponseYou've already chosen the best response.2and did u get correct answer for the problem posted ? n=7,r=1,p=0.1,q=0.9

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2all n, r selections \(\huge \checkmark\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yaaay!!!!! For the first problem I got a.) 0.478 b.) 0.522 c.) 0.974

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2sorry, i didn't do calculation part, i thought u had answers to verify, method is correct

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did have the answers =)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you once again my friend!

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.2so did they match? welcome ^_^
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.