anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@hartnn
anonymous
  • anonymous
@mathslover ?
hartnn
  • hartnn
are u sure about n=7, r= 0 ?

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hartnn
  • hartnn
out of 7, any'one' has done a one.... so do u feel r should be 1 ?
hartnn
  • 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
  • hartnn
just clear your confusion about, n and r
anonymous
  • anonymous
oh sorry I was gone for while. Let's see here. Give me a second to read everything. Sorry :S
anonymous
  • 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
  • anonymous
I meant n=# of trials
hartnn
  • hartnn
number of trials or events or no. of creatures/people on which experiment is performed = n =7 here, \(\checkmark\)
hartnn
  • hartnn
r is the number of people 'chosen' here probability that anyONE has done fling implies, r=1 make sense? or more explanation?
anonymous
  • anonymous
so when we say r=0...
hartnn
  • hartnn
almost never.....when we don't perform experiment.....maybe there are some special case for this
hartnn
  • hartnn
make sure u understand why n=7, r=1 here
anonymous
  • anonymous
When no one has done a fling wouldn't that be r=0?
hartnn
  • hartnn
umm, yes i think so....(0.9)^7 seems right
anonymous
  • 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
  • hartnn
b)P( atleast 1 person) =1-P (no one )
hartnn
  • hartnn
no more = less r<=2
anonymous
  • anonymous
oh yes. sorry.
hartnn
  • hartnn
exactly 1---->r=1 atleast 1 = 1- none
hartnn
  • hartnn
atmost 1 --> r<=1
hartnn
  • hartnn
make sense ?
anonymous
  • anonymous
Yes, I think so.
anonymous
  • 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
  • hartnn
i will. yes show
anonymous
  • 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
  • hartnn
and did u get correct answer for the problem posted ? n=7,r=1,p=0.1,q=0.9
anonymous
  • anonymous
and for (c) n=10 r<5
hartnn
  • hartnn
all n, r selections \(\huge \checkmark\)
anonymous
  • anonymous
Yaaay!!!!! For the first problem I got a.) 0.478 b.) 0.522 c.) 0.974
hartnn
  • hartnn
sorry, i didn't do calculation part, i thought u had answers to verify, method is correct
anonymous
  • anonymous
I did have the answers =)
anonymous
  • anonymous
Thank you once again my friend!
hartnn
  • hartnn
so did they match? welcome ^_^
anonymous
  • anonymous
yes sir!
hartnn
  • hartnn
ok :)

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