## MathSofiya 3 years ago 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?

1. MathSofiya

@hartnn

2. MathSofiya

@mathslover ?

3. hartnn

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

4. hartnn

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

5. 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 $$\checkmark$$

6. hartnn

7. MathSofiya

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

8. MathSofiya

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

9. MathSofiya

I meant n=# of trials

10. hartnn

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

11. hartnn

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

12. MathSofiya

so when we say r=0...

13. hartnn

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

14. hartnn

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

15. MathSofiya

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

16. hartnn

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

17. MathSofiya

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?

18. hartnn

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

19. hartnn

no more = less r<=2

20. MathSofiya

oh yes. sorry.

21. hartnn

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

22. hartnn

atmost 1 --> r<=1

23. hartnn

make sense ?

24. MathSofiya

Yes, I think so.

25. MathSofiya

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

26. hartnn

i will. yes show

27. MathSofiya

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]

28. hartnn

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

29. MathSofiya

and for (c) n=10 r<5

30. hartnn

all n, r selections $$\huge \checkmark$$

31. MathSofiya

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

32. hartnn

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

33. MathSofiya

I did have the answers =)

34. MathSofiya

Thank you once again my friend!

35. hartnn

so did they match? welcome ^_^

36. MathSofiya

yes sir!

37. hartnn

ok :)

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