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Probablity :confused totally

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#1
Oct23-06, 10:39 AM
P: 3
Hi,Can anyone help me with this one?

Probability of an event occuring is 0.6

Find,
1)Probability of 1 of such events occurring out of total 5?

2)and 4 of such event occurring out of total 5.?



Answers given are:

1) 0.768
2) 0.2592


Please help me out by giving and explaining this one?..
thnks in advance
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radou
#2
Oct23-06, 11:10 AM
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You may want to use the binomial distribution [tex]f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k}[/tex], where n is the number of trials, and k the number of times an event whose probability is 0.6 occured.
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#3
Oct23-06, 11:33 AM
P: 3
Quote Quote by radou
You may want to use the binomial distribution [tex]f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k}[/tex], where n is the number of trials, and k the number of times an event whose probability is 0.6 occured.

I tried..but still not getting the ans..dunno where am going wrong..could u pls solve it and show me...
thnk u very much!....i guess i am making the same mistake again n again..but cant see thro ' it..will really appreciate if u solve it and explain...

U r right tht we have to use binomial distribution..is there any other way also to solve it?

thnks

CRGreathouse
#4
Oct23-06, 11:44 AM
Sci Advisor
HW Helper
P: 3,684
Probablity :confused totally

Complete words, please!

Quote Quote by RSS
I tried..but still not getting the ans..dunno where am going wrong..could u pls solve it and show me...
thnk u very much!....i guess i am making the same mistake again n again..but cant see thro ' it..will really appreciate if u solve it and explain...

U r right tht we have to use binomial distribution..is there any other way also to solve it?
I'm not sure at all where you're going wrong. You have the formula -- just plug in the appropriate values and you have the answer. What are you getting, and how?
HallsofIvy
#5
Oct23-06, 04:34 PM
Math
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Sci Advisor
Thanks
PF Gold
P: 39,491
You have already been told to use
[tex]f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k}[/tex]

When p= 0.6, n= 5, P(1)= [itex]\frac{5!}{(4!)(1!)}.6^1 .4^4[/itex]
[itex]= 5(.6)(0.025)= 0.0768[/itex]. Was it the arithmetic you had trouble with?

To answer (2) take k= 4 rather than 1.
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#6
Oct23-06, 07:08 PM
P: 3
sorry frnds..was making a very silly mistake with decimals..I got it after radou's 1st reply..was just trying to work out if there is any other way to solve besides using the binomial formula..
anyways, i have stuck with wht u all suggest..thnks again all of you!!..appreciate it!


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