
#1
Oct2306, 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 



#2
Oct2306, 11:10 AM

HW Helper
P: 3,225

You may want to use the binomial distribution [tex]f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1p)^{nk}[/tex], where n is the number of trials, and k the number of times an event whose probability is 0.6 occured.




#3
Oct2306, 11:33 AM

P: 3

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 



#4
Oct2306, 11:44 AM

Sci Advisor
HW Helper
P: 3,680

probablity :confused totally
Complete words, please!




#5
Oct2306, 04:34 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

You have already been told to use
[tex]f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1p)^{nk}[/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. 



#6
Oct2306, 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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