# Probablity :confused totally

Tags: confused, probablity, totally
 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
 HW Helper P: 3,220 You may want to use the binomial distribution $$f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k}$$, where n is the number of trials, and k the number of times an event whose probability is 0.6 occured.
P: 3
 Quote by radou You may want to use the binomial distribution $$f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k}$$, 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

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,491 You have already been told to use $$f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k}$$ When p= 0.6, n= 5, P(1)= $\frac{5!}{(4!)(1!)}.6^1 .4^4$ $= 5(.6)(0.025)= 0.0768$. Was it the arithmetic you had trouble with? To answer (2) take k= 4 rather than 1.