## probablity :confused totally

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.?

1) 0.768
2) 0.2592

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 Recognitions: Homework Help 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.

 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

Recognitions:
Homework Help

## probablity :confused totally

 Recognitions: Gold Member Science Advisor Staff Emeritus 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.