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
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
Complete words, please! 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?
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.
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!