# Probablity :confused totally

1. ### RSS

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. ### radou

3,215
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.

3. ### RSS

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. ### CRGreathouse

3,682
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?

5. ### HallsofIvy

40,918
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.

6. ### RSS

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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