Question

Assume that the procedure yields a binomial distribution with trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial.
n=5, X=2 , p=0.45

Thanks!

Answer 1

As I said, I know little about probability calculations. Perhaps you should ask @bhaskarbabu, as he claims to have a firm grasp on it.

Answer 2

I am asking you a favor not him! So don't worry about it!

Answer 3

Can you help me, Plzzz!!!

Answer 4

Okay, write out your formula. Determine what n, X, and p represent.

Answer 5

ok! hold on!

Answer 6

my compu keep frezing! not sure whats the problem.

Answer 7

I had to change the question..? I got it wrong and they gave the new.

Answer 8

That's fine. The equation stays the same.

Answer 9

Here is what I got!
p(3)=(5!)/(5-3)!3!= 0.65^(3)*0.35^(3)

Answer 10

Then (65)(0.2746)(0.042875) Then I got stuck here.

Answer 11

Okay. I'm not sure about your use of factorials, but that's not the formula I'm familiar with.

Answer 12

I know of:
\[P = \left(\begin{matrix}n \\ k\end{matrix}\right) p^{k} q^{n-k}\]

Answer 13

No, it's different, they gave me the formula! you want me to give to you?

Answer 14

Please.

Answer 15

P(x)=(n!)/((n-x)!x!)*p^(x)*q^(n-x)

Answer 16

Let me make sure I understand you. Is it:

\[P_{x} = \frac{n!}{((n-x)! x!)} p^{x} q^{n-x}\]
or
\[P_{x} = \frac{n!}{((n-x)! x!) p^{x} q^{n-x}}\]

Answer 17

Yes!

Answer 18

Which one? They're very different.

Answer 19

the first one..

Answer 20

Okay. What does q stand for?

Answer 21

q=1-p

Answer 22

in this case new question it will be 1- 0.15=0.85

Answer 23

Am I correct?

Answer 24

Okay, so you should be able to plug in all your values.

Earlier, you wrote p(3)=(5!)/(5-3)!3! but you seem to have forgotten to multiply it by p^x q^(n-x)

Answer 25

Ok, I figured it out! But how do I multiply.

Answer 26

The way that I did I got 30..

Answer 27

p(2)=(5!)/(5-2)!2!

Answer 28

Go back to your formula. You're missing everything after the fraction.
\[P_{x} = \frac{n!}{((n−x)!x!)} p^{x} q^{n−x}\]

Answer 29

P(2)=(5!)/((5-2)!2!)*0.15^(2)*0.85^(5-2)

Answer 30

@geoffb Here is what I got. Can you explain me why they the right answer is 0.138??
Thank you!

Answer 31

Is this how you wrote out your formula on paper?
\[P = \frac{5!}{3! \times 2!} (0.15)^{2} (0.85)^{3}\]

Answer 32

How you calculate the fraction only?

Answer 33

I got 120/8 is that correct?

Answer 34

Nope. Your numerator is right, but not the denominator. What is 3!? What is 2!? What is the product of the two?

Answer 35

Ohh 12

Answer 36

Good. Now are you able to get the answer?

Answer 37

Nooo!! I have an idea but nop..

Answer 38

So what I do is divide the fraction after in this case will be (10).

Answer 39

Okay, so now you have this, right? :

\[P = \frac{120}{12} (0.15)^{2} (0.85)^{3}\]

Answer 40

(10)(0.0225)(0.7225)

Answer 41

Yes!

Answer 42

No, 0.85^3 is not 0.7225.

Answer 43

0.614

Answer 44

Yes.

Answer 45

ok! let me see.

Answer 46

I got it right now..

Answer 47

My mistake was leaving the whole fraction without realizing It was missing..I am dummi!! jejeeje!!

Answer 48

No problem. You seem to understand how the calculation works now. :)

Answer 49

Yes, Thank you sooo muchhh!!!! :)