# Independent probabilities question

• Feb 3rd 2010, 01:38 AM
cheema
Independent probabilities question
Dear All,

I am new to probability theory and it would be a great help if you can help me answering this question.

There are N balls. There is a circle in the space. The probability of each ball to fall in the circle is p. What is the probability that there are *at least* k out of N balls that fall in the circle? I came up with the probability of at least one out of N balls falling in the circle and that is ( 1 - (1 - p)^N). Can someone please guide me how to do for k.

Thanks ,
Cheema
• Feb 3rd 2010, 05:41 AM
Hello cheema

Welcome to Math Help Forum!
Quote:

Originally Posted by cheema
Dear All,

I am new to probability theory and it would be a great help if you can help me answering this question.

There are N balls. There is a circle in the space. The probability of each ball to fall in the circle is p. What is the probability that there are *at least* k out of N balls that fall in the circle? I came up with the probability of at least one out of N balls falling in the circle and that is ( 1 - (1 - p)^N). Can someone please guide me how to do for k.

Thanks ,
Cheema

Looking at the title of your post, I'm assuming that the ways in which the balls fall are independent of each other, so this is an example of a Binomial Distribution.

The probability of each trial resulting in a success is $\displaystyle p$, where a 'trial' is a ball falling and a 'success' is the ball falling into the circle. The number of trials is $\displaystyle N$, and we want the probability that the number of successes is greater than or equal to $\displaystyle k$.

The formula that gives the probability of getting exactly $\displaystyle r$ successes in N trials is:
$\displaystyle \binom Nrp^r(1-p)^{N-r}$
and we shall need the sum of all the terms like this, for values of $\displaystyle r$ from $\displaystyle k$ to $\displaystyle N$; in other words:
$\displaystyle \sum_{r=k}^N\binom Nrp^r(1-p)^{N-r}$
And that's it.