# Amazingly simple probability question

1. Feb 21, 2005

### msmith12

I know this is a really easy problem, but i am confusing myself with it...

there are three people who each have two possible actions-pay, or not pay (with symmetric probabilities of p for pay, and 1-p for not pay).

what is the probability that at least two people pay?

I first thought that it was just p^2, but that can't be right because that doesn't take into account the third person... so I then thought that it was 3p^2, because there are three possible combinations of 2 people paying. Is this all I need to do?

thanks

~feeling stupid

2. Feb 21, 2005

### bomba923

Just remember binomial probability. The probability that @least 2people pay means either 2 or 3 people pay. Thus, apply the binomial probability formula:
(where p=probability of paying)

[3!/(2!*1!)]*(p^2)*(1-p) + [3!/(3!*0!)]*(p^3)
(As you can see, I don't know LaTex)

If the probability of paying is 50%, then the above expression should amount to 0.5,
meaning that there is a 50%chance that @least two people will pay.

3. Feb 21, 2005

### arildno

Just elaborating on bomba's reply:
$$1=1^{3}=((1-p)+p)^{3}=\sum_{i=0}^{3}\binom{3}{i}(1-p)^{i}p^{3-i}$$
Hence, your probability is:
$$\sum_{i=0}^{1}\binom{3}{i}(1-p)^{i}p^{3-i}$$

4. Feb 24, 2005

### bomba923

Hey, can you teach me LaTex? That'd be great :rofl:

5. Feb 24, 2005

### arildno

Click on the images to see how the code is written.

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