Amazingly simple probability question

[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

 

[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.

 

[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}

 

[bomba923]

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}

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

 

[arildno]

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

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