Calculating Probability of Receiving $1 from Multiple People

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To calculate the probability of receiving at least $1 from Bob, Ted, and Frank, each with a 30% chance of giving a dollar, the formula is 1 minus the probability that none will give a dollar. The probability that none will give a dollar is calculated as 0.7 (the chance of not receiving a dollar from each person) multiplied together: 0.7 x 0.7 x 0.7. This results in a 65.7% chance of receiving at least one dollar, confirming the calculation. Understanding this approach allows for solving similar probability problems effectively.
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Should be simple enough... unfortunately I'm 20 years removed from college and can't seem to find an equation so I thought I'd post it here.

If there is a 30% chance that Bob will give me a $1, and a 30% chance that Ted will give me a $1 and a 30% chance that Frank will give me a $1. What are the chances that at least one of them gives me a $1?

Very hypothetical, but if someone could give me the equation to solve it then I can figure some other problems out on my own.

Thanks,
Dale
 
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The chance that at least one will give you a dollar is 100% minus the chance that none will give you a dollar. The chance that none will give you a dollar is the chance that the first won't give you a dollar times the chance the second won't times the chance the third won't.
 
So... if I get this straight

1 - (.7 x .7 x .7) = 65.7% ?
 
Yup, or you can look at all the combinations that can occur, e.g. Bob gives you a dollar and Ted and Frank don't, and so on, and then sum the appropriate probabilities.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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