How Do You Calculate the Probability of Getting at Most One Brown M&M?

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To calculate the probability of getting at most one brown M&M from a selection of seven, where 30% are brown, the binomial distribution is used. The relevant parameters are n=7 (the number of trials), p=0.3 (the probability of selecting a brown M&M), and q=0.7 (the probability of selecting a non-brown M&M). The formula for calculating this probability involves summing the probabilities of getting zero and one brown M&M, specifically P(X=0) and P(X=1). The calculations involve using q^7 for P(X=0) and n*p*q^6 for P(X=1). This approach provides the correct method for determining the desired probability.
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Okay so I did this problem and got it wrong but I get one more chance to get it right. I tried using Binomial Dist to solve it but I failed.

30% of all M&Ms are brown. If 7 M&Ms are randomly selected, what is the probability that at most 1 is brown?

I thought I would use 0 and 1 but I guess not. Can someone help me?
 
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p0=.37
p1=7*.36*.7

Add them up.
 
Sorry, got the probs reversed. switch .3 and .7.

p0=.77
p1=7*.76*.3

Add them up.
 
For Binomial Distribution,
take n=7, p=0.3, q=1-p=0.7.

P(X<=1)=P(X=0)+P(X=1)=q^7+n*p*q^6

thts it gys...
enjy
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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