Binomial Distrubition/Die rolling Question.

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    Binomial Rolling
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SUMMARY

The discussion focuses on calculating the probability of rolling at least 2 "sixes" in 6 rolls of a balanced die using the Binomial Formula. The relevant formula is P(X = r) = nCr * p^r * (1-p)^(n-r), where p represents the probability of rolling a six (1/6), n is the number of trials (6), and r is the number of successes (2). Participants clarified the correct interpretation of the formula, emphasizing the need to compute both success and failure probabilities accurately.

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How to Find the Probability of rolling at least 2 "sixes" in 6 rolls of a balanced die.

I am trying to solve using the Binominal Formula, P(X = r) = nCr p r (1-p)n-r

But am not really sure what the probability rates for success and failure should be or how to compute it.

Any advice?

Thanks.
 
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derek10rr2 said:
How to Find the Probability of rolling at least 2 "sixes" in 6 rolls of a balanced die.

I am trying to solve using the Binominal Formula, P(X = r) = nCr p r (1-p)n-r

But am not really sure what the probability rates for success and failure should be or how to compute it.

Any advice?

Thanks.

p(S)=1/6 n=6 r=2 I assume in your formula you mean: [tex]p^r (1-p)^{n-r}[/tex]
 
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