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If an event has an 85% chance of success, and you attempt the event 4 times, what are your chances of 3 or more successes?

If these were coin flips, I know there are $2^4$ possibilities. Of these $16$ possibilities, $4$ would show $3$ heads, and $1$ tail. Only $1$ possibility gives $4$ heads, so chances are $\frac{5}{16}$ for $4$ events like this. When trying to calculate when percent success as given, I'm a bit lost.

(Disclosure, this question was part of a practice exam for MTEL, a proficiency test to teach HS math in my state. The practice test came with no explanations.)

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Because it hasn't been mentioned: this is a 'Binomial distribution' type of question. –  Ronald Jun 12 '13 at 10:45
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2 Answers

up vote 3 down vote accepted

There is just one way to get $4$ successes: you must succeed each time, and the probability of that is $0.85^4$. There are $4$ ways to get exactly $3$ successes: you can fail the first time, the second time, the third time, or the fourth time. Each of those $4$ ways has the same probability, $0.85^3\cdot0.15$, since the probability of a failure on any given trial is $1.00-0.85=0.15$. Thus, the desired probability is

$$0.85^4+4\cdot0.85^3\cdot0.15=0.89048125\;.$$

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Much thanks. 89% it was. Now I see how it was calculated, beautiful. –  JoeTaxpayer May 9 '13 at 4:14
    
@JoeTaxpayer: You’re welcome. –  Brian M. Scott May 9 '13 at 4:15
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If $S$ means success and $F$ means failure, you want to calculate the probability of any of the following: $SSSF, SSFS, SFSS, FSSS, SSSS$. These are disjoint events, so you may calculate the probability of each and then add.

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