Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)

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


share|cite|improve this answer
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.