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Supplementary information The binomial distribution represents the probability distribution of the number of successes in a sequence of n independent experiments. Each experiment is a Bernoulli trial, which means that in each experiment, there are only two possibilities: a success or a failure. The probability to have a success in an experiment is noted p (so the probability of a failure is 1-p). For n experiments, the probability p of success is the same. Let X the random variable representing the number of success among the n experiments. X follows a binomial distribution with parameters n and p. The expected number of X is E(X)=np. The probability of getting exactly k successes (and so n-k failures) is n! P X k C nk p k (1 p ) n k where C nk (k! is the factorial of k and is equal to k!(n k )! k! 1 2 3 ... (k 1) k (with the exception 0!=1) In this paper, the fact that an individual presents (or not) a de novo mutation can be considered as a Bernoulli trial with a probability of success of 2.5x10-5. In the studied family, there are 92 relatives, and the fact that an individual presents a de novo mutation is independent of the fact that another individual presents (or not) a de novo mutation. The total number of de novo mutations among the 92 relatives follows a binomial distribution with n=92 and p=2.5x10-5. In a given family of 92 relatives, the expected number of de novo mutation is np=0.00125 (on average, one mutation for 800 families). The probability to observe none de novo mutation is: 0 92 92 92! 0 P X 0 C 92 p 0 (1 p) 920 2.5 10 5 1 2.5 10 5 1 2.5 10 5 0.997703 0!92! The probability to observe one de novo mutation is: 1 91 91 92! 1 P X 1 C92 p 1 (1 p) 921 2.5 10 5 1 2.5 10 5 922.5 10 5 1 2.5 10 5 0.002295 1!91! The probability to observe two de novo mutations is: 2 P X 2 C 92 p 2 (1 p) 92 2 92! 2.5 10 5 2!90! 1 2.5 10 2 5 90 4186 2.5 10 5 1 2.5 10 2 5 90 So the probability to observe at least three de novo mutations is: P X 3 1 P( X 3) 1 P( X 0) P( X 1) P( X 2) 1 1 2.5 10 5 92 2.5 10 5 1 2.5 10 5 91 4186 2.5 10 5 1 2.5 10 2 5 90 2.0 10 9 This probability represents one family among about 511 millions of families of 92 individuals. 2.6 10 6