Question 939927
  
Given:
Assume class is a random sample with respect to left-handedness.
population is 10% left-handed, i.e. probability p=0.10 is constant.
The number of steps of the experiment is known, n=30.
Each step is a Bernoulli experiment, i.e. with one of two possible outcomes.
  
Under these conditions, the binomial distribution applies.
  
Let P(X=r) be the probability of having r successes, then 
P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
where 
P(X=r) = {{{C(n,r)*p^r*(1-p)^(n-r)}}}
C(n,r) = {{{n!/((n-r)!*r!)}}} is combination of r objects out of n
n = 30 = number steps of experiment
p = 0.1 = probability of success (left-handed)
  
P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
= {{{C(30,0)*0.1^0*(0.9)^(30) + C(30,1)*0.1^1*(0.9)^(29) + C(30,2)*2.1^0*(0.9)^(28) + C(30,3)*0.1^3*(0.9)^(27)}}}
= {{{0.0424 + 0.1413 + 0.2277 + 0.2361}}}
= {{{0.6474}}}
  
Answer: the probability of seeing up to 3 left-handed students is 0.6474