Of 10 fish in a pond 4 are tagged. If three fish are caught and not thrown back into the pond what is the probability that fewer than two of them are tagged? Solution Binomial Distribution PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k Where k = number of successes in trials n = is the number of independent trials p = probability of success on each trial tagged (p) = 4/10 = 0.4 P( X < 2) = P(X=1) + P(X=0) = ( 3 1 ) * 0.4^1 * ( 1- 0.4 ) ^2 + ( 3 0 ) * 0.4^0 * ( 1- 0.4 ) ^3 = 0.648.