Of 10 fish in a pond 4 are tagged If three fish are caught a

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