A box contains 6 red,4 white and 5 black balls.A person draws 4 balls at random.Find the probability that among the 4 balls drawn there is atleast 1 ball of each color.
Please check my solution and point out my mistake.
total no. of selections possible= 15C4
since there has to be at least 1 ball of each color,i first select 1 ball of each color,which can be done in 6C1 * 4C1 * 5C1 ways.
now out of the remaining 12 balls i can select any ball,the condition that 1 ball of each color has to be selected will be satisfied everytime.
therefore total number of favorable cases = 6C1 * 4C1 * 5C1 * 12C1
so probability = (6C1 * 4C1 * 5C1 * 12C1 )/15C4 which comes out to be 96/91 (not possible)
so where am i wrong?
the correct answer is 48/91
Reference: This post
Here also this looks like a problem related to order.
When combinations is under consideration then order can create over counting as order is a friend of permutations :-). The problem is not with sample space but is with the no. of favourable events. Making cases to compute no. of favourable events should do away with this problem.