Question

An urn contains 9 balls two of whigh are red, three blue and four black. Three balls are drawn at random. The probability that they are of same colour is​

Answer 1

Answer:

Given:

Firstly the total no of ways of choosing 3 balls from the given set of balls is 9C3 = 84 (2 red+ 3 blue + 4 black =9 balls). Same colour can mean 2 red or 2 blue or 2 black balls. Notice that I use the word “or” here and hence I ought to use addition rule. My first case is choosing 2 reds;

2C2 x 7C1=7 (why? 2 reds from the given reds and the third ball from the rest of the balls (9–2=7)). My second case is choosing 2 blues;

3C2 x 6C1=18 (2 blues from the given 3 blues and 1 from the rest). My third case’d be choosing 2 blacks;

4C2 x 5C1=30 (2 blacks from the given 4 blacks and 1 from the rest).

Now I add all the three cases and not MULTIPLY them as I don’t need all the three cases to be true at the same time (only one case is taken at a time). Adding so and dividing it by the total no of ways will give me:

Probability=55/84.

PS: Combination is used in this problem over permutation because you are only concerned about choosing the balls and not the order in which you choose them. If you aren’t comfortable that way, take to the “boxes” method for clear understanding.