When we are selecting more than one items, we have to use combination formula to find probability.


Combination


Combination is the number ways of selecting r items from a set of n.  It is denoted by  nCr
nCr= n!/r!(n-r)!
n! = 1 x 2 x ... x n
0! = 1
1! = 1


Problems based on Combinations

Let us work on  a few examples based on combination probability.

Example 1 : -
A bag contains 9 white and 6 black balls.  What is the probability of selecting
(i) 2 white balls
(ii) 3 white balls
(iii) 4 black balls
(iv) 1 white and 3 black balls
(v) 4 white and 5 black balls
Solution: -
We know that the combination formula for selecting r items from n is
nCr= n!/r!(n-r)!
Also we know that probability of an event A is
P(A) = Number of favorable outcomes/ Total number of outcomes
(i) We have to find the probability of selecting 2 white balls.
So favorable cases will be obtained when we select 2 balls from 9 white balls.  This can be done in 9C2 ways.
Since there are 9 + 6 = 15 balls, the total number of outcomes will be obtained by selecting 2 balls from the 15 balls.  This can be done in 15C2 ways.
So the required probability is 9C2/15C2
(ii)We have to find the probability of selecting 3 white balls.
So favorable cases will be obtained when we select 3 balls from 9 white balls.  This can be done in 9C3 ways.
Since there are 9 + 6 = 15 balls, the total number of outcomes will be obtained by selecting 3 balls from the 15 balls.  This can be done in 15C3 ways.
So the required probability is 9C3/15C3
(iii) We have to find the probability of selecting 4 blackballs.
So favorable cases will be obtained when we select 4 balls from 6 black balls.  This can be done in 6C4 ways.
Since there are 9 + 6 = 15 balls, the total number of outcomes will be obtained by selecting 4 balls from the 15 balls.  This can be done in 15C4 ways.
So the required probability is 6C4/15C4
(iv) We have to find the probability of selecting 1 white and  3 black balls.
So favorable cases will be obtained when we select 1 white from the 9 white balls and 3 black balls from the 6 black balls.  This can be done in 9C1 x 6C3 ways.
Since there are 9 + 6 = 15 balls, the total number of outcomes will be obtained by selecting 4 balls from the 15 balls.  This can be done in 15C4 ways.
So the required probability is 9C1 x 6C3 /15C4
(v) We have to find the probability of selecting 4 white and  5 black balls.
So favorable cases will be obtained when we select 4 white from the 9 white balls and 3 black balls from the 5 black balls.  This can be done in 9C4 x 6C5 ways.
Since there are 9 + 6 = 15 balls, the total number of outcomes will be obtained by selecting 9 balls from the 15 balls.  This can be done in 15C9 ways.
So the required probability is 9C4 x 6C5 /15C9

Example 2: -
There are 4 men and 5 women.  Find the probability of selecting 3 of which
(i) exactly two are women
(ii) no woman
(iii) at least one women
(iv) at most one women
(v) no men
Solution: -
We know that the combination formula for selecting r items from n is
nCr= n!/r!(n-r)!
Also we know that probability of an event A is
P(A) = Number of favorable outcomes/ Total number of outcomes
(i) We have to find the probability of selecting exactly two women.
Since we have to select three people, exactly two women means two women and one man
So favorable cases will be obtained when we select 2 women from the 5 women and 1 man from the 4 men.  This can be done in 5C2 x 4C1 ways.
Since there are 4 + 5 = 9 people, the total number of outcomes will be obtained by selecting 3 people from the 9 people.  This can be done in 9C3 ways.
So the required probability is 5C2 x 4C1/9C3
(ii) We have to find the probability of selecting no woman.
Since we have to select three people, no woman means all three are men.
So favorable cases will be obtained when we select 3 men from the 4 men.  This can be done in  4C3 ways.
Since there are 4 + 5 = 9 people, the total number of outcomes will be obtained by selecting 3 people from the 9 people.  This can be done in 9C3 ways.
So the required probability is  4C3/9C3
(iii) We have to find the probability of selecting at least one woman.
Since we have to select three people, at least one women means 1 women or 2 women or 3 women.  That is 1 woman and 2 men or 2 women and 1 man or all three woman.
So favorable cases will be 5C1 x 4C2 + 5C2 x 4C1 + 5C3 ways.
Since there are 4 + 5 = 9 people, the total number of outcomes will be obtained by selecting 3 people from the 9 people.  This can be done in 9C3 ways.
So the required probability is (5C1 x 4C2 + 5C2 x 4C1 + 5C3)/9C3
(iv) We have to find the probability of selecting at most one woman.
Since we have to select three people, at most one women means 0 women or 1 women.  That is 0 woman and 3 men or 1 woman and 2 men.
So favorable cases will be  4C3 + 5C1 x 4C2 ways.
Since there are 4 + 5 = 9 people, the total number of outcomes will be obtained by selecting 3 people from the 9 people.  This can be done in 9C3 ways.
So the required probability is (4C3 + 5C1 x 4C2 )/9C3
(ii) We have to find the probability of selecting no men.
Since we have to select three people, no men means all three are women.
So favorable cases will be obtained when we select 3 women from the 5 women.  This can be done in  5C3 ways.
Since there are 4 + 5 = 9 people, the total number of outcomes will be obtained by selecting 3 people from the 9 people.  This can be done in 9C3 ways.
So the required probability is  5C3/9C3