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

## Combination

_{r}

nC

n! = 1 x 2 x ... x n_{r}= n!/r!(n-r)!0! = 1

1! = 1

_{}

## Problems based on Combinations

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

nC

_{r}= 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 9C

_{2}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 15C

_{2}ways.

So the required probability is

**9C**

_{2}/15C_{2}(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 9C

_{3}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 15C

_{3}ways.

So the required probability is

**9C**(iii)

_{3}/15C_{3}**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 6C

_{4 }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 15C

_{4}ways.

So the required probability is

**6C**

_{4}/15C_{4}(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 9C

_{1}x 6C

_{3 }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 15C

_{4}ways.

So the required probability is

**9C**

_{1}x 6C_{3 }**(v)**

_{}/15C_{4}**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 9C

_{4}x 6C

_{5 }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 15C

_{9}ways.

So the required probability is

**9C**

_{4}x 6C_{5 }**/15C**

_{9}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

nC

_{r}= 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 5C

_{2}x 4C

_{1}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 9C

_{3}ways.

So the required probability is

**5C**

_{2}x 4C_{1/}**9C**(ii) We have to find the probability of selecting no woman.

_{3}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 4C

_{3}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 9C

_{3}ways.

So the required probability is

**4C**

_{3/}**9C**

_{3}(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 5C

_{1}x 4C

_{2}+ 5C

_{2}x 4C

_{1 }+ 5C

_{3}

_{}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 9C

_{3}ways.

So the required probability is

**(5C**

_{1}x 4C_{2}+ 5C_{2}x 4C_{1 }+ 5C_{3})/**9C**(iv) We have to find the probability of selecting at most one woman.

_{3}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 4C

_{3}+ 5C

_{1}x 4C

_{2 }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 9C

_{3}ways.

So the required probability is

**(4C**

_{3}+ 5C_{1}x 4C_{2 }**)/**

**9C**(ii) We have to find the probability of selecting no men.

_{3}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 5C

_{3}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 9C

_{3}ways.

So the required probability is

**5C**

_{3/}**9C**

_{3}