### Introduction

The combinations formula is very useful in calculating probability in many problems. It helps you easily calculate the number of favorable events and the total number of events in the sample space. Thus it saves you the time of manually listing the sample space and counting the number of total and number of favorable events.

### Formula

The formula for combinations is as follows.

`C_r^n = {n!}/{r!(n-r)!}`

This formula helps you calculate the number of favorable outcomes and the total number of outcomes in the given problem, which then helps you calculate the probability. The solved examples below will help you understand how this formula is applied to probability problems.

### Solved Examples

#### Problem 1

*There are three red, five blue and two orange marbles in a bag. Two marbles are selected from it without replacement. What is the probability that both the marbles will be red?*

In this problem we will use the combinations formula to calculate the number of favorable outcomes and the total number of outcomes. The number of favorable outcomes are all those outcomes in which two red marbles are selected. Since there are a total of three red marbles in the bag, hence the different ways in which two red marbles can be selected from three red marbles are given by

`C_2^3 = (3!)/(2!(3-2)!) = 3`Therefore the number of favorable outcomes is 3. Now we will calculate the total number of outcomes. Total number of ways in which two balls (of any color) can be selected from 10 balls is given by

`C_2^{10} = (10!)/(2!(10-2)!) = 45`Now that we have the number of favorable and the total number of outcomes, we calculate the probability by the help of the formula for theoretical probability.

`P(A) = "Number of favorable outcomes"/"Total number of outcomes"`

`P("two red marbles") = 3/45 = 1/15`Therefore the probability of selecting two red marbles out of the bag is `1/15`.

#### Problem 2

*There are twenty boys and ten girls in a class. What is the probability that two of the selected students will both be boys.*

Calculate the total number of possible outcomes. The total number of different ways in which two students can be selected from a total of 20 + 10 = 30 students are given by

`C_2^{30} = (30!)/(2!(30 - 2)!) = 435`Now calculate the number of favorable outcomes. Since a favorable outcome is selecting two boys and there are twenty boys in the class, hence the total number of favorable outcomes is given by

`C_2^{20} = (20!)/(2!(20-2)!) = 190`Therefore the total number of possible outcomes is 435 while the number of favorable outcomes is 190. Applying the formula for theoretical probability,

`P(A) = "Number of favorable outcomes"/"Total number of possible outcomes"`

`P("two boys are selected") = 190/435 = 38/87`

#### Problem 3

*Out of five hundred students in a school, two hundred opted to study Math, two hundred opted to study English and one hundred opted to study Biology. No student can opt for more than one subject. What is the probability that five student chosen at random will all be studying Math?*

Find the total number of possible outcomes

*.*There are a total of 500 students and the different number of ways in which 5 are chosen are

`C_5^{500} = (500!)/(5!(500-5)!) = 255244687600`Find the number of favorable outcomes. Choosing five students all studying math will give a favorable outcome. Since 200 students opted for math, therefore the different number of ways in which five students can be selected from the 200 students is given by

`C_5^{200} = (200!)/(5!(200-5)!) = 2535650040`Now applying the formula for theoretical probability,

`P("all five students study math") = 2535650040/255244687600 = 0.01`

#### Problem 4

*There are ten apples in the box out of which five are green and five are red. What is the probability that two apples selected from the box will have one red and one green apple?*

Find the total number of possible outcomes. The total number of ways of selecting two apples from ten are given by

`C_2^{10} = (10!)/(2!(10-2)!) = 45`There are five red and five green apples. The number of ways of selecting one red apple out of five are given by `C_1^5=5` and those of selecting one green apple out of five are given by `C_1^5 = 5`. Therefore the total number of ways of selecting one red and one green apple (by the fundamental principle of counting) are `5 times 5 = 25`. Thus, total number of favorable outcomes is 25.

Now applying the formula for theoretical probability,

P(one red and one green) = `"Number of favorable outcomes"/"Total number of outcomes"`

P(One red and one green) = `25/45 = 5/9`Therefore the probability that the two apples selected will be one green and one red is `5/9`.

#### Problem 5

*Seventy people were surveyed for their favorite outdoor games. Twenty said football, thirty said baseball, ten said basketball and ten said lawn tennis. What is the probability that three people selected from the seventy people will all like basketball?*

Find the total number of possible outcomes. The total number of different ways of selecting three out of seventy people are given by

`C_3^{70} = (70!)/(3!(70-3)!) = 54740`Find the number of favorable outcomes. A total of ten people play basketball. The total number of different number of ways of selecting three people out of the ten are given by

`C_3^{10} = (10!)/(3!(10-7)!) =120`Now apply the formula for theoretical probability,

P(A) = `"Number of favorable outcomes"/"Total number of possible outcomes"`

P(all three people like basketball) = `120/54740 = 6/2737`Thus the probability that all the three people selected will like basketball is `6/2737`.

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