**Combinatorics and Probability**

Combinatorics is quite useful in the computation of probabilities of events, as it can be used to determine exactly how many outcomes are possible in a given event.

__Worked Example__:

Question: At a school, learners each play 2 sports. They can choose from netball, basketball, soccer, athletics, swimming, or tennis. What is the probability that a learner plays soccer and either netball, basketball or tennis?

Answer

Step 1: Identify what events we are counting

We count the events: soccer and netball, soccer and basketball, soccer and tennis. This gives three choices.

Step 2: Calculate the total number of choices

There are 6 sports to choose from and we choose 2 sports. There are

choices.

Step 3 : Calculate the probability

The probability is the number of events we are counting, divided by the total number of choices.

Probability 3/15=⅕=0.2

The problems of restricted **permutation** or **combination** are convertible into problems of probability.

__For example:__

How many numbers of four distinct digits can be made with the digits 1,2,3 and 4 which begin with 4? It is a problem on restricted pemiutation. A similar problem in probability can bc as following:

“A four digit number of distinct digits is written down at random by using digits 1,2,3 and 4. What is the probability (chance) that the number begins with the digit 4?”

answer:

Here n(S)=total number of numbers of four distinct digits that can be made with 1,2,3 and 4 (without restriction) =^{4}P_{4}=4!

n(E)=the number of numbers of four distinct digits beginning with 4 that can be made with 1,2,3 and 4 is equal to 1×^{3}P_{3}=3!

Hence, required probability

Q1. If Maryam buys 5 tickets in a raffle where 95 tickets are sold altogether,what is the probability of her winning both first and second prizes?

solution:

Maryam has 5 chances out of a total of 95 to win first prize. If she wins first prize, she only has 4 tickets left as the winning ticket has already been drawn out. This only leaves 94 tickets altogether.

Using combination.

Q2. A box contains 18 CDs, four of which are defective. If two CDs are selected andtested, find the probability that both are defective.

solution:

Since there are four defective CDs out of 18, the probability that the first CD is defective is 4/18. Since the second CD is selected from the remaining 17and there are three defective CDs left, the probability that it is defective is 3/17. Hence, the probability that both CDs are defective is

Using combination.

Q3. In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint order of the numbers is not important.]

solution:

Total number of ways in which one can choose six different numbers from 1 to 20

Hence, there are 38760 combinations of 6 numbers.

Out of these combinations, one combination is already fixed by the lottery committee.

∴Required probability of winning the prize in the game

Q4. Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (i) all Kings (ii) 3 Kings (iii) atleast 3 Kings.

Solution: Total number of possible hands =

^{52}C

_{7}

(i). Number of hands with 4 Kings =

^{4}C

_{4}×

^{48}C

_{3}(other 3 cards must be chosen from the rest 48 cards). Hence P(a hand will have 4 Kings)

(ii). Number of hands with 3 Kings and 4 non-King cards =

^{4}C

_{3}×

^{48}C

_{4}. Therefore P(3 Kings)

(iii). P(atleast 3 King)=P(3 Kings or 4 Kings)=P(3 Kings)+P(4 Kings)

**A few ways to approach a probability problem**

There are a few typical ways that you can use for solving probability questions. Let’s consider example, how it is possible to apply different approaches:

**Example #1**

Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel?

**solution**:

1)

**combinatorial approach**:

The total number of possible committees is

*N*=

^{8}C

_{2}. The number of possible committee that includes both Bob and Rachel is in

*n*=1.

2)

**reversal combinatorial approach**:

Instead of counting probability of occurrence of certain event, sometimes it is better to calculate the probability of the opposite and then use formula

*p=1-q*. The total number of possible committees is

*N*=

^{8}C

_{2}. The number of possible committee that does not includes both Bob and Rachel is:

*m*=

^{6}C

_{2}+2∙

^{6}C

_{1}

That

^{6}C

_{2}: the number of committees formed from 6 other people,

2∙

^{6}C

_{1}: the number of committees formed from Rob or Rachel and one out of 6 other people.

3) probability approach:

The probability of choosing Bob or Rachel as a first person in committee is 2/8. The probability of choosing Rachel or Bob as a second person when first person is already chosen is 1/7. The probability that the committee includes both Bob and Rachel is.

4) reversal probability approach:

We can choose any first person. Then, if we have Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have neither Rachel nor Bob as first choice, we can choose any person out of remaining 7 people. The probability that the committee includes both Bob and Rachel is.

**Example #2**

Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?

**solution**:

1)

**combinatorial approach**:

^{5}C

_{3}: we choose 3 couples out of 5 couples.

^{2}C

_{1}: we chose one person out of a couple.

(

^{2}C

_{1})^3: we have 3 couple and we choose one person out of each couple.

^{10}C

_{3}: the total number of combinations to choose 3 people out of 10 people.

2)

**Reversal combinatorial approach**:

In this example reversal approach is a bit shorter and faster.

^{5}C

_{1}: we choose 1 couple out of 5 couples.

^{8}C

_{1}: we chose one person out of remaining 8 people.

^{10}C

_{3}: the total number of combinations to choose 3 people out of 10 people.

3)

**probability approach**:

1st person: 10/10=1: we choose any person out of 10.

2nd person: 8/9: we choose any person out of 8=10-2 (one couple from previous choice).

3rd person: 6/8: we choose any person out of 6=10-4 (two couples from previous choices).

Let’s read Explanation, Examples and Solutions on Combinations