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An Introduction to Probability and Theory of Counting

Sayan Mukhopadhyay
17 Aug 0 0

In this lesson, we introduce the concept of probability with high school mathematics as a prerequisite. Before we start, I want to make you familiar with some standard terms in probability theory.

 

The first concept is the Random Experiment. If an experiment is conducted, and its result is unpredictable, i.e., the outcome can be one of many issues. Such an experiment is called a Random Experiment. Each outcome of a random experiment is an Event. Set of all possible outcomes is called the Sample Space. For example, when you are tossing a coin H(Head), and T(Tail) are two possible outcomes or event and { H, T } is the Sample Space of the Random Exprimenttossing coin.

There are two special kinds of events. Let A and B are two events. We call them Mutually Exclusive Events if A ∩ B = Φ, and we call them Mutually Exhaustive Events if A ∪ B = S. For example, in a random experiment, we are choosing a student to form a class. A is an event the student is a boy, and B is an event where the student is a girl. A and B are Mutually Exhaustive as a student is either a boy or girl there is no other option. And A and B are Mutually Exclusive because a student can not be a boy and girl both same time.

 

Probability of an event E = P(E) = (Number of outcomes in favor of E) / (Total number of outcomes)

 

For example, when you are tossing a coin probability of getting head is 1/2. Similarly, when you are throwing a dice probability of getting five is 1/6. As in the first case, the total number of outcome is 2 - head and tail, and the number of favourable results is one only head. Similarly, in the second case, the total number of results is 6 - 1 to 6, and the desired outcome is only 5.

 

So we have seen to calculate the probability we need to count the number of outcomes of an experiment. So it is crucial to know the theory of computing to work in probability. The fundamental theory of counting is that if you can do one work in m ways and another job in n ways, then you can do both tasks together in mxn ways. You can visualize this theory by imagining there is a closed room which has m doors to enter and n doors to exit. So a person can enter the room in m ways, and for each entry, he can exit in n ways. So he can enter and exit the room in mxn ways.

 

Problem 1: A polygon has 44 diagonals. What is the number of sides?
Solution:
If the polygon has n sides, we can choose the first end of a diagonal in n ways and second point in (n-3) ways excluding the point, and it's adjacent two which form side of the polygon. So both can be selected in n*(n-3) ways. Now in choosing the point, we are selecting each diagonal for twice. Each end is counted as fist and second end. So the number of diagonal = n*(n-3)/2 = 44 => n*(n-3) = 88 = 11x8.
So n = 11 (Ans)

 

Problem 2: Find the number of devisor in the form 4n + 2 (n>1) of integer 210.
Solution:
The divisor is in the form = 2*(2n+1) form. Factor of 210 in 2n+1 (n>1) form is 1,3,5. So we have to find the all possible combination of 3, 5, and 7.
Let generalized the problem assuming we have n different elements x1, x2 .. Xn and we have to find the all possible set can form by xi. So we can choose the first element in 2 ways either we can choose it or not choose it. Similarly, the second element we can also choose in 2 ways. So the first two elements we can choose in 2x2 ways. Similarly the first 3 elements we can choose 2^3 ways. So all element we can choose it in 2^n ways. So with n distinct elements, we can form 2^n different sets. There is one set which has no element ie, an empty set. So the number of the nonempty set is 2^n -1.
In our case, n=3 so answer is 7. Numbers are 3, 5, 7, 15, 21, 35, 105. (Ans)

 

Now we introduce the concept of Permutation and Combination. The permutation is the number of arrangement we can make by some elements, and the combination is the number of sets we can make by those elements. Difference between permutation and combination is that permutation consider the order or arrangement of the members which combination does not. Suppose a & b are two elements then {ab} and {ba} are two different permutations, but the same combination as the combination does not consider the arrangement.

Let, nPr is the number of permutations of r objects selected from n different objects. We can imagine there are n different balls and r basket and nPr is the number of ways we can fill the boxes by the given balls. So the first box we can fill in n ways as there are n balls. Then the second balls can be filled in (n-1) ways as one ball is already allocated in the first box. So according to the theory of counting 2 boxes can be filled in n*(n-1) ways. In same way r box can be filled in n*(n-1)*(n-2)...(n-r+1) ways.
So nPr = n*(n-1)*(n-2)....(n-r+1)
Now we are defining factorial of n as n!=n*(n-1)*(n-2).....3*2*1
Then nPr = n!/(n-r)!
The thing to be noted, nPn = n! as 0! =1 as it indicates an empty set or permutation.
Similarly, nCr is the number of combination of r objects selected from n different objects. Now you take one combination. It has r differents. So you can r! Permutation from it. Now if you take each combination and create r! Permutation from it, you get all possible permutation of r elements from n distinct elements.
So nCr * (r!) = nPr
=> nCr = (n!)/(n-r)! * (r!)

 

Problem 3: Find the sum of the digits in the unit place of all numbers forms with the help of 3,4,5,6.

Solution:

So the number of all the number's last digit where 3 is the last is the number of all permutation of rest three numbers = 3P3
So the sum of all the number's last digit where 3 is the last = 3P3 x 3
Similarly, the sum of all the number's last digit where 4 is the last = 3P3 x 4
So the sum of the digits in the unit place of all numbers forms with the help of 3,4,5,6 = 3P3x(3+4+5+6)(Ans)

 

Problem 4: Find the number of parallelograms that can form from a set of 4 parallel lines intersecting another 3 parallel lines.

Solution:

Answer is 4C2 x 3C2 ( An explanation will be provided on Request )

 

Problem 5: If two verticals are selected form a polygon of n side in random, what is the probability that it is diagonal?

The number of pairs of vertices = nC2 (number of ways 2 points selected from n points )
The number of diagonal = n(n-3)/2 ( see problem 1)

So, the answer = n(n-3)/2*4C2

Some properties of combinations:

1. nCr = nCn-r

Every time we choose r points from a set of n points there is a set of n-r points remain in the original set.  So nCr = nCr-1

2. (n+1)Cr = nCr + nCr-1

(n+1)Cr = number of ways r elements can be choose from n+1 elements.

Let, one element keeps apart from n+1 elements. Then number of combination = nCr

So (n+1)Cr = nCr + number of set can be formed with one seperated elements.

We can put the special element in one place of r place then

number of set can be formed with special element = from n element how to choose element for r-1 place = nCr-1

So  (n+1)Cr = nCr + nCr-1

3. nC0 + nC1 + nC2 + ....... + nCn = 2^n

See problem 2

Problem 6: Tn = number of triangles drawn from a n side polygon. Tn+1 - Tn =21. n=?

Solution:

(n+1)C3 - nC3 = 21

=> nC3 + nC2 -nC3 = 21

=> nC2 = 21

=> n(n-1) = 42 = 6x7 => n = 7 (Ans)

 

Back to Probability:

If A` is event for not happenning A then  P(A`) = 1- P(A)

If A and B are independent event then

                P(A∩B) = P(A)*P(B)

                P(A∪B) = P(A) + P(B) - P(A∩B)

 

Problem 7: Probability of A and B will die within a year is p and q. What is the probability that only one of them will alive at the end of the year.

Ans: p*(1-q) + q*(1-p)

Problem 8: Probabilty that a person will hit a target is 0.3. He shoots 10 times. What is the probability, he hits the target?

Ans: 1- (0.7)^10

Problem 9: In a throw of dice find the probability of getting one in even number of throw.

Ans:

(1/6)*(5/6) + (1/6)*(5/6)^3 + (1/6)*(5/6)^5 +   .... ∝

=(5/36) [ 1+ (5/6)^2 + (5/6)^4 + (5/6)^6 + .... ∝ ]

= (5/36)[ 1/(1-25/36)] = 5/36 * 36/11 = 5/11 ( Ans )

Problem 10: What is the probability that a leap year selected in random has 53 Sunday.

Ans: 364 days has 52 Sundays. 

So the probability of there is one Sunday in remaining two consecutive days =1/7 + 1/7 = 2/7 ( Ans )

Problem 11: E and F are two independent events. Probability of both events happen togethre is 1/12 and neither will happen is 1/2. P(E) = ? and P(F) = ?

Ans: P(E) * P(F) = 1/12

       1 - P(E or F) = 1/2 => P(E) + P(F) -1/12 = 1/2 => P(E) + P(F) = 7/12

You have two equation p+q = 7/12 and pq = 1/12. Solve them and get p and q.

 

Baye's Theorem:

 

 

Problem 12: A box contains N coins, m of which are fair and rest are biased. The probability of getting head of biased coin is 2/3. A coin is drawn from the box at random and tossed twice. First time it shows head and second time tail. What is the probability the coin is fair.

 

Problem 13: In a multiple choice question there are m options. A student know the answer, that's probability is p. He answered the question correctly. What is the probability that he knows the answer?

 

Some harder problems:

 

Problem 14: X is a set of n distinct elements. If A and B are two subsets of X, picked at random. What is the probability that A and B have same number of element?

 

Problem 15: A box contains ticket number from 1 to N. n tickets are drawn with replacement. Find the probability that the largest number in on drawn ticket is k.

 

 

 Now I will explain how a bank calculates the probability of a cardholder to be a fraud. They select some crucial features like age, income .. of the people. Let say a card holder's age is 40 and income INR 50,000 from the historical data they calculate the probability of a person become fraud when his age is 40 from the ratio of the total number of person of age. 40, and the number of people among them are a fraud. Similarly, they can calculate the probability of becoming fraud with income 40000. The final probability is the product of each feature probability. This is known as the Naive Bayes Classifier.

 

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