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Modern Math
PROBABILITY  1
PROBABLITY
The chapter on Probability forms the foundation of the advanced mathematics that involves using probabilistic methods in deploying chances of success of an event. Probability simply means the chances of occurrence of an event or its non occurrence. Probability is a value between 0 and 1 where 0 is the probability of an impossible event and 1 is a probability of occurrence of a sure event. It finds its application in statistics, physics, gaming, artificial intelligence, machine learning and in game theory. Now, before we further delve into solving problems, there are a few concepts and terms important for us to understand. Note that the knowledge of Permutations and Combinations is a prerequisite.
Basic concepts and terms used in probability

Experiment: An operation that has as its result some well defined outcomes is called an experiment.

Random Experiment: An experiment whose outcome cannot be predicted with certainty. If we perform same experiment several times under the same conditions and outcome were not the same, then experiment is said to be random.

Sample Space: The set of all possible outcomes of a random experiment is called sample space and denoted by ‘S’.
Example: When we toss a coin, Head and Tails are the outcomes. The sample space is { H, T}

Event: A subset of sample space is called event.
When a coin is tossed, the sample space S = {H, T}. The subset {H} denotes the event of occurrence of heads in the toss of a coin.

Simple Event: An event is called a simple event if it is a singleton subset of the sample space.
Example: toss of a coin, sample space: {H, T} and Simple event : {H}

Compound Event: A subset of sample space S, which contains more than one element is called a compound event.
Example: For the throw of a die, the sample space is {1,2,3,4,5,6}. The occurrence of odd number {1, 3, 5} is a compound event.

Equally likely Events: A event is said to be equally likely event when outcomes are equally probable.
Example: occurrence of Head or Tail in tossing of a coin is equally likely event.

Mutually Exclusive Events: Two events are called mutually exclusive if both cannot occur together.
Example: In a toss of coin, occurrence of head or tail is mutually exclusive event.

Probability: it is defined as number of favorable outcomes divided by total number of outcomes.
Example: in the toss of a coin, probability of occurrence of head = ½
Example 1: A coin is tossed successively three times. Find the probability of getting exactly one head or two heads.
Solution:
Total number of outcomes= 2*2*2 = 8
Favorable outcomes (exactly one head or two head) = {HTT, THT, TTH, HHT, HTH, THH}
Required probability = 6/8 = 3/4
Example 2: A leap year is selected random, what is the probability of having 53 Mondays?
Solution:
A leap year has 366 days out of which 52 weeks and 2 extra days. The leap year will have 53 Mondays if out of two consecutive days one is Monday.
Required probability = 2/7
Example 3: If 6 men and 6 women sit in a row randomly, find the probability of getting all 6 men together?
Solution:
Total number of ways of seating 6 men and 6 women in a row = 12!
Number of ways in which 6 men can be together = 7! * 6!
Required probability =
Example 4: A box contains 8 pairs of shoes. If 4 shoes are selected at random then find the probability that exactly one pair of shoe is selected.
Solution:
Total number of ways in which 4 shoes can be selected out of 8 pairs (16 shoes) = 16C4
The number of ways in which 1pair of shoe can be selected out of 8 pairs = 8C1. From the remaining 7 pairs we have to select two shoes so that no pair is formed = 7C2 * 22
Required probability:
Example 5: A bag contains 36 tickets, numbered from 0 to 35. Three of tickets are drawn at random. Find the probability of the sum of the numbers in the three tickets to 36.
Solution:
The number of ways in which 3 tickets can be selected out of 36 = 36C3
The favorable number of ways in which sum of numbers in the three tickets is 36 =
The coefficient of x36 in the expansion of
Required probability =
Example 6: From a pack of 52 playing cards, three cards are drawn at random. Find the probability of drawing a king, a queen and a jack.
Solution:
The number of ways in which three cards can be drawn out of 52 cards = 52C3
The number of ways in which a king, a queen and a jack can be selected = 4C1* 4C1*4C1
The required probability =
Example 7: A four digit number of distinct digits is written down at random by using digits 3,4, 5 and 6. Find the probability that the number

Starts with 6

Is divisible by 4
Solution:

The total number of numbers of 4 digits can be made is = 4C4 * 4!
The number of ways in which a number of 4 digit can be made which starts with 6 is = 3C3*3!
Required Probability =

The number of ways in which a number of 4 digit can be made which is divisible by 4 is (last two digits must be divisible by 4 i.e. 36, 56, 64) = 2! *3 = 6
Required probability =
Example 8: Two distinct numbers x and y are chosen at random from the set {1, 2, 3……..3n} . Find the probability that x2y2 is divisible by 3
Solution:
The number of ways in which 2 numbers can be chosen out of 3n =
Dividing all the numbers from 1 to 3n into 3 sets:
B1= {3, 6, 9………3n) = 3k type
B2= {1, 4, 7, 11……..3n2) = 3k+1 type
B3= { 5, 8, 11, ………3n1) = 3k+2 type
Now (x2y2) will be divisible by 3 if (xy) (x+y) is divisible by 3. For this, x and y should come from B1 or B2 or B3 or one is selected from B2 and other from B3.
The number of required ways =
The required probability =
Example 9: A point is chosen at random inside a circle. Find the probability that point is closer to the center of circle than to its circumference.
Solution:
Let S denotes the set of points inside the circle having radius R.
E represents set of points inside the circle having radius R/2.
Here, any point of E is closer to the center of the circle to its circumference.
Required probability = Area of E / Area of S = π (R/2)^{2} /πR^{2} = 1/4
Example 10: Out of 2n+1 ticket consecutively numbered, three are drawn at random. Find the probability that the numbers on them are in A.P.
Solution:
The number of ways in which 3 tickets can be selected from 2n is =
Let the three numbers drawn be x, y , z where x<y<z
Difference 
Possible triplets 
Number of possible triplets 
1 
(1,2,3) (2,3,4)…(2n1, 2n, 2n+1) 
2n1 
2 
(1,3,5) (2,4,6)….(2n3, 2n1, 2n+1) 
2n3 
…. 

….. 

n1 
(1,n , 2n1), (2, n+1, 2n), ( 3, n+2, 2n+1) 
3 
n 
(1, n+1, 2n+1) 
1 
The possible ways to select three tickets in A.P = 1+3+5….. +2n1 = n2
Required Probability =