What are the Different Types of Probability

What are the Different Types of Probability

Probability is simply the possibility of the happening of an event. There are three types of probabilities as you have already mentioned in your question.

1. Classical – There are ‘n’ number of events and you can find the probability of the happening of an event by applying basic probability formulae. For example – the probability of getting a head in a single toss of a coin is 1/2. This is Classical Probability.
2. Empirical or Experimental – This type of probability is based on experiments. Say, we want to know that how many times a head will turn up if we toss a coin 1000 times. According to the Traditional approach, the answer should be 500. But according to Empirical approach, we’ll first conduct an experiment in which we’ll toss a coin 1000 times and then we can draw our answer based on the observations of our experiment.
3. Subjective – This is solely based on the intuition of a person. It is vague and rarely accurate. For example – on a particular day, a person might feel that there is a 40% probability that it will rain on that day. There’s no formula to calculate it. It’s simply based on that person’s intuition.

Experimental Probability : Let there be n trials of an experiment and A be an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by
P(A) =  \(\frac{m}{n}\)
=  \(\frac{{No.\,of\,favourable\,\,cases\,\,to\,\,A}}{{No.\,of\,exhaustive\,\,cases\,\,to\,\,A}}\)
Note : It is obvious that 0 £ m £ n. If an event A is certain to happen, then m = n thus P (A) = 1.
If A is impossible to happen then m = 0 and so P (A) = 0. Hence we conclude that
0 ≤ P (A) ≤ 1
Further, if \(\bar A\) denotes negative of A i.e. event that A doesn’t happen, then for above cases m, n ; we shall have
P (\(\bar A\)) = \(\frac{{n – m}}{n} = 1 – \frac{m}{n}\) = 1– P (A)
P (A) + P (\(\bar A\)) = 1
Playing Cards :
(i)   Total : 52 (26 red, 26 black)
(ii)  Four suits : Heart, Diamond, Spade, Club – 13 cards each
(iii) Court Cards : 12 (4 Kings, 4 queens, 4 jacks)
(iv) Honour Cards: 16 (4 aces, 4 kings, 4 queens, 4 jacks)

Types of Experimental probability

1. Deterministic : Deterministic experiments are those experiments which when repeted under identical conditions produce the same result or outcome. For example, if we mark head (H) on both sides of a coin and it is tossed, then we always get the same outcome assuming that it does not stand vertically.
2. Random or probabilistic : If an experiment, when repeated under identical conditions, do not produce the same outcome every time but the outcome in a trial is one of the several possible outcomes, then it is known as a random or probabilistic experiment. For example, in the tossing of a coin one is not sure if a head (H) or tail (T) will be obtained, so it is a random experiment. Similarly, rolling an unbiased die is an example of a random experiment.

Types of Probability Example Problems with Solutions

Example 1:     A coin is tossed 500 times with the following frequencies of two outcomes :
Head : 240 times, tail : 260 times
Find the probability of occurrence of each of these event.
Solution:     It is given that the coin is tossed 500 times.
∴    Total number of trials = 500
Let us denote the event of getting a head and of getting a tail by A and B respectively. Then,
Number of trials in which the event
A happens = 240.
and, Number of trials in which the event
B happens = 260.
∴ P(A) = \(\frac{{Number\;of\;trials\;in\;which\;the\;event\;A\;happens}}{{Total\;number\;of\;trials}}\)
= \(\frac{{240}}{{500}}\) = 0.48
∴ P(B)  = \(\frac{{Number\;of\;trials\;in\;which\;the\;event\;B\;happens}}{{Total\;number\;of\;trials}}\)
= \(\frac{{260}}{{500}}\) = 0.52
Note : We note that P(A) + P(B) = 0.48 + 0.52. Therefore, A and B are the only two possible outcomes of trials.

Example 2:     A die is thrown 1000 times with the following frequency for the outcomes 1, 2, 3, 4, 5 and 6 as given below :
Outcome :  1      2       3      4       5       6
Frequency: 179  150    157    149   175   190
Find the probability of happening of each outcome.
Solution:     Let A_i denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6. Then,
P(E₁) = Probability of getting outcome 1
= \(\frac{{Frequency\;of\;1}}{{Total\;number\;of\;times\;the\;die\;is\;thrown}}\)
= \(\frac{{179}}{{1000}}\)  = 0.179
P(E₂) = Probability of getting outcome 2
= \(\frac{{Frequency\;of\;2}}{{Total\;number\;of\;times\;the\;die\;is\;thrown}}\)
= \(\frac{{150}}{{1000}}\) = 0.15
Similarly, we have,
P(E₃) = \(\frac{{157}}{{1000}}\) = 0.157,
P(E₄) = \(\frac{{149}}{{1000}}\) = 0.149,
P(E₅) = \(\frac{{175}}{{1000}}\) = 0.175
and, P(E₆) = \(\frac{{190}}{{1000}}\) = 0.19

Example 3:     The percentage of marks obtained by a student in the monthly unit tests are given below :

Find the probability that the student gets :
(i)   a first class i.e. at least 60 % marks
(ii)  marks between 70 % and 80 %
(iii) a distinction i.e. 75 % or above
(iv) less than 65 % marks.
Solution:     Total number of unit tests held = 5
(i)   Number of unit test in which the student gets a first class i.e. at least 60 % marks = 4.
∴ Probability that the student gets a first class
= \(\frac{4}{5}\) = 0.8
(ii)  Number of tests in which the student gets between 70 % and 80 % = 1.
∴    Probability that a student gets marks between 70 % and 80 % = \(\frac{1}{5}\) = 0.2.
(iii) Number of tests in which the student gets distinction = \(\frac{2}{5}\) = 0.4
(iv) Number of tests in which the student gets less than 65 % marks = 3
∴    Probability that a student gets less than 65 % marks = \(\frac{3}{5}\) = 0.6.

https://www.youtube.com/watch?v=5SwdIrhmCoE

Example 4:     On one page of a telephone directory, there were 200 telephone numbers. The frequency distribution of their unit place digit (for example, in the number 25828573, the unit place digit is 3) is given in the table below :
Digit :        0     1    2    3   4   5   6    7    8    9
Frequency: 22   26   22    22  20  10  14   28   16   20
A number is chosen at random, find the probability that the digit at its unit’s place is :
(i)   6
(ii)  a non-zero multiple of 3
(iii) a non-zero even number
(iv) an odd number.
Solution:     We have,
Total number of selected telephone numbers  = 200
(i)   It is given that the digit 6 occurs 14 times at unit’s place.
∴    Probability that the digit at unit’s place is  6 = \(\frac{{14}}{{200}}\) = 0.07
(ii)  A non-zero multiple of 3 means 3, 6 and 9.
Number of telephone number in which unit’s digit is either 3 or 6 or 9 = 22 + 14 + 20 = 56.
∴    Probability of getting a telephone number having a multiple of 3 at unit’s place = \(\frac{{56}}{{200}}\)  = 0.28
(iii) Number of telephone number having an even number (2 or 4 or 6 or 8) at unit’s place
= 22 + 20 + 14 + 16 = 72
∴ Probability of getting a telephone number having an even number at units place = \(\frac{{72}}{{200}}\) = 0.36
(iv) Number of telephone number having an odd digit (1 or 3 or 5 or 7 or 9) at units’ place
= 26 + 22 + 10 + 28 + 20 = 106
∴    Probability of getting a telephone number having an odd numbr at unit’s place = \(\frac{{106}}{{200}}\) = 0.53

Example 5:     A tyre manufacturing company kept a record of the distance covered before a tyre to be replaced. Following table shows the resuts of 1000 cases.

If you buy a tyre of this company, what is the probability that :
(i)   it will need to be replaced before it has covered 400 km ?
(ii)  it will last more that 900 km ?
(iii) it will need to be replaced after it has covered somewhere between 400 km and 1400 km ?
(iv) it will not need to be replaced at all ?
(v)  it will need to be replaced ?
Solution:     We have,
(i)   The number of trials = 1000
∴    Probability that a tyre will need to be replaced before it has covered 400 km = \(\frac{{210}}{{1000}}\) = 0.21
(ii)  The number of tyres that last more than 900 km = 385 + 80 = 465
∴    Probability that a tyre will last more than 900 km = \(\frac{{465}}{{1000}}\) = 0.465
(iii) The number of tyres which require replacement after covering distance between 400 km and 1400 km = 325 + 385 = 710.
∴    Probability that a tyre require replacement 400 km and 1400 km = \(\frac{{710}}{{1000}}\) = 0.71
(iv) The number of tyres that do not need to be replaced at all = 0
∴    Probability that a tyre does not need be replaced = \(\frac{{0}}{{1000}}\) = 0
(v)  Since all the tyres we have considered to be replaced, so
Probability that a tyre needs to be replaced = \(\frac{{1000}}{{1000}}\) = 1

Example 6:     Fifty seeds were selected at random from each of 5 bags of seeds, and were kept under standardised conditions favourable to germination. After 20 days the number of seeds which had germinated in each collection were counted and recorded as follows :

What is the probability germinated of :
(i)   more than 40 seeds is a bag ?
(ii)  49 seeds in a bag ?
(iii) more than 35 seeds in a bag ?
(iv) at least 40 seeds in a bag ?
(v)  at most 40 seed in a bag ?
Solution:     Total number of bags = 5
(i)   Number of bags in which more than 40 seeds germinated out of 50 seeds = 3.
∴    Probability of germinated of more than 40 seeds in a bag = \(\frac{3}{5}\)
(ii)  Number of bags in which 49 seeds germinated = 0.
∴ Probability of germination of 49 seeds = \(\frac{0}{5}\) = 0
(iii) Number of bags in which more than 35 seeds germinated = 5.
∴ Probability of germination of more than 35 seeds = \(\frac{5}{5}\) = 1.
(iv) Numbr of bags in which at least 40 seeds germinated = 4
∴ Probability of germination of at least 40 seeds = \(\frac{4}{5}\)
(v)  Number of bags in which at most 40 seeds germinated = 2.
∴ Probability of germination of at most 40 seeds = \(\frac{2}{5}\)

Example 7:     The distance (in km) of 40 female engineers from their residence to their place of work were found as follows –

Find the probability that an engineer lives :
(i)   less than 7 km from her place of work ?
(ii)  at least 7 km from her place of work ?
(iii) within \(\frac{1}{2}\) km from her place of work ?
(iv) at most 15 km from her place of work ?
Solution:     Total number of female engineers = 40
(i)   Number of female engineers living at a distance less than 7 km from their place of work = 10.
∴ Probability that a female engineer lives at a distance less than 7 km from her place of work = \(\frac{10}{40}\) = \(\frac{1}{4}\) = 0.25
(ii)  Number of female engineers living at least 7 km away from her place of work = 30
∴ Probability that a female engineer lives at least 7 km away from her place of work = \(\frac{30}{40}\) = 0.75
(iii) Since there is no engineer living at a distance less than \(\frac{1}{2}\) km from her place of work.
∴ Probability that an engineer within \(\frac{1}{2}\) km from her place of work = \(\frac{0}{40}\) = 0.
(iv) Number of engineers living at a distance of 15 km or less away from her place of work = 30.
∴ Probability that an engineer lives at most 15 km away from her place of work = \(\frac{30}{40}\) = 0.75

Example 8:     An insurance company selected 2000 drivers at random in a particular city to find a relationship between age and accidents. The data obtained are given in the following table:

Find the probabilities of the following events for a driver chosen at random form the life city:
(i)   being 18-29 years of age and having exactly 3 accidents in one year.
(ii)  being 30-50 years of age and having one or more accidents in a year.
(iii) having no accidents in one year.
Solution:     Total number of drivers = 2000
(i)   The number of drivers who are 18-29 years old and have exactly 3 accidents in one year is 61.
∴ Probability of a driver being 18-29 years of age and has exactly 3 accidents = \(\frac{{61}}{{2000}}\) = 0.0305
(ii)  The number of drivers 30-50 years of age and having one or more accidents in one year = 125 + 60 + 22 + 18 = 225.
∴ Probability of a driver being 30-50 years of age and having one or more accidents = \(\frac{{225}}{{2000}}\) = 0.1125
(iii) The number of drivers having no accidents in one year = 440 + 505 + 360 = 1305
∴ probability of a driver having no accident in one year = \(\frac{{1305}}{{2000}}\) = 0.653

Example 9:     Find the probability that a number selected at random from the numbers 1 to 25 is not a prime number when each of the gievn number is equally likely to be selected.
Solution:     Here S = {1, 2, 3, 4, …., 25}
Let E = event of getting a prime number = {2, 3, 5, 7, 11, 13, 17, 19, 23}.
Then,    n (E) = 9
∴    P(E) = \(\frac{{n(E)}}{{n(S)}}\) = \(\frac{9}{{25}}\)
Required probability
= 1 – P(E) = \(\left( {1 – \frac{9}{{25}}} \right) = \frac{{16}}{{25}}\)

Example 10:     Eleven bags of wheat flour, each marked 5 kg. actually contained the following weights of flour (in kg.) :
4.97  5.05  5.08  5.03  5.00  5.06  5.08  4.98  5.04 5.07  5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:     Total number of bags = 11
Number of bags containing more than 5 kg of flour = 7
Therefore, probability of bags containing more then 5 kg of flour
= \(\frac{{Number\;of\;bags\;containing\;more\;than\;5kg\;flour}}{{Total\;number\;of\;bags}}\)
= \(\frac{7}{11}\)

Example 11:     The record of a weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times.
(i)   What is the probability that on a given day it was correct ?
(ii)  What is the probability that it was not correct on a given day ?
Solution:     The total number of days for which the record is available = 250
(i)   P(correct forecast)
= \(\frac{{Number\;of\;days\;when\;the\;forecast\;was\;correct}}{{Total\;number\;of\;days\;for\;which\;the\;record\;is\;available}}\)
= \(\frac{175}{250}\) = 0.7
(ii)  The number of days when the forecast was not correct = 250 – 175 = 75.
P(not correct forecast) = \(\frac{75}{250}\) = 0.3

Example 12:     A box contains 20 balls bearing numbers, 1, 2, 3, 4, … 20. A ball is drawn at random from the box. What is the probability that the number on the balls is
(i) An odd number      (ii) Divisible by 2 or 3
(iii) Prime number      (iv) Not divisible by 10
Solution:     Total number of possible outcomes = 20
Probability = \(\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}\)
(i)   Number of odds out of first 20 numbers = 10
Favourable outcomes by odd = 10
P(odds)
= \(\frac{{Favourable\,\,outcomes\,\,of\,\,odd}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}\)
= \(\frac{{10}}{{20}} = \frac{1}{2}\)
(ii)  The numbers divisible by 2 or 3 are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20.
Favourable outcomes of numbers divisible  by 2 or 3 = 13
P (numbers divisible by 2 or 3)
= \(\frac{{Favourable\,\,outcomes\,\,of\,\,divisible\,\,by\,\,2\,\,or\,\,3}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}\)
= \(\frac{{13}}{{20}}\)
(iii) Prime numbers out of first 20 numbers are 2, 3, 5, 7, 11, 13, 17, 19
Favourable outcomes of primes = 8
P(primes)
= \(\frac{{Favourable\,\,outcomes\,\,of\,\,primes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}\)
= \(\frac{{8}}{{20}} = \frac{2}{5}\)
(iv) Numbers not divisible by 10 are 1, 2, .. 9, 11, …19
Favourable outcomes of not divisible by 10 = 18
P(not divisible by 10)
= \(\frac{{Favourable\,\,outcomes\,\,of\,\,not\,\,divisible\,\,by\,\,10}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}\)
= \(\frac{{18}}{{20}} = \frac{9}{10}\)