TOPICS
Unit-15(Examples)

Example-1 :-  A coin is tossed 1000 times with the following frequencies:
Head : 455, Tail : 545
Compute the probability for each event.

Solution :-
 Since the coin is tossed = 1000 times, 
 The total number of trials = 1000. 
 Let H = Event of outcomes Head, T = Event of outcomes Tail
 Then, the number of times H happens, i.e., the number of times a head come up, is 455.
 
 So, P(H) = (no. of Heads)/(Total no. of trials)
     P(H) = 455/1000
     P(H) = 0.455

 The number of times T happens, i.e., the number of times a tail come up, is 545.
 Similarly,
     P(T) = (no. of Tails)/(Total no. of trials) 
     P(T) = 545/1000
     P(T) = 0.545
    

Example-2 :-  Two coins are tossed simultaneously 500 times, and we get
Two heads : 105 times
One head : 275 times
No head : 120 times
Find the probability of occurrence of each of these events.

Solution :-
  Let us denote the events of getting two heads, one head and no head by E₁, E₂ and E₃, respectively. So,
  Total outcomes = 105 + 275 + 120 = 500
  P(E₁) = 105/500 = 0.21
  P(E₂) = 275/500 = 0.55
  P(E₃) = 120/500 = 0.24
    

Example-3 :-  A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table : Find the probability of getting each outcome.

Solution :-
  Let Eᵢ denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6. 
  Total outcomes = 179 + 150 + 157 + 149 + 175 + 190 = 1000.
  
  P(E₁) = 179/1000 = 0.179
  P(E₂) = 150/1000 = 0.150
  P(E₃) = 157/1000 = 0.157
  P(E₄) = 149/1000 = 0.149
  P(E₅) = 175/1000 = 0.175
  P(E6) = 190/1000 = 0.190
    

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 Table: Without looking at the page, the pencil is placed on one of these numbers, i.e., the number is chosen at random. What is the probability that the digit in its unit place is 6?

Solution :-
  Let E be the event that the digit 6 being in the unit place.
  No. of frequency of 6 = 14
  Total no. of telephone numbers = 200
  P(E) = 14/200
  P(E) = 0.07
    

Example-5 :-  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) Number of days when the forecast was correct = 175   
  P(the forecast was correct on a given day) = 175/250 = 0.7

(ii) Number of days when the forecast was not correct = 250 - 175 = 75 
  P(the forecast was not correct on a given day) = 75/250 = 0.3 
    

Example-6 :- A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results 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 4000 km?
(ii) it will last more than 9000 km?
(iii) it will need to be replaced after it has covered somewhere between 4000 km and 14000 km?

Solution :-
(i) The total number of trials = 1000. The frequency of a tyre that needs to be replaced before it covers 4000 km is 20.
  So, P(tyre to be replaced before it covers 4000 km) = 20/1000 = 0.02

(ii) The frequency of a tyre that will last more than 9000 km is 325 + 445 = 770
  So, P(tyre will last more than 9000 km) = 770/1000 = 0.77

(iii) The frequency of a tyre that requires replacement between 4000 km and 14000 km is 210 + 325 = 535.
  So, P(tyre requiring replacement between 4000 km and 14000 km) = 535/1000 = 0.535
    

Example-7 :- The percentage of marks obtained by a student in the monthly unit tests are given below: Based on this data, find the probability that the student gets more than 70% marks in a unit test.

Solution :-
  The total number of unit tests held is 5. 
  The number of unit tests in which the student obtained more than 70% marks is 3.
  So, P(scoring more than 70% marks) = 3/5 = 0.6
    

Example-8 :-  An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) 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 from the 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.
  So, P (driver is 18-29 years old with exactly 3 accidents) = 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 .
  So, P(driver is 30-50 years of age and having one or more accidents) = = 225/2000 = 0.1125

(iii)  The number of drivers having no accidents in one year = 440 + 505 + 360 = 1305.
  Therefore, P(drivers with no accident) = 1305/2000 = 0.653 
    

Example-9 :-  Consider the frequency distribution table, which gives the weights of 38 students of a class.
(i) Find the probability that the weight of a student in the class lies in the interval 46-50 kg.
(ii) Give two events in this context, one having probability 0 and the other having probability 1.

Solution :-
(i) The total number of students is 38, and the number of students with weight in the interval 46 - 50 kg is 3. 
  So, P(weight of a student is in the interval 46 - 50 kg) = 3/38 = 0.079

(ii) For instance, consider the event that a student weighs 30 kg. 
  Since no student has this weight, the probability of occurrence of this event is 0. 
  Similarly, P (a student weighing more than 30 kg) = 38/38 = 1
    

Example-10 :-  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 of germination of
(i) more than 40 seeds in a bag?
(ii) 49 seeds in a bag?
(iii) more that 35 seeds in a bag?

Solution :-
  Total number of bags is 5. 
(i) Number of bags in which more than 40 seeds germinated out of 50 seeds is 3.
  P(germination of more than 40 seeds in a bag) = 3/5 = 0.6

(ii) Number of bags in which 49 seeds germinated = 0.
  P(germination of 49 seeds in a bag) = 0/5 = 0

(iii) Number of bags in which more than 35 seeds germinated = 5.
  So, the required probability = 5/5 = 1
    
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