Geometric Probability: Formula, Illustration, Model

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When we measure and define the uncertainty of the occurrence of an event using numbers, this is what we call Probability in mathematics. The scale which we use to express the chance that whether an event will occur or will not occur ranges between 0-1. 

Read Also: Events in Probability

We can also represent probability of an event as a percentage also where 0% denotes an impossible event and 100% implies an event surety. In our daily life, we come across multiple events which define probability, and those events are as follows: It will probably rain today, I am not sure whether she will qualify this exam or not, most probably, Suman will come second this year in final examination. The concept of probability is extensively used in a variety of fields such as Physical Sciences, Commerce, Biological Science, Medical Science, Weather Forecasting.


Basic Terms related to Probability

  • The result of a random experiment is an Outcome.

For E.g. When we roll a dice, the probability of getting six is an outcome.

  • The set of outcomes is termed as an Event.

For E.g. When we roll dice, the probability of getting a number less than 4,5 is an event.

Note: An Event can only have a single outcome.

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Geometric Probability

If we talk about Geometric Probability, then it is the likelihood or chance that one will hit the particular area of a figure.

Formula for Geometric Probability = Desired Outcome/Total Outcome

Example: Geometric Probability Here Probability would be 1/4 = 25%

There are infinite outcomes when it comes to the Geometric Probability Concept.

Geometric Probability provides a useful approach by allowing us to transform probability problems into geometric problems. The process on which geometric distribution is based upon is Binomial Process (a series of independent trials with two possible outcomes). This concept is used to determine the probability that a specified number of trials will take place before the first success occurs. Otherwise it can also be used to figure out the probability that a specified number of failures will occur before the first success takes place.


Geometric Probability Formulas

  • To find out the probability that a given number of trials takes place until the first success occurs,

P(X=x) = (1-P)x-1 P for x=1, 2,3…….

Here x can be any whole number(integer).

X is a geometric random variable, x is the number of trials required until the first success occurs and P is the probability of success on a single trial.

  • To find out the probability that a given number of failures occur before the first success, the formula will be:

P(X=x) = (1-P)X P where x = 0,1,2,3…..

X represents the no. of failures that occur before the first success.

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Geometric Probability Illustrations

  1. A person looking for a new job that is both challenging and fulfilling. What is the probability that he will quit his job zero times, one time, two times and so on till he finds a perfect job?
  2. A pharmaceutical company is working on a new drug to treat a certain disease having minimal side effects. What is the probability that zero drugs fail the test, one drug fails the test, two drugs fail in the test and so on till they can design an ideal drug for this disease?

Geometric Probability Categorization

There are 3 types of geometric probabilities, one for each of the commonly used dimension of space including length, area and volume.

  • 1D P(Target) = Target length/ Total length
  • 2D P(Target) = Target Area/ Total Area
  • 3D P(Target) = Target Volume/ Total Volume

Read More: Events and their Types in Probability

Geometric Probability Model:

Geometric Probability
Model Length Angle Measure Area
Example length Angle Measure GP Area
Sample Space All Points on AD All points in the circle All points in the rectangle
Event All points on BC All points in the shaded region All points in the triangle
Probability P = BC/AD P = Measure of angle/360° P = Area of triangle/Area of Rectangle

Things to Remember

  • For an event to fall under the category of geometric probability, following 4 conditions must be satisfied.
  1. Each observation is one of the two possibilities – either a success or a failure.
  2. All observations are independent.
  3. The probability of success (P) is the SAME for each observation.
  4. The variable of the internet is the number of trials required to get the FIRST success.
Empirical Probability Theoretical Approach of Probability
Value of Cos 180 Cosine Rule

Sample Questions

Ques. A person is throwing dice and will stop once he gets 5 since there are 6 possible outcomes, find out the probability of the first 3 trials?

Sol. There are 6 possible outcomes, hence

Probability of success, P = 1/6 => 0.17

Probability of success, P = 1- 0.17 => 0.83

The person gets number 5 for the first time,

No. of failures before the first success is zero.

Thus, x = 0 and k = 1

We know that, P(x=k) = (1-P)k-1 P

After substituting all the corresponding values, P(x=0) = (0.83)0* 0.17

= 01.7

The person gets number 5 for the second time,

No. of failures before the first success is 1, thus x = 1 and k = 2

Hence, P(x=1) = (1 – 0.17)2-1 *(0.17)

= (0.83)*(0.17) => 0.14

Ques. Find the probability of randomly choosing a point in Region B.

Probability of randomly choosing a point in Region B

Sol. As we know that, Geometric Probability = Desired Area/Total Area

                                                                       = Area B/ Total Area (for this figure)

Area of Region B = Length* Breadth = 8*15 => 120 cm2

Total Area = Length* Breadth = 20*35 => 700

After putting all the values in the given formula,

P = 120/700 = 6/35 => about 17%

Check Important Notes for Class 12 Probability

Ques. You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you more likely to get 10 points or 0 points?

Probability of getting 10 Points or 0 Points

Sol. Area of Red Circle = \(\Pi\)r2 = 3*3* => 9\(\Pi\)

Area of Blue region = (18*18) - 81\(\Pi\)

Therefore, Probability (10 points) = Area of Red Circle/ Area of Board

= 9 \(\Pi\)/ 18*18 => 0.0873 (approx.)

And Probability (0 points) = Area of Blue region/ Area of Board

= ((18*18) – (81\(\Pi\)))/ (18*18)

= 0.215 (Approx)

Ques. If a point is selected at random in the circle, calculate what will be the probability that it lies-

  • In the red sector
  • In the green sector
  • In Area except the green sector

Sector having Green and Red Colour

Sol. For the red sector, Probability will be 

P = Area of red shaded region/Total Area

= 180° / 360° => 50%

Similarly, For the green sector probability will be

P = Area of green shaded region/Total Area

= 120°/360° => 1/3 or 33%

And lastly for the area except the green sector, probability will be

P = (Area of red shaded region+ Area of Blue shaded region) /Total Area

= (180°+60°)/ 360° => 67% (approx.)

Read More: What is a Set?

Ques. Find the probability that a point will be chosen randomly inside the rectangle is in each given shape. Round to the nearest hundredth.

Probability that a point will be chosen randomly inside the rectangle

Sol. For the equilateral triangle, The area of triangle would be,

A = ½ aP = ½ (6) (36√3) => 187 m2

Area of rectangle is A = bh = 45*20 => 900 m2

The probability is P = 187/900 => 0.21 (approx.)

For the Trapezoid

The area of trapezoid is A = ½ (b1+b2)h

= ½ (3+12) (10) => 75 m2

Area of rectangle is A = bh = 45*20 => 900 m2

The probability is P = 75/900 => 0.08 (approx.)

For the Circle,

The area of circle is A = \(\Pi\)r2 = \(\Pi\)(62) = 36\(\Pi\) => 113.1 m2

Area of rectangle is A = bh = 45*20 => 900 m2

The probability is P = 113.1/900 => 0.13 (approx.)

Also Read:

CBSE X Related Questions

1.

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them

Monthly consumption 
(in units)

 Number of consumers

65 - 85 

4

85 - 105

5

105 - 125

13

125 - 145

20

145 - 165

14

165 - 185

8

185 - 205

4

      2.
      Find the sums given below :
      1. \(7 + 10\frac 12+ 14 + ....... + 84\)
      2. \(34 + 32 + 30 + ....... + 10\)
      3. \(–5 + (–8) + (–11) + ....... + (–230)\)

          3.

          The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

          Length (in mm)

          Number of leaves

          118 - 126

          3

          127 - 135 

          5

          136 - 144

          9

          145 - 153

          12

          154 - 162

          5

          163 - 171

          4

          172 - 180

          2

          Find the median length of the leaves. 
          (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

              4.

              Solve the following pair of linear equations by the substitution method. 
              (i) x + y = 14 
                  x – y = 4   

              (ii) s – t = 3 
                  \(\frac{s}{3} + \frac{t}{2}\) =6 

              (iii) 3x – y = 3 
                    9x – 3y = 9

              (iv) 0.2x + 0.3y = 1.3 
                   0.4x + 0.5y = 2.3 

              (v)\(\sqrt2x\) + \(\sqrt3y\)=0
                  \(\sqrt3x\) - \(\sqrt8y\) = 0

              (vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
                  \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

                  5.
                  Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) \(x + y = 5\),\( 2x + 2y = 10\) (ii)\( x – y = 8 , 3x – 3y = 16\) (iii) \(2x + y – 6 = 0\) , \(4x – 2y – 4 = 0\) (iv) \(2x – 2y – 2 = 0,\) \( 4x – 4y – 5 = 0\)

                      6.
                      Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

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