Concept of Probability and Probability Distribution

Dr. Abhijit Kar Gupta, Phys. Dept, Panskura Banamali College, WB, India, e-mail: [email protected] [email protected]

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The Concept of Probability and Probability Distribution

The concept of probability applies to a random event . It is associated with chance and its  prediction in mathematical terms. Example: The outcome of a coin tossing, the throwing of dice in the game of Ludo, hitting a target with a gun, whether it will rain today etc. If a similar event occurs many times, sometimes the outcome o f the events is favorable (that means that I am looking for) and other times it is not. For example, if we toss a coin the outcome will be either Head or Tail. Tail. Suppose, I want Head, then one of the two outcomes is favourable here. In case ca se of Dice throwing, if I want six, then one of the 6  possible outcomes is favourable. Thus if the dice is face up with ‘six’, it is favourable otherwise it is not. We can calculate the relative frequency of the favouravle events by taking the ratio of  number of favourable events with the total number of events. This ratio can be different if  the number of events or trials are different. When the total number of events (or trials) is made very large, the relative frequency tends towards a fixed value. This can then be said to be the probability of occurrence of that event. It is a positive value, v alue, a fraction between 0 and 1 by definition. Definition of Probability: Probability (p) is the ratio of the number of favourable events (n) to the total number of  events when the total number of events (N) is made very large. n  p = , where  N  is very large.  N  Thus if  n  N  , all the events are favourable,  p = 1 . When no events are favourable, n 0 ,  p = 0 . We have 0 <  p < 1 . =

=

Suppose, m = the number of events that are not favourable. We We can write, n + m =  N  . ∴ We can write, the probability of not occurring the particular outcome is m  N  − n n q= = = 1 − = 1 −  p .  N   N   N  Thus we have  p + q = 1. This says, in any trial, something must happen, either in favou r  or not in favour. favour. Example: The probability of occurring Head in a coin toss is  p = 1 / 2 . The probability of not occurring Head (or occurring Tail) Tail) is q = 1 / 2. Thus p = q = 1 / 2 + 1 / 2 = 1 . In other way 1 we can say that the probability of occurring Head  p H  = and the probability of  2 1 occurring Tail  pT  = , so that  p H  +  pT  = 1 . 2

Dr. Abhijit Kar Gupta, Phys. Dept, Panskura Banamali College, WB, India, e-mail: [email protected] [email protected]

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Thus the sum of the probabilities of all the events is = 1. This is also called   Normalization. For a dice throw, among 6 different outcomes, the probability of occurring one, is  p1 , the probability of occurrence of two is  p 2  p1

=

1 6

6

6

6

1 6

and so on. Thus

+  p 2 +  p3 +  p 4 +  p5 +  p6 = 1 + 1 + 1 + 1 + 1 + 6

=

6

1 =1. 6

Problem: A box contains 4 identical balls, one red and three blue. What is the probability of  drawing a red ball from the box? Disjoint events: If two random events A and B do not occur at the same time, that is, either A occurs or B and one is not dependent on other, they are called disjoint (mutually exclusive) events.

Addition of Probabilities: Suppose in a given experiment, a random event A occurs with probability  p ( A) and another with probability p ( B) . If the two events are disjoint or mutually mutuall y exclusive, then the joint probability of the two events, that is, the probability that either the event A or B will take place, is given by  P (A  P (A or B) = P(A) + P(B). Two events are complimentary when no other events occur other than these. In that case P(A) + P(B) = 1. In this case it is said that the events A and B form a complete group. group. If the random events A, B, C, D form a complete group then P(A)+P(B)+P(C)+P(D)=1. Example: In the case of coin tossing, P(H or T) = P(H) + P(T) = ½ + ½ =1. In the case of Dice throwing, the probability of occurring ‘six’ or ‘one’, P(6 or 1) = P(6) + P(1) =1/6 + 1/6 = 2/6. Example: Two fair dice are thrown. What will be the probability that the total score is either 10 or each score on each dice is more than 4?

The event A: (5,5), (4,6), (6,4) ∴ P(A) = ∴

P(B) =

P(A or B)

4 36

≠

12

. The event B: (5,5), (5,6), (6,5), (6,6)

9 36 6 12 9 P(A) + P(B). The events A and B are not mutually exclusive. exclusive.

36

P(A or B) =

=

1

1

7

∴

6

36

=

1

=

1

3

. However, P(A) + P(B) =

1

+ =

. Thus

Dr. Abhijit Kar Gupta, Phys. Dept, Panskura Banamali College, WB, India, e-mail: [email protected] [email protected]

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Multiplication of Probabilities: If the two events A and B are independent then the joint probability of occurring the two events at the same time is  P (A  P (A and B) = P(A) + P(B). Example:

If we toss two coins together, then there will be 4 kinds of events (HH, HT, TH, TT). For  any of the events to occur, o ccur, the probability is  P   HH  =  P   HT  =  P  TH  =  P  TT  = 1 / 4 . Again, we may see that  P  may say that the probability of occurring  HH  + P   HT  + P  TH  + P  TT  = 1 . So we may either HH or HT or TH or TT is given by the rule of addition of probabilities. Also we see that the probability probab ility of any of the events, say P   HH   P   HT 

1

1

1

1

1

4

2

2

= = × = P  H  × P  H 

and

1

= = × =  P  H  × P T 

etc. Note that the event of head or tail in two coins happens 4 2 2 the same time and they are independent. The joint probability of any number of  independent events taking place at the same time is the product of the probabilities for the individual events.

Conditional Probability: If A and B are two events eve nts and the event B happens h appens only when A happens, the conditional  probability of B given A is denoted by P(B/A). ∴

The probability that both the events happen together is P(A and B) = P(B/A)

× P(A).

Problem: In a population of 100, there are 60 men and 40 women. Among the men 20 are graduates and among the women 10 are graduates. What is the probability of finding a graduate woman if we pick a person randomly from the population?

If we choose from the women only on ly,, the probability that the woman will be 10 1 = . However, the probability that the chosen person will be a graduate is P(G/W) = 40 4 40 4 = . Thus the probability that the chosen person woman is P(W) = person will be woman 100 10 1 4 1 = . and a graduate is P(W and G) = P(G/W).P(W) = × 4 10 10

 Solution.

Find out the probability that the randomly chosen person will be graduate man.

Dr. Abhijit Kar Gupta, Phys. Dept, Panskura Banamali College, WB, India, e-mail: [email protected] [email protected]

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Probability of compatible events: Random events A and B are compatible if in a given trial or experiment both the events can occur. The joint probability of two compatible events is calculated from the formula: P(A or B) = P(A) + P(B) – P(A and B). Problem: The probability of hitting a target by a gun A is 8/10 and by another gun B is 7/10. Two guns are fired together. What is the probability that the target can be destroyed?

If any of the guns hits the target, it will be destroyed. Also, Also, both the guns can 8 7 8 7 94 + − . = hit the target. Thus the probability of destroying the target is = = 10 10 10 10 100 0.94.

 Solution:

How do we calculate average in a probabilistic event?  Normally,  Normally, we calculate the average (or mean) of a variable by summing up the values given and dividing up by the number :  x

=

 x1 + x 2

+ .... + x n n

. This is arithmetic average.

In a probabilistic event, the average ave rage is calculated in the following way:  x

= ∑ xi P i =  x1 × P 1 + x 2 × P 2 + ......... i

For the average of square:  x 2

= ∑ xi 2 P i =  x1 2 × P 1 + x 2 2 × P 2 + ......... i

In the previous case of arithmetic average it may be assumed that the events occur at a 1 constant probability, probability,  P = . This means any of the events is equally equa lly probable. n

What is a probability distribution?  A probability distribution is a set of probabilities describing the chan ces of events to occur. Any frequency distribution can be turned into an empirical probability distribution. When we calculate the distribution d istribution from a finite sample, it is a frequency distribution. How is it calculated?

Dr. Abhijit Kar Gupta, Phys. Dept, Panskura Banamali College, WB, India, e-mail: [email protected] [email protected]

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Let us take some data. In order to obtain a distribution, we divide the data into different classes or groups (like different age groups in a class or at a place). We We check how many man y data fall into a class or group. The ratio of the number in a class (class frequency) to the total number of data is called the relative frequency. The relative frequency is assumed to be the empirical probability. probability. W Wee generally represent them graphically through a bar  diagram. diagram. This empirical probability tends towards actual probability  P ( x ) at some point  x when we take the number of data to be very large. The distribution curve is a plot of  P ( x ) against  x where  P ( x ) is the probability for the data point  x to occur. There are some well known probability distributions: (i) Normal or Gaussian distribution, (ii) Binomial distribution, (iii) Poisson distribution. Normal or Gaussian distribution is that which normally occurs for random events. The distribution curve looks like a bell shaped curve, symmetrically spread around the mean value  x .

The spread of the distribution is given by the variance following formula: σ  

2

=

1 n

∑ ( x − x ) i

i

2

σ  

2

(sigma-square) by the

2 =  x 2 − x , for a sample of size n ( n numbers of data points). The

square root of  σ  2 is called the standard deviation, SD =

σ  

.

When the value of  σ  2 is small, the spread of the distribution is small. When the value of  2 is large, the spread of the distribution is also large. So, the variance gives the σ   dispersion of data points from the mean value. va lue. The mathematical formula for the above normal curve is  P ( x ) =

1

[

(

)

2

]

exp −  x − x / 2σ  2 , where "exp" denotes the

2π  σ   exponential function, e = 2.71828. The distribution is symmetric and the peak (most  probable value) of a normal distribution occurs at  x =  x (mean value).

Dr. Abhijit Kar Gupta, Phys. Dept, Panskura Banamali College, WB, India, e-mail: [email protected] [email protected]

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 P ( x )

 Normal Distribution  x

 x

 Approximately 95% of area under curve falls within 2 SDs on either side of mean, and  about 68% of the area falls within 1 SD from mean. Total Total area under the curve is 100%.

Obtaining Normal Distribution: Throwing of a dice. A dice has 6 faces. As we throw the dice randomly any of the scores 1, 2, 3, 4, 5, 6 will appear with equal probability  P  = 1 / 6 . If we go on throwing the dice randomly and indefinitely, indefinitely, we may ma y generate a large number n umber of scores. We may think that this set of  large number of scores is our population. Now we want to take samples from this  population to calculate the properties of this population. Sample size 1: (taking one data at a time)

For a single throw of the dice, the probability distribution is given by the following table.

x

1

2

3

4

5

6

P(x)

1/6

1/6

1/6

1/6

1/6

1/6

The mean value of the distribution  x =

The variance, 2

1

1

σ  

1

2

= ∑ xi 2 P i − x

× +2 × +3 6

2

6

2

1 6

1

2

I

1+ 2 + 3 + 4 + 5 + 6 6

6

2

2

= 1

+4 × +5 × +6 2

6

Sample size 2: (taking two data at a time)

2

2

7   35 × −      = . 6  2   12 1

7

= = 3.5 .

Dr. Abhijit Kar Gupta, Phys. Dept, Panskura Banamali College, WB, India, e-mail: [email protected] [email protected]

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If we throw two dice together, we shall obtain (1,1), (1,2), (2,1), (2,2) ….and so on. There are 36 possible outcomes. We now consider the mean value of the two data as our new variable. The probability table is as follows.

 x =

 x1 + x 2

1

1 .5

2

2.5

3

3 .5

4

4. 5

5

5.5

6

2

P(x)

1

2

3

4

5

6

5

4

3

2

1

36

36

36

36

36

36

36

36

36

36

36

If we plot  P ( x) against  x , we get a symmetrical distribution with the peak at  x

=

3.5 .

We can check c heck that  x = 3.5 from the symmetry of the distribution. The variance,

σ  

2

= {12 ×

1 36

+ (1.5) 2 ×

2 36

+ ........} − (3.5) 2 =

35 24

.

In this way if we check with sample size 3 (that means, throwing 3 dice together) and more, we obtain a clear c lear symmetric distribution with mean value at  x = 3.5 . The spread of the distribution (variance) becomes consistently smaller as we go on taking larger  samples. The distribution takes a definite symmetrical shape around the mean value for  large sample sizes. This distribution is normal or Gaussian distribution.

⇑

This lecture note is compiled from many sources (Books and websites) as an introduction to the subject. This is intended to be distributed privately among students. Books: 1. Statistics 1 & 2 (Advanced Level Leve l Mathematics), Steve Dobbs and Jane Miller. Cambridge University Press. 2. Differential and Integral Calculus –II , N. Piskunov, Piskunov, MIR   pub., Moscow some websites.

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