RRB ALP Maths Trade Notes on Statistics and Probability

Study Notes on Statistics

Example 1:

The measurements of heights of all creatures in the world are data. All numerical characteristics are called variables. A large number of observations on a single variable can be summarized in a table of frequencies. Any particular pattern of variation is termed as distribution.

Frequency is the number of times a particular value occurs in a set of data. Usually, we would record the frequency of data in a frequency table.

In statistics, mode, median, mean and range are typical values to represent a pool of numerical observations. They are calculated from the pool of observations.

Measures of Central Tendency

Mode

Mode is the most common value among the given observations. For example, a person who sells ice creams might want to know which flavor is the most popular.

Median

Median is the middle value, dividing the number of data into 2 halves. In other words, 50% of the observations are below the median and 50% of the observations are above the median. If the number of observations is odd, then the median is  observation. If the number of observations is even, then the median is the mean of  observations.

Mean

Mean is the average of all the values in the set. Its value is given by . For example, a teacher may want to know the average marks of a test in his class.

Example 2:

Find the mean, mode and median of the following set of points:

15, 14, 10, 8, 12, 8, 16, 13

Solution:

First arrange the point values in ascending order (or descending order).

8, 8, 10, 12, 13, 14, 15, 16

Mean = (8+8+10+12+13+14+15+16)/8 = 96/8

= 12.

Mode = 8 (since it has maximum frequency)

The number of point values is 8, an even number. Hence the median is the average of the 2 middle values.

Median 

Skewness

In a normal distribution, the mean, median, and mode are all the same value. In various other symmetrical distributions it is possible for the mean and median to be the same even though there may be several modes, none of which is at the mean. By contrast, in asymmetrical distributions the mean and median are not the same. Such distributions are said to be skewed, i.e., more than half the cases are either above or below the mean.

Example 3:

The distribution of salaries shown in Figure has a pronounced skew distribution.

Figure: A distribution with a very large positive skew

Table shows the measures of central tendency for these data. The large skew results in very different values for these measures. No single measure of central tendency is sufficient for data such as these. There is no need to summarize a distribution with a single number. When the various measures differ, our opinion is that you should report the mean, median, and either the tri-mean or the mean trimmed 50%. Sometimes it is worth reporting the mode as well. In the media, the median is usually reported to summarize the center of skewed distributions. You will hear about median salaries and median prices of houses sold, etc. This is better than reporting only the mean, but it would be informative to hear more statistics.

Table: Measures of central tendency

Measures of Dispersion:

Dispersion measures the degree of scatteredness of the variable about a central value. The following are the measure of dispersion:

Range

Range is the difference between the maximum and minimum values in the set. It is the simplest measure of variation to find.

RANGE = MAXIMUM VALUE - MINIMUM VALUE

Example 4:

Ten students were given a mathematics test. Time taken by them to complete the test is listed below. Find the range of these times

8 12 7 11 12 9 8 10 8 13 (in min.)

Solution:

It can be seen that maximum time taken by a student to complete the test is 13 min and minimum time taken is 7 min. So,

Range = Max. value - Min. value = 13 - 7 = 6 min

Mean Deviation

It is the arithmetic mean of the absolute deviations of the terms of the distribution from its statistical mean. It is least about median.

Mean Deviation for Ungrouped Data:

Let x₁, x₂, x₃ ....x_n are n values of variable X and k be the statistical mean (A.M, median, mode) about which we have to find the mean deviation. The mean deviation about k is given by

Mean Deviation for Grouped data

a) Discrete Frequency Distribution:

where d_i = x_i - k, f_i be frequencies and N = total frequency.

The mean of given discrete frequency distribution is given by

The median of given discrete frequency distribution is found out by arranging the observations in ascending order and then calculating the cumulative frequency. The observation, whose cumulative frequency is equal to or just greater than N/2, is the required median.

b) Continuous Frequency Distribution:

The mean of a continuous frequency distribution is calculated with the assumption that the frequency in each class is centered at its mid-point.

 where di=xi-k, fi be frequencies and N=total frequency

Arithmetic mean

a = assumed mean, h = common factor and N = total frequency.

where,

l = lower limit of median class

f = frequency of the median class

h = width of the median class

c = cumulative frequency of the class just preceding the median class.

Example 5:

50 villages are inspected for calculating total number of towers. List below shows the distribution in number. Find the mean deviation of the distribution.

No. of towers 5, 6, 7, 8, 9, 10

No. of villages 8, 12, 9, 15, 4, 2

Solution:

Mean can be calculated from the formula given above as

Mean distribution is calculated as

Variance

The variance of a variate is the arithmetic mean of the squares of all deviations from the mean.

Standard Deviation

It is the proper measure of dispersion about the mean of a set of observations and is expressed as positive square root of variance.

Example 6:

In the previous example we have calculated the mean deviation of the data.

Using the same data calculate variance and standard deviation.

Solution:

Variance is given as:

Now, Standard Deviation is

Analysis of Frequency Distribution (Measures of Variability)

In order to compare the variability of two series with same mean, which are measured in different units, merely calculating the measures of dispersion are not sufficient, but we require such measures which are independent of the units. The measure of variability which is independent of units is called coefficient of variation (C.V.) and defined as

Example 7:

From the data in the above two examples, find the coefficient of variation (C.V.) of no. of towers.

Solution:

From the formula above C.V. can be calculated as

C.V. = 0.6145 (which is the S.D.)/7.02 (which is the mean) × 100 = 8.7535

Example 8:

Given sets

A = {0,5,10,15,25,30,40,45,50,71,72,73,74,75,76,77,78,100} and

B = {0,22,23,24,25,26,27,28,29,50,55,60,65,70,75,80,85,90,95,100}.

Here simple inspection will indicate very different distribution, however, it is found that

Min. = 0, Max. = 100

So range = 100,

Median = 50,

Middle of the bottom half of the set (Q1) = 25,

Middle of the upper half of the set (Q3) = 75 is the same in both cases.

Hello Friends,

Today we are presenting an article on “How to Solve Probability Questions in competitive examination. Probability questions are an important part of Quantitative aptitude section of most competitive exams like SBI, IBPS, PO/Clerk, LIC-AAO etc. These questions are asked frequently so it becomes really relevant to know the right technique of solving these questions.

Study Notes on Probability

What is Probability?

‘Probability’ in simple terms tell us about the chance of something occurring. The probability of an event happening ranges between 0 to 1. That means the value of probability can never be a negative number or a number greater than 1.

Consider this, if it’s cloudy outside then two things can happen. First, either it will rain or second, it won’t rain. So, the total events are ‘2’ (raining or not raining). And, the probability of raining is 1/2

So, Probability of an event happening = Concerned Events / Total Events

Probability of an event happening is denoted by P(E)
Probability of an event not happening is denoted by P(Ē).

And, P(E) + P(Ē) = 1

Types of Events:
1. If ‘And’ event is given then we multiply or count events together.
2. If ‘Or’ event is given then we ‘add’ the two or more events.

Classical Cases:
The classical cases discussed in probability are based on the following points:

1. Dice:

The dice used here is the one used to play ‘Ludo’. A typical dice has numbers 1, 2, 3, 4, 5 and 6 are written over its six faces as shown below.

When a dice is thrown the number that appears on upper face is the concerned event. The questions based on dice are mainly of two types(not exhaustive):

(I) When Only One dice is thrown once:
In such cases, the number rolling on playing dice is either 1 or 2 or 3 or 4 or 5 or 6.
Here, the concerned event of rolling out ‘1’ is 1 only because 1 is written only one face. And, total events = 6 because total different numbers written different faces are six in total.
So, Probability of rolling number 1 = 1/6

Similarly, Probability of rolling number 2 ( or any number from 3 to 6) = 1/6

Question asked:
What is the probability of getting an even number on rolling a dice?

Now, the concerned event should have an even number which are 2, 4 and 6
So, Concerned Event = 3
Total Event = 6
So, Probability = 3/6 = 1/2

(II) When Two dices are thrown:
In such cases, either of the two things happen either two dices are thrown simultaneously and the numbers appearing on top faces of both dices are noted and summed up; or one dice is thrown two times in a row and the numbers appearing on the top faces in the two times are noted and summed up. Whatever is done, the treatment is same in either the cases, two dices at once or rolling one dice twice. So, this summed up number is the concerned event in such question.

The various combinations of numbers that can turn up on throwing two dices (or one dice twice) can be listed as below –

For example: (1, 6) shows that ‘1’ would turn up on dice 1 and ‘6’ would turn up on dice 2. Here, total outcome is the total number of combinations stated above: (1,1,); (1,2) ……. (6,5); (6,6) = 36

The questions asked can be of following type:
Qs. 1 – What is the probability of getting a combination of ‘5’ and ‘3’ on throwing two dices?

Solutions – Now the concerned event should have number ‘5’ and ‘3’ so concerned events = 2{(3,5) and (5,3)}
Total events = 36
⇒ Probability = 2/36 = 1/18

Qs. 2 –  What is the probability of getting a sum of ‘10’ on rolling a dice twice?

Solutions – Now, the concerned events should have sum of 10 i.e. No. on dice 1 +No. on dice 2 = 10  
This can be seen in these cases: (4,6); (5,5) and (6,4)
So, Concerned Events = 3
And, we know Total Events are always ‘36’
So, Probability  = 3/36 = 1/12

2. Coins:

Coin is a currency token which has two faces, one is head and other is tail. So, when throw a coin in air and when it lands it might have either a head or tail. Coin questions can be three types as shown below:

I. One Coin once:
When a coin is tossed is only once then there can be two outcomes either a head or a tail. In such cases, total events = 2
Question: What is the probability of getting a head in a toss?
Solutions – Concerned event = 1(One head)
Total Event = 2
⇒ P(E) = 1/2
II. Two Coins or One Coin Twice:
When two coins are tossed together or one coin is tossed in twice then following outcomes can be obtained:

Here, ‘H’ = Head; ‘T’ = Tail.

(H,T) shows that on coin 1 it’s Head while on coin 2 it’s Tail. Here, we can see that Total Event = 4

Question: What is the Probability of getting at most one head on tossing a coin?
Solutions – At most one head means there can be 0 head or there can be 1 head.
So, Concerned Event = 3 {(H,T) (T,H) (T,T)}
Total Events = 4
⇒ P(E) = 3/4

III. Three Coins or One Coin Thrice:

When three coins are tossed together or one coin is tossed in thrice then following outcomes can be obtained:
Here, ‘H’ = Head; ‘T’ = Tail.
(HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT)
Here, (HTH) shows that coin 1 has a head, coin 2 has a Tail while coin 3 has a head.
In such cases, Total Events = 8

Question: Three unbiased coins are tossed. What is the probability of getting at least 2 heads?
Here, at least heads means there can be 2 heads and 3 heads.
So, Concerned Events = 4 {(HHH), (HHT), (HTH), (THH)}
Total Events = 8
⇒ P(E) =4/8 =1/2

3. Cards:

There are four kinds of symbol used in playing cards. The etymology for different symbols is as below:
i) Spade ⇒ ♠ ⇒ Black in color (13 in number)
ii) Club ⇒ ♣ ⇒ Black in color (13 in number)
iii) Heart ⇒ ♥ ⇒ Red in color (13 in number)
iv) Diamonds ⇒ ♦ ⇒ Red in color (13 in number)

Each of these 4 variants have 13 numbers each as 1, 2, 3 …. 10 and, Jack, Queen, King and Ace. There are
1) 26 red cards and 26 black cards.
2) 4 cards each of 1, 2, 3 …. 10 and, Jack, Queen, King and Ace.
3) 13 cards each of Spade, Heart, Club and Diamond.
So, in total there are 13 × 4 = 52 cards.

Types of questions asked:

1. One card drawn:
In such types of question a card is drawn from the pack of cards. Here, the Total Events = 52

Question: What is the probability of getting a King of Spade or Queen of Heart in one draw?

Solutions – Here, ‘King of Spade or Queen of Heart’ means that either the card can be the Spade King or Heart Queen. Clearly, there is only one King of Space and only one Queen of Heart.
So, Concerned Event = 2
Total Event = 52
⇒ P(E) = 2/52 = 1/26

2. More than One Card drawn:
In such questions when more than One card is drawn we use the concept of Combination formula. For example the question below:

Question – Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart is ____ ?
Here, the ‘one spade’ card has to be drawn from 13 spade cards, so its event = ¹³C₁
And, ‘one heart’ card has to be drawn from 13 heart cards, so its event  = ¹³C₁
So, Concerned Event = ¹³C₁×¹³C₁ 
Total Events (as two cards are to be drawn from 52) = ⁵²C₂
So, P(E) = (¹³C₁×¹³C₁)÷ ⁵²C₂  =  (13 x 13) x 2 / 52 x 51 = 13 / 102

Question –  Two cards are drawn together from a pack of 52 cards. The probability that either both are red or both are Kings ____ ?
Here, the ‘both red’ cards have to be drawn from 26 red cards, so its event = ²⁶C₂ 
Or, ‘both king’ cards have to be drawn from 4 King cards, so its event  = ⁴C₂
But there are Two Red Kings which are common in both Red cards & King cards so they have been double counted. And, they will be taken out of two red king cards only so they’ll be deducted from concerned events.
So, Concerned Event = ²⁶C₂  + ⁴C₂ – ²C₂
Total Events (as two cards are to be drawn from 52) = ⁵²C₂
So, P(E) = (²⁶C₂  + ⁴C₂ – ²C₂)÷ ⁵²C₂  = [ (26 x 25) + (4 x 3) – 1 ] / 52 x 51 = 330 / 1326 = 55 / 221

4. Balls:

In such questions, a bag contains certain balls and some ball(s) is(are) drawn.

I. One ball drawn:

Question  – In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked at random. What is the probability that it is neither red nor green?

Solutions – Total Events = 8 (red) + 7 (blue) + 6 (green) = 21
Since, the selected ball has to be neither red nor green then it’d be Blue and blue balls are 7.

So, Concerned Event = 7 ⇒ P(E) = 7/21 = 1/3 

II. More than One ball drawn without replacement:

Question – A box contains 10 black or 10 white balls. The probability of drawing two balls of same colors?

Solution – Total Events (as 2 balls are drawn) = ²⁰C₂
The balls drawn can either be both black color or both white color and for ‘OR’ event we add up the two events.
So, Concerned Event = ¹⁰C₂ (if black) + ¹⁰C₂ (if white)
⇒ P(E) = (¹⁰C₂ + ¹⁰C₂)÷ ²⁰C₂ = 10 x 9 + 10 x 9 / 20 x 19 = 9 / 19

Question – A box contains 10 black and 10 white balls. The probability of drawing two balls of same colors?

Solution – Total Events (as 2 balls are drawn) = ²⁰C₂
The balls drawn can be both black color or both white color. for ‘And’ event we multiply the two events.

First case – If both ball is white –

So, Concerned Event = ¹⁰C0 (if black) x ¹⁰C₂ (if white)

⇒ P(E) = (¹⁰C0 x ¹⁰C₂)÷ ²⁰C₂ = 1 x 10 x 9 / 20 x 19 = 9 / 38

Second case – If both ball is black –

So, Concerned Event = ¹⁰C2 (if black) x ¹⁰C0 (if white)

⇒ P(E) = (¹⁰C2 x ¹⁰C0)÷ ²⁰C₂ = 10 x 9 x 1 / 20 x 19 = 9 / 38

Other Miscellaneous questions:

Qs 1. Four persons are chosen at random from a group of 3 men, 2 women and 4 children. What is the probability of exactly two of them being children?

Solution – Total People = 3 + 2 + 4 = 9
Since, 4 people are chosen from 9 so Total Event = ⁹C₄
And, since 2 of the chosen people have to be children so these 2 person have to be from 4 children so Concerned Event for children= ⁴C₂
And, other 2 people will be from Men & Women (3+2 = 5), their Concerned event = ⁵C₂
⇒ Concerned Event (for all 4 people) = ⁴C₂ ×⁵C₂
⇒ P(E) = (⁴C₂ ×⁵C₂) _÷ ⁹C_{4 = (4 x 3 x 5 x 4) x (2 x3) / 9 x 8 x 7 x 6 = 10 / 21} 

Qs. 2. A and B give exam. Chance of husband’s selection is 1/7 and of wife’s selection is 1/5. Find probability of only one of them is selected.

Solution – P( husband selecting) = 1/7
P (husband not selecting) = 1 – 1/7 = 6/7
P( wife selecting) = 1/5
P (wife not selecting) = 1 – 1/5 = 4/5
Probability of only one of them is selected = P( husband selecting)× P (wife not selecting) + P( wife selecting)× P (husband not selecting) =  (1/7 x 4/5) + (1/5 x 6/7) = 10/35 = 2/7.

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