Showing posts with label Constructing sample spaces. Show all posts
Showing posts with label Constructing sample spaces. Show all posts

Sunday, 19 February 2012

Constructing sample spaces

Constructing sample spaces:
Sample space is generally denoted by S, Ω or U. Fundamentally it’s the set of all random trials; like in tossing a coin the sample space is (head, tail). If we toss two coins then sample space for this random trial is (head, head), (head, tail), (tail, head), (tail, tail). In the trial of tossing a single sided die sample space s is (1, 2, 3, 4, 5, and 6). It’s not necessary that one trial should have only one sample space; like in a trial of drawing a card from a standard deck of 52 playing cards one probability of outcome is could be the rank (ace through king) and the another sample space could be suits i.e. (club, diamond, hearts or spades). The subsets of sample space are called event. Problems would be critical after increasing the number of sample spaces. So it would be problematic when there are infinite sample spaces means there are infinite events.
Sample space construction:
Sample space is the set of all possible outcomes and it’s necessary to consider all possibilities. It may be a difficult task and for this purpose counting principle can be used. If there is more than one event it’s important to determine all possibilities that exist. It can be stated that:
“If there are A ways of an event to occur and B ways for occurrence of second event then there are A.B ways for both to occur”. This is the concept of the counting principle.
An example is given below which can help to understand sample space better. (know more about cbse sample papers, here)
Example: one jar contains 1 red, 3 green, 2 blue or 4 yellow balls. Then what would be the probability of each outcome if a single ball is chosen from the jar.
Solution: Sample space for this random trial is
S = (red, green, blue, yellow)
Probability can be calculated as:
Probability:  P (red ball) = 1/10
P (green ball) = 3/10
P (blue ball) = 2/10 = 1/5
P (yellow ball) = 4/10 = 2/5
In the next session we will discuss about Sampling errors. 

Wednesday, 15 February 2012

Proportions

Here we are going to learn about Proportions (some portion of this topic taken from ICSE class 10 syllabus).
When any two ratios are equal, we say that the two ratios are in proportion. It is the relationship between two ratios whose output is same and constant.
It is represented by a/b  :: c/d
or
a : b :: c : d
Here the ratio of a: b is proportion to c: d. Both symbols :: and = are used to represent the proportionality of two ratios.
If we have a : b : : c : d , it is read as a is to b as  c is to d
In the above statement we have 'a' as First  term, 'b' as second term , 'c' as third term and  'd' as the fourth term.
In this a and d are called extreme terms or extremes and b and c are middle terms and are also called means
If the given four numbers are in proportion, then the product of means is equal to the product of extremes.
In order to check that the two ratios are in proportion, we simply check if the product of extremes and the product of means are equal.
Let us see how to solve proportion with the help of an example:
Example: solve Proportion problem  60 : 105  :: 84 :147 .
Sol: We first take the product of means ie 105 * 84 = 8820
now we take the product of extremes i.e. 60 * 147 = 8820
Here we observe that the product of means = product of extremes. So the two ratios are in proportion.
This can also be checked by converting both ratios in lowest terms, if both the values are same, they are in proportion
Let us try it for the same data:
60 / 105  ,we divide numerator and denominator by 5 and get
= 12 /21 , again dividing by 3 we get
= 4 / 7
Similarly we write 84/147 , dividing & multiplying  by 3 we get
= 28 / 49
Now dividing & multiplying by 7 we get
= 4 /7
So they are in proportion.

In next post we will talk on Estimation of Solutions in Grade VI. For more information on Substitution Method, you can visit our website