When learning about Independent event, one will understand the importance of two events in one sample space. Subjective probability is based on a personal evaluation in favor of that independent favor when trying to measure various probabilities. Subjective probabilities may not cooperate with third axiom of probability.
So here we’ll suppose things are positively working out in favor of A and positioning a : b condition: P(A) = a/(a+b). If the opposite situation occurs, where things are positioning a negative circumstance and quite against A with a : b then we’ll get P( A) = b/(a+b).
Let us consider this following example and get the measure the subjective probability:
Six men in a company of 15 are engineers. If 3 men were picked out of the 15, then calculate probability of at least one engineer?
Here 3 men were picked out of 15 where if we’ll separate minimum one engineer in each group and then measure we’ll get these three cases:
- 1 engineer and 2 non-engineers
- 2 engineers and 1 non-engineer
- And 3 engineers and 0 non-engineer
So we’ll get these three separate cases of probabilities:
(6/1) (9/2)/ (15/3) = 216/455, (6/2) (9/1)/ (15/3) = 135/455, (6/3) (9/0)/ (15/3) = 20/455
Hence we’ll get the answer:
P ( at least one engineer) = 216/455 + 135/455 + 20/455 = 371/455
Here is another example to consider:
An article manufactured by a company consists of two parts I and II. In the process t1 manufacture of part L, 9 in the total of 100 are stated as defective. Similarly, 5 in total of 100 are stated as defective in the manufacture of part II. Calculate probability that assembled article will not have any defective pieces.
The first thing to point out is that there is a mention of defective part both in part I and II and we have to calculate the part that’ll not be defective.
So, first part I: P( defective part I) = 9/100
Second part II: P (non-defective part I) = 1-(9/100) = 0.91
P (defective part I) = 5/100
P (non-defective part II) = 1- (5/100) = 0.95
Now comes the part where we describe Part I and II as independent events then
P = (non-defective part I). (non-defective part II)
So we’ll place: