In a random experiment, out of ‘n’ exhaustive, mutually exclusive, equally likely, independent events if ‘m’ of them are favorable to the occurrence of an event, say, ‘A’, then the probability of an event ‘A’, denoted by P(A) is

According to this definition, P(A) is found a priori without actual experimentation. For instance, if we keep using examples like fair coins, unbiased dice and standard deck of cards, we can state the answer in advance (a priori) without tossing a coin, or rolling a die or drawing a card.
Sometimes P(A) can be expressed by saying that “the odds in favour of A are m : (n-m) or the odds against A are (n-m) : m
P(A) +P(Ac) = 1 or P(Ac) = 1-P(A)

Limitations of classical definition probability

The classical definition of probability has certain drawbacks and fails at times in different situations as described below.
1.This definition emphasizes that the
events must be equally likely. Thus,
it fails when various outcomes of a
trail are not equally likely.
For example if a die is biased that gives numbers greater than 3 more often than the numbers less than 3, then the occurrence of numbers on the die is not equally probable.
Similarly, the definition also fails when we have to find the probability of an operation to be successful as the event of success or failure are not equally probable.
2. This definition is useful only when
we deal with card games, dice
games, coin tossing and the like. It
fails in situations when we try to
apply it to less orderly decision
problem we encounter in
management.
3. It does not consider those
situations that are unlikely but that
could conceivably happen. Like the
occurrence of a coin landing on its
edge or our room burning down
while watching TV etc., which are
extremely unlikely but not
impossible.
4. In case the exhaustive number of
cases in a trial is infinite, the
definition fails to give the required
probability.

5. In some situation there may be a
difference of opinion with respect to
the ways of forming the possible
and favourable outcomes.

Examples

1. Three light bulbs are selected at random from 15 bulbs of which 5 are defectives. Find the probability that
(i) None is defective
(ii) Exactly one is defective
(iii) At least one is defective
(iv) At most one is defective
Solution:
(i) - Exhaustive number of cases
15c3 = 455 ways
- Favourable cases
10c3 = 120 ways
Let A1 be an event that none of the bulb chosen is defective, then
P(A1)=120/455=0.26
(ii) - Exhaustive number of cases
15c3 = 455 ways
- Favourable cases
5c1 x 10c2 = 5 x 45 = 225 ways
Let A2 be an event that exactly one bulb chosen is defective, then
P(A2)=225/455=0.49
(iii) Let A3 be an event that at least one
bulb chosen is defective, then
P(A3)= 1-P(A1) = 1-0.26=0.74

(iv) Let A4 be an event that at most
one bulb chosen is defective, then
P(A4) = P(A1) +P(A2) = 0.26 + 0.49 =0.75

Posted in Posted by waytofeed at 8:49 AM