Statistical Methods

1.7.1.Theorem on probability - Multiplication theorem

Unit 1 - Basic concepts of probability

1.7.1.Theorem on probability
II.Multiplication Theorem
Let A and B be two events with probability P (A) and P (B) respectively. Let P (B/A) denote the conditional probability of event B, given that event A has happened P (A/B) the conditional probability of event A, given that event B has happened. Then the probability of occurrence of both the events A and B denoted by P (AB) is given by,

P(AB) =P(A). P(B/A) =P(B). P(A/B)

If events A and B are independent, then,

P(AB) = P(A). P(B) = P(B). P(A)

Example 9
In a pond containing 100 fishes, 35 are marked. If two are caught one after another and without replacement, what is the probability that both the fishes caught are marked?
Answer
Let A denote the event of catching marked fish in the first draw and B denote the event of catching marked fish in the 2nd draw, then
P(A) =35/100
The probability of drawing marked fish in the 2nd draw, given that the first fish caught was marked is,
P(B/A) =34/99
Hence, P (both the fish caught are marked)= P(AB)= P(A). P(B/A) = (35/100)(34/99)=1190 /9900=0.12
Example 10
A pond contains 200 fishes of which 40 are marked. A second pond contains 300 fishes of which 50 are marked. One fish is drawn from each of the ponds. What is the probability that the fishes drawn are both marked?
Answer
Let A denote the 2nd event of catching marked fish from 1st pond, B denote the event of catching marked fish from 2nd pond.
Hence,

As the events A B are independent,

Example 11
An urn contains 7 white and 8 black pomfrets. A second urn contains 5 white 9 black pompfrets. One pompfrets is taken out at random from the first urn and put into the second urn without noticing its colour. A fish is then drawn at random from the second urn. What is the probability that it is a white pompfrets?
Answer
Two cases arise here
Case (i) Pompfrets taken first urns is white
Let A denote the event of drawing white pomfrets from first and let B denote the event of drawing white pompfrets from second urn.
Here, P(A) =7/15, P(B/A)=6/15
Hence, P(AB) =P(A).P(B/A) = (7/15) (6/15)=42/225=0.1867

Case (ii) Pompfret taken out from first urn is black
Let A denote drawing black pompfrets from first urn and let B denote drawing white pompfret from the second urn
Here, P(A) =8/15, P(B/A) =5/15
Hence, P(AB) =P(A).P(B/A) = (8/15) (5/15)=40/225
Therefore, required probability
= P(Case i) + P(Case ii)
= (42/225) + (40/225)
= 82/225
= 0.36