The condition of two events is explained with the help of the multiplication rule probability.
Assume two events J and K that are associated with a sample space S. When both the J and K events occur, it is denoted by the set J ∩ K. Hence, the occurrence of both the events J and K are represented by ( J ∩ K ). This event can also be written as JK and using the properties of conditional probability you can obtain the probability of event JK. In this article, you will also learn the multiplication theorem of probability.
Let us now learn the multiplication law of probability.
We know that the conditional probability of an event J given the fact that K has occurred is represented by P ( J | K ), and the formula goes by:
P ( J | K ) = \[\frac{P(J\bigcap K)}{P(K)}\]
Where P ( K ) ≠ 0
P ( J ∩ K ) = P ( K ) × P ( J | K ) ……………………………………..( A )
P ( K | J ) = P ( K ∩ J ) / P ( J )
Where P ( J ) ≠ 0.
P ( K ∩ J ) = P ( J ) × P ( K | J )
Since, P ( J ∩ K ) = P ( K ∩ J )
P ( J ∩ K ) = P ( J ) × P ( K | J ) ………………………………………( B )
From (1) Jnd (2), we get:
P ( J ∩ K ) = P ( K ) × P ( J | K ) = P ( J ) × P ( K | J ) where,
P ( J ) ≠ 0, P ( K ) ≠ 0.
The result that is obtained above is known as the multiplication rule of probability.
For independent events J and K, P( K | J ) = P ( K ). The equation (2) can be modified into,
P( J ∩ K ) = P ( K ) × P ( J )
Now that we have learned about the multiplication rules that are implemented in probability, such as;
P ( J ∩ K ) = P ( J ) × P ( K | J ) ; if P ( J ) ≠ 0
P ( J ∩ K ) = P ( K ) × P ( J | K ) ; if P ( K ) ≠ 0
Now, let us focus on learning about the multiplication theorems for independent events of J and K.
The probability of simultaneous occurrence of two independent events will be equal to the product of the probabilities of the events if J and K are two independent events for a random experiment. Hence,
P ( J ∩ K ) = P ( J ) . P ( K )
Now, from understanding the multiplication rule, we know that;
P ( J ∩ K ) = P ( J ) × P ( K | J )
Since J and K events are independent here, we get;
P ( K | J ) = P ( K )
On solving again, we get;
P ( J ∩ K ) = P ( J ) . P ( K )
Hence, proved.
Question 1: A bag contains 15 red and 5 blue balls. Without the replacement of the balls, two balls are drawn from a bag one after the other. What is the probability of picking both the balls as red?
Solution:
Let J and K represent the events that the first ball is drawn and the second ball drawn is the red balls.
Now, we have to find P ( J ∩ K ) or P ( JK ).
P ( J ) = P ( the red balls drawn in the first draw ) = 15 / 20
Now, only 14 red balls and 5 blue balls are the ones left in the bag. The conditional probability is that of a Probability of drawing a red ball in the second draw too, where the drawing of the second ball depends on the drawing of the first ball.
Therefore, the conditional probability of K on J will be,
P ( K | J ) = 14 / 19
By using the multiplication rule of probability,
P ( J ∩ K ) = P ( J ) × P ( K | J )
P ( J ∩ K ) = \[(\frac{15}{20})\] * \[(\frac{14}{19})\] = 21 / 38
1) State and Prove the Multiplication Theorem of Probability or Explain the Proof of Multiplication Rule Probability.
We know that the conditional probability of an event J given the fact that K has occurred is represented by P ( J | K ), and the formula goes by:
P ( J | K ) = [{P (J ∩ K)}/P(K)]
Where P ( K ) ≠ 0
P ( J ∩ K ) = P ( K ) × P ( J | K ) ……………………………………..( A )
P ( K | J ) = P ( K ∩ J ) / P ( J )
Where P ( J ) ≠ 0.
P ( K ∩ J ) = P ( J ) × P ( K | J )
Since, P ( J ∩ K ) = P ( K ∩ J )
P ( J ∩ K ) = P ( J ) × P ( K | J ) ………………………………………( B )
From (1) Jnd (2), we get:
P ( J ∩ K ) = P ( K ) × P ( J | K ) = P ( J ) × P ( K | J ) where,
P ( J ) ≠ 0, P ( K ) ≠ 0.
The result that is obtained above is known as the multiplication rule of probability.
For independent events J and K, P( K | J ) = P ( K ). The equation (2) can be modified into,
P( J ∩ K ) = P ( K ) × P ( J ).
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