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The situation between two events is explained by the multiplication rule of probability. Assume you are playing with two dice with your friends. You make a list of the sample space's different possible outcomes. When one of your mates rolls the dice, he keeps the outcome a secret. Instead, he stated that adding the two numbers gives an even number. Are you able to predict the outcome of his throw at this point? It'll be interesting to see how this new information affects the probabilities of different outcomes. This is the fundamental principle behind the multiplication theorem on probability. Let's get to know it a little better.
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Key Terms: Conditional probability,mathematical theorem on probability,multiplication theorem,independent events.
Definition
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In conditional probability, we know that since some of the potential events have already happened, the probability of occurrence of another event is affected. When we know that a specific event B has occurred, we focus on B rather than S when measuring the probability of event A occurring provided B.
Using the above example of two dice, the possible outcomes are as follows:
S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}.
The sample space S contains 36 elements. The probability of occurrence of any of the possible outcome is P(Ei) = 1/36. We don't know what the outcome of the friend's dice throw was. We do, however, know that the sum of the numbers is even. Let's see how this knowledge influences the outcome's probability.
The sum of the numbers is an even number, as shown by Event A. A = (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6), (6, 2), (6, 4), (6, 6). This means that out of 36 outcomes these 18 outcomes are now only possible and the remaining are not.
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The probability for each of these outcomes is P(A |Ei) = 18⁄36 = ½.This example illustrates how additional information can alter the probability of an event happening.
Multiplication Theorem on Probability Video Explanation
1st Multiplication Theorem on Probability
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For two events A and B,
P(A ∩ B) = P(A) P(B | A), P(A) > 0.
or, P(A ∩ B) = P(B) (A | B), P(B) > 0.
Here, P(B | A) represents the conditional probability of the event B happening if the event A has already occurred. When the event B has already occurred, P(A | B) represents the conditional probability of A occurring.
Proof: From the concept of conditional probability, we have
P(A | B) = P(A ∩ B) ⁄ P(B).
Rewriting the preceding, we now have, P(A ∩ B) = P(B) P(A | B).
Similarly, P(B | A) = P(A ∩ B) ⁄ P(A).
⇒ P(A ∩ B) = P(A) P(B | A).
The probability of two events A and B occurring simultaneously is equal to the product of the probability of one of these events and the conditional probability of the other, given that the first has occurred, according to a mathematical theorem on probability.
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2nd Multiplication Theorem on Probability
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For two events A and B such that P(B) > 0, P(A | B) ≤ P(A).
Proof: The number of common outcomes in A and B is obviously less than or equal to the number of outcomes in either of the events.
n(A ∩ B) ≤ n(A) … (i),
and, n(B) ≤ n(S) … (ii)
Dividing (i) and (ii), we get,
n(A ∩ B) ⁄ n(B) ≤ n(A) ⁄ n(S)
⇒ P(A | B) ≤ P(A).
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Multiplication Theorem for Independent Events
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For independent cases, the multiplication theorem on probability can be expanded. From the theorem, we have, P(A ∩ B) = P(A) P(B | A). If the events A and B are independent, then, P(B | A) = P(B). The above theorem reduces to
P(A ∩ B) = P(A) P(B).
This shows that the probability of any of these happening at the same time is the product of their respective probabilities.
Extension of Multiplication Theorem of Probability to n Events
For n events A1, A2, … , An, we have
P(A1 ∩ A2 ∩ … ∩ An) = P(A1) P(A2 | A1) P(A3 | A1 ∩ A2) … × P(An |A1 ∩ A2 ∩ … ∩ An-1)
For n independent events, the multiplication theorem reduces to
P(A1 ∩ A2 ∩ … ∩ An) = P(A1) P(A2) … P(An).
For more reference, check this.
Read More: Sum of Squares
Sample Questions on Multiplication Theorem on Probability
Question: What is the multiplication rule?(1 mark)
Answer: According to the multiplication theorem, "the probability of occurrences of given 2 events, or, in other words, the probability of the intersection of 2 given events, is equal to the product obtained by finding the product of the probability of occurrence of both events."
Question: What are the rules for probability?(1 mark)
Answer: The addition, multiplication, and complement rules are the three fundamental rules that are associated with chance. ‘P(A or B) = P(A) + P(B) – P(A B)' is how we express them.
Question: What are the 3 axioms of the probability?(1 mark)
Answer: The following are the three axioms of probability:
- In an event A, ‘P(A) ≥ 0’. In English, that’s ‘For an event A, the probability of ‘A’ is superior or equal to zero (0)’.
- When ‘S’ is the sample space in an experiment i.e. the set of all possible results, ‘P(S) = 1’.
- In case ‘A’ and ‘B’ are commonly exclusive products, ‘P(A ∪ B ) = P(A) + P(B)’.
Question: A box contains 5 brown and 7 black pebbles. What is the probability of drawing a brown pebble if the first pebble drawn is balck? The balls drawn are not replaced in the box.
(frac {5}{11})
(frac {8}{1})
(frac {4}{18})
(frac {14}{11}) (1 mark)
Answer: (a)
Explanation: Total number of balls = 5 = 7 = 12
The probability of the first ball to be blue = (frac {7}{12})
The probability of drawing red ball as the second ball out without replacement = (frac {5}{11}).
Questions. Find the probability of drawing a diamond card in each of the two consecutive draws from a well shuffled pack of cards, if the card drawn is not replaced after the first draw?(2 marks)
Answer: Let A be the event of drawing a diamond card in the first draw and B be the event of drawing a diamond card in the second draw. Then,
After drawing a diamond card in first draw, 51 cards are left out of which 12 are diamond cards.
Therefore, P(B/A) = Probability of drawing a diamond card in the second draw when a diamond has already been drawn in the first draw
= 12/51 = 4/17
The required probability = 
P (B/A) = ¼ x 4/17 = 1/17
Question: A bag contains 19 tickets that are numbered from 1 to 19. A ticket is drwn and then another ticket is drwn without replacement. Find the probability that both tickets will show even numbers. (2 marks)
Answer: Let A be the event of drawing an even numbered ticket in the first draw and B be the event of drawing an even numbered ticket in the second draw.
Then, the required probability =
P (B/A) …. (i)
As there are 19 tickets numbered 1 to 19 in a bag out of which 9 are even numbered viz. 2, 4, 6, 8, 10, 2, 14, 16, 18.
therefore, P(A) = 9/19
As the ticket drawn in the first draw is not replaced, therefore the second ticket drawn is from the remaining 18 tickets out of which 8 are even numbered.
therefore, P(B/A) = 8/18 = 4/9
Substituting these values in equation i, we get
Required probability =
P(B/A) = 9/19 X 4/9 = 4/ 19.
Question: Which of this represents the multiplication theorem of probability?
a) P(A ∩ B) = P(B) P(B/A)
b)P (A ∩ B) = P (A) P(B/A)
c)P (A ∩ B) = P (A) P(B/B)
d)P (A ∩ B) = P(A) P(A/A)(1 mark)
Answer: (b)
Explanation: The multiplication theorem of probability states that if A and B are two events of a random experiment with P(A) > 0 and P(B) > 0, then P(A ∩ B) = P(A) P(B/A) or P(A ∩ B) = P(B) P(A/B).
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