Elementary Statistics and Computer Application

Theorems of Probability

There are two important theorems of probability namely,

1. The addition theorem on probability
2. The multiplication theorem on probability

I. Addition Theorem on Probability

• (i) Let A and B be any two events which are not mutually exclusive
• P(A or B) = P(AUB) = P(A + B) = P(A) + P(B) – P(AUB) (or)
• = P(A) + P(B) – P(AB)
Proof





• (ii) Let A and B be any two events which are mutually exclusive
• P(A or B) = P(AUB) = P(A + B) = P(A) + P(B)
Proof:



We know that, n(AUB) = n(A) + n(B)
P(AUB) =
=
=
P(AUB) = P(A) + P(B)
Note:

• (i) In the case of 3 events, (not mutually exclusive events)
• P(A or B or C) = P(AUBUC) = P(A + B + C)
• = P(A) + P(B) + P(C) – P(AUB) – P(BUC) – P(AUC) + P(AUBUC)
• (ii) In the case of 3 events, (mutually exclusive events)
• P(A or B or C) = P(AUBUC) = P(A + B + C) = P(A) + P(B) + P(C)
Example:

• Using the additive law of probability we can find the probability that in one roll of a die, we will obtain either a one-spot or a six-spot. The probability of obtaining a one-spot is 1/6. The probability of obtaining a six-spot is also 1/6. The probability of rolling a die and getting a side that has both a one-spot with a six-spot is 0. There is no side on a die that has both these events. So substituting these values into the equation gives the following result:


• Finding the probability of drawing a 4 of hearts or a 6 or any suit using the additive law of probability would give the following:
• There is only a single 4 of hearts, there are 4 sixes in the deck and there isn't a single card that is both the 4 of hearts and a six of any suit.


• Now using the additive law of probability, you can find the probability of drawing either a king or any club from a deck of shuffled cards. The equation would be completed like this:


• There are 4 kings, 13 clubs, and obviously one card is both a king and a club. We don't want to count that card twice, so you must subtract one of it's occurrences away to obtain the result.

II. Multiplication Theorem on Probability

• (i) If A and B be any two events which are not independent, then (i.e.) dependent.
• P(A and B) = P(AUB) = P(AB) = P(A) . P(B/A)→ (I)
• = P(B) . P(A/B) →(II)
• Where P(B/A) and P(A/B) are the conditional probability of B given A and A given B respectively.
• Proof
• Let n is the total number of events
• n(A) is the number of events in A
• n(B) is the number of events in B
• n(AUB) is the number of events in (AUB)
• n is the number of events in
=







(ii) If A and B be any two events which are independent, then,
P(B/A) = P(B) and P(A/B) = P(A)
P(A and B) = = P(AB) = P(A) . P(B)

Note
(i) In the case of 3 events, (dependent)
= P(A) . P(B/A) . P(C/AB)

(ii) In the case of 3 events, (independent)
= P(A) . P(B) . P(C)
Example
So in finding the probability of drawing a 4 and then a 7 from a well shuffled deck of cards, this law would state that we need to multiply those separate probabilities together. Completing the equation above gives:



Given a well shuffled deck of cards, what is the probability of drawing a Jack of Hearts, Queen of Hearts, King of Hearts, Ace of Hearts, and 10 of Hearts?

In any case, given a well shuffled deck of cards, obtaining this assortment of cards, drawing one at a time and returning it to the deck would be highly unlikely (it has an exceedingly low probability).