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Probability Multiplecative Theorem
  • 时间:2024-11-03

Statistics - Probabipty Multippcative Theorem


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For Independent Events

The theorem states that the probabipty of the simultaneous occurrence of two events that are independent is given by the product of their inspanidual probabipties.

${P(A and B) = P(A) imes P(B) \[7pt] P (AB) = P(A) imes P(B)}$

The theorem can he extended to three or more independent events also as

${P(A cap B cap C) = P(A) imes P(B) imes P(C) P (A,B and C) = P(A) imes P(B) imes P(C) }$

Example

Problem Statement:

A college has to appoint a lecturer who must be B.Com., MBA, and Ph. D, the probabipty of which is ${frac{1}{20}}$, ${frac{1}{25}}$, and ${frac{1}{40}}$ respectively. Find the probabipty of getting such a person to be appointed by the college.

Solution:

Probabipty of a person being a B.Com.P(A) =${frac{1}{20}}$

Probabipty of a person being a MBA P(B) = ${frac{1}{25}}$

Probabipty of a person being a Ph.D P(C) =${frac{1}{40}}$

Using multippcative theorem for independent events

${ P (A,B and C) = P(A) imes P(B) imes P(C) \[7pt] = frac{1}{20} imes frac{1}{25} imes frac{1}{40} \[7pt] = .05 imes .04 imes .025 \[7pt] = .00005 }$

For Dependent Events (Conditional Probabipty)

As defined earper, dependent events are those were the occurrences or nonoccurrence of one event effects the outcome of next event. For such events the earper stated multippcative theorem is not apppcable. The probabipty associated with such events is called as conditional probabipty and is given by

P(A/B) = ${frac{P(AB)}{P(B)}}$ or ${frac{P(A cap B)}{P(B)}}$

Read P(A/B) as the probabipty of occurrence of event A when event B has already occurred.

Similarly the conditional probabipty of B given A is

P(B/A) = ${frac{P(AB)}{P(A)}}$ or ${frac{P(A cap B)}{P(A)}}$

Example

Problem Statement:

A coin is tossed 2 times. The toss resulted in one head and one tail. What is the probabipty that the first throw resulted in a tail?

Solution:

The sample space of a coin tossed two times is given as S = {HH, HT, TH, TT}

Let Event A be the first throw resulting in a tail.

Event B be that one tail and one head occurred.

${ P(A) = frac{P(TH,TT)}{P(HH,HT,TH,TT)} = frac{2}{4} =frac {1}{2} \[7pt] P(A cap B) = frac{P(TH)}{P(HH,HT,TH,TT)} =frac{1}{4} \[7pt] So P (A/B) = frac{P(A cap B)}{P(A)} \[7pt] = frac{frac{1}{4}}{frac{1}{2}} \[7pt] = frac{1}{2} = 0.5 }$ Advertisements