# statistics sampling methods

## For Independent Events

The theorem states that the probability of the simultaneous occurrence of two events that are independent is given by the product of their individual probabilities.

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

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

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

### Example

Problem Statement:

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

Solution:

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

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

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

Using multiplicative theorem for independent events

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

## For Dependent Events (Conditional Probability)

As defined earlier, dependent events are those were the occurrences or nonoccurrence of one event effects the outcome of next event. For such events the earlier stated multiplicative theorem is not applicable. The probability associated with such events is called as conditional probability 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 probability of occurrence of event A when event B has already occurred.

Similarly the conditional probability 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 probability 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.

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

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