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Statistics Examples

P (A) = 0.1

$P (A) = 0.1$ ,

P (B) = 0.13

$P (B) = 0.13$ ,

P (B g i v e n A) = 0.13

$P (B g i v e n A) = 0.13$

Step 1

Two events are independent events when the occurrence of one does not affect the probability of the other.

P (A | B) = P (A)

$P (A | B) = P (A)$ and

P (B | A) = P (B)

$P (B | A) = P (B)$ .

P (A | B) = P (A)

$P (A | B) = P (A)$

P (B | A) = P (B)

$P (B | A) = P (B)$

Step 2

P (B | A)

$P (B | A)$ should equal to

P (B)

$P (B)$ because the occurrence of

A

$A$ should not effect the probability of

B

$B$ for independent events

A

$A$ and

B

$B$ . In this case,

P (B | A) = P (B) = 0.13

$P (B | A) = P (B) = 0.13$ .

P (B | A) = P (B) = 0.13

$P (B | A) = P (B) = 0.13$

Step 3

Find

P (A | B)

$P (A | B)$ using Bayes's rule.

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Step 3.1

Using the Bayes' rule,

P (A | B) = \frac{P (B | A) P (A)}{P (B)}

$P (A | B) = \frac{P (B | A) P (A)}{P (B)}$ .

P (A | B) = \frac{P (B | A) P (A)}{P (B)}

$P (A | B) = \frac{P (B | A) P (A)}{P (B)}$

Step 3.2

Substitute the given values

P (A) = 0.1

$P (A) = 0.1$ ,

P (B) = 0.13

$P (B) = 0.13$ , and

P (B | A) = 0.13

$P (B | A) = 0.13$ in Bayes' rule.

P (A | B) = \frac{(0.13) \cdot (0.1)}{0.13}

$P (A | B) = \frac{(0.13) \cdot (0.1)}{0.13}$

Step 3.3

Cancel the common factor of

0.13

$0.13$ .

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Step 3.3.1

Cancel the common factor.

$P (A | B) = \frac{0.13 \cdot 0.1}{0.13}$

Step 3.3.2

Divide

$0.1$ by

$1$ .

$P (A | B) = 0.1$

$P (A | B) = 0.1$

$P (A | B) = 0.1$

Step 4

$P (A | B)$ should equal to

$P (A)$ because the occurrence of

$B$ should not effect the probability of

$A$ for independent events

$A$ and

$B$ . In this case,

$P (A | B) = P (A) = 0.1$ .

$P (A | B) = P (A) = 0.1$

Step 5

$P (A | B) = P (A)$ and

$P (B | A) = P (B)$ , which means that

$A$ and

$B$ are independent events.

A and B are independent events

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$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$