Statistics Examples

$P (A) = 0.1$ , $P (B) = 0.13$ , $P (B g i v e n A) = 0.13$

Two events are independent events when the occurrence of one does not affect the probability of the other. $P (A | B) = P (A)$ and $P (B | A) = P (B)$ .

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

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

$P (B | A)$ should equal to $P (B)$ because the occurrence of $A$ should not effect the probability of $B$ for independent events $A$ and $B$ . In this case, $P (B | A) = P (B) = 0.13$ .

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

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

Tap for more steps...

Using the Bayes' rule, $P (A | B) = \frac{P (B | A) P (A)}{P (B)}$ .

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

Substitute the given values $P (A) = 0.1$ , $P (B) = 0.13$ , and $P (B | A) = 0.13$ in Bayes' rule.

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

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

Tap for more steps...

Remove parentheses.

Tap for more steps...

Multiply $0.13$ by $0.1$ .

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

Multiply $0.13$ by $0.1$ .

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

Reduce the expression by cancelling the common factors.

Tap for more steps...

Cancel the common factor.

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

Divide $0.1$ by $1$ .

$P (A | B) = 0.1$

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

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

$P (A | B) = P (A)$ and $P (B | A) = P (B)$ , which means that $A$ and $B$ are independent events.

A and B are independent events

Enter YOUR Problem