Question

for each set of probabilities, determine whether the events a and b are independent or dependent.
probabilities
(a) $p(a)=\frac{1}{7}; p(b)=\frac{1}{3}; p(b|a)=\frac{1}{3}$
(b) $p(a)=\frac{1}{3}; p(b)=\frac{1}{8}; p(a|b)=\frac{1}{3}$
(c) $p(a)=\frac{1}{3}; p(b)=\frac{1}{6}; p(a \text{ and } b)=\frac{1}{8}$
(d) $p(a)=\frac{1}{3}; p(b)=\frac{1}{2}; p(a \text{ and } b)=\frac{1}{6}$
independent
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dependent
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Explanation:

Step1: Recall independence rule

Two events $A$ and $B$ are independent if $P(A \text{ and } B) = P(A) \times P(B)$, or equivalently $P(B|A)=P(B)$ / $P(A|B)=P(A)$.

Step2: Check case (a)

Verify $P(B|A) = P(B)$:
$P(B|A)=\frac{1}{3}$, $P(B)=\frac{1}{3}$. Since $\frac{1}{3}=\frac{1}{3}$, events are independent.

Step3: Check case (b)

Verify $P(A|B) = P(A)$:
$P(A|B)=\frac{1}{3}$, $P(A)=\frac{1}{3}$. Since $\frac{1}{3}=\frac{1}{3}$, events are independent.

Step4: Check case (c)

Calculate $P(A) \times P(B)$:
$P(A) \times P(B) = \frac{1}{3} \times \frac{1}{6} = \frac{1}{18}$. Given $P(A \text{ and } B)=\frac{1}{8}$. Since $\frac{1}{18}
eq \frac{1}{8}$, events are dependent.

Step5: Check case (d)

Calculate $P(A) \times P(B)$:
$P(A) \times P(B) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$. Given $P(A \text{ and } B)=\frac{1}{6}$. Since $\frac{1}{6}=\frac{1}{6}$, events are independent.

Answer:

(a) Independent
(b) Independent
(c) Dependent
(d) Independent