Examples

Independence and complements

Prove that these are logically equivalent statements:

$A$ and $B$ are independent
$A$ and $B^{c}$ are independent
$A^{c}$ and $B^{c}$ are independent

Make sure you demonstrate both directions of each equivalency.

Solution

(1) Assume $P [A B] = P [A] \cdot P [B]$ and show $P [A B^{c}] = P [A] \cdot P [B^{c}]$ :

Divide $A$ into the $B$ cases:
$P [A] = P [A B] + P [A B^{c}]$
Apply the assumption:
$≫ ≫ P [A] = P [A] \cdot P [B] + P [A B^{c}]$
Algebra:
$≫ ≫ P [A] (1 - P [B]) = P [A B^{c}]$
Negation rule:
$≫ ≫ P [A] \cdot P [B^{c}] = P [A B^{c}]$

(2) Assume $P [A B^{c}] = P [A] \cdot P [B^{c}]$ and show $P [A B] = P [A] \cdot P [B]$ :

Reason the same way as above, but apply the new assumption to break up the second term instead of the first.

(3) Show that $P [A B^{c}] = P [A] \cdot P [B^{c}]$ and $P [A^{c} B^{c}] = P [A^{c}] \cdot P [B^{c}]$ are equivalent:

To do this, simply notice that in the first equivalence we can replace $A$ with $B^{c}$ and $B$ with $A$ . Use $A B = B A$ too.

Independence by hand: red and green marbles

A bin contains 4 red and 7 green marbles. Two marbles are drawn.

Let $R_{1}$ be the event that the first marble is red, and let $G_{2}$ be the event that the second marble is green.

(a) Show that $R_{1}$ and $G_{2}$ are independent if the marbles are drawn with replacement.

(b) Show that $R_{1}$ and $G_{2}$ are not independent if the marbles are drawn without replacement.

Solution

(a) With replacement.

Identify knowns:

Know: $P [R_{1}] = \frac{4}{11}$
Know: $P [G_{2}] = \frac{7}{11}$

Now compute both sides of independence relation:

P [R_{1} G_{2}] = P [R_{1}] \cdot P [G_{2}]

The right side is $\frac{4}{11} \cdot \frac{7}{11}$ .

For $P [R_{1} G_{2}]$ , we have $4 \cdot 7$ ways to get $R_{1} G_{2}$ , and $11^{2}$ total outcomes. So left side is $\frac{4 \cdot 7}{11^{2}}$ , which equals the right side.

(b) Without replacement. This is a bit harder.

(1) Identify knowns:

Know: $P [R_{1}] = \frac{4}{11}$ and therefore $P [R_{1}^{c}] = \frac{7}{11}$

We seek: $P [G_{2}]$ and $P [R_{1} G_{2}]$

(2) Find $P [G_{2}]$ using Total Probability (Division into Cases):

\begin{matrix} G_{2} = G_{2} \cap R_{1} ⋃ G_{2} \cap R_{1}^{c} \\ ≫ ≫ P [G_{2}] = P [R_{1}] \cdot P [G_{2} | R_{1}] + P [R_{1}^{c}] \cdot P [G_{2} | R_{1}^{c}] \end{matrix}

Find RHS factors by counting, then compute:

≫ ≫ P [G_{2}] = \frac{4}{11} \cdot \frac{7}{10} + \frac{7}{11} \cdot \frac{6}{10} ≫ ≫ \frac{70}{110}

(3) Find $P [R_{1} G_{2}]$ using multiplication rule:

P [R_{1} G_{2}] = P [R_{1}] \cdot P [G_{2} | R_{1}] ≫ ≫ \frac{4}{11} \cdot \frac{7}{10} ≫ ≫ \frac{28}{110}

(4) Compare both sides:

Left side: $P [R_{1} G_{2}] = \frac{28}{110}$ .

Right side:

P [R_{1}] \cdot P [G_{2}] = \frac{4}{11} \cdot \frac{70}{110} = \frac{28}{121}

But $\frac{28}{110} \neq \frac{28}{121}$ so $P [R_{1} G_{2}] \neq P [R_{1}] \cdot P [G_{2}]$ and they are not independent.

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Examples

Independence and complements

Independence by hand: red and green marbles

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Table of Contents

Examples

Independence and complements

Independence by hand: red and green marbles

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