0
$\begingroup$

Assume $3$ elements: $\left\{A,B,C\right\}$.

Find the probability that the first element is NOT $A$, and the second element is NOT $B$ given that the first element is NOT $A$.

Answer

$P(\text{not } A \text{ as first}) \times P(\text{not }B\text{ as second }|\text { not }A\text{ as first}) = P(\text{not }A\text{ as first AND not }B\text{ as second})$


Permutations outcome: $\{AB, AC, BA, BC, CA, CB \}$

NOT $A$ as first element $\Rightarrow$ $\{ BA, BC, CA, CB \}$

NOT $B$ as second element $\Rightarrow$ $\{ AC, BA, BC, CA \}$

Intersecting given that the first element is NOT $A$, and the second element is NOT $B$:

positive outcomes from possible outcomes: $\{BA, BC, CA\}$

$\Rightarrow$ $P(\text{not } B\text{ as second }|\text{ not }A\text{ as first}) = \frac34$

$\Rightarrow$ $P(\text{not }A\text{ as first AND not }B\text{ as second}) = \frac23 \times \frac34 = \frac12$


which matches the sets:

$6$ outcomes $\Rightarrow$ $\{AB, AC, BA, BC, CA, CB\}$ of which $\{BA, BC, CA\}$ match both conditions
$\Rightarrow$ $\frac36 = \frac12$

My attempt

I cannot understand what the $4$ (in $\frac34$) represents when considering that there are only $3$ items $\{A,B,C\}$ in total.


Initially I thought about it in this way:
$\frac23 \times \frac12 = \frac13$

which seems wrong.



$\Rightarrow$ I have $2$ elements $\{B,C\}$ which are NOT $A$, multiplied by the one element that is left $\{A\}$, which gives $\{BA, CA\}$
$\Rightarrow$ $\frac26 = \frac13$,

which again seems wrong.



How can I write out the permutations that finally give me $\frac23 \times \frac34 = \frac12$?

Thank you for your help!

1 Answers 1

0

Assuming I understand what you are asking correctly:

Permutations outcome: $\{AB, AC, BA, BC, CA, CB \}$

NOT $A$ as first element $\Rightarrow$ $\{ BA, BC, CA, CB \}$

NOT $B$ as second element $\Rightarrow$ $\{ AC, BA, BC, CA \}$

positive outcomes from possible outcomes: $\{BA, BC, CA\}$

These are the sets you are getting the probabilty fractions from. The $\frac{3}{4}$ is the probabilty $P(\text{not } B\text{ as second }|\text{ not }A\text{ as first})$. Because it is given that not $A$ as first is given, your possible outcomes are $\{ BA, BC, CA, CB \}$. You positive outcomes from those are $\{BA, BC, CA\}$. That is 3 our of 4, so the probability is $\frac34$.