PROBABILITY

[CAMELSMOM]

i HAVE A STATS QUESTION AND I JUST NEED HELP GETTING STARTED can anyone help me????
. _
P(AorB)=.85
. _
P(Aand B)=.41
. _____
P(Aor B)=.21
........... _
Find P(A/B)

please ignore periods above... this is the only way I can figure to get the lines where I need them

 

[soroban]

Hello, CAMELSMOM!

I have a solution . . . I hope I can explain it clearly.
And I sincerely hope that someone has a more elegant solution . . .

\(\displaystyle \text{Given: }\;\begin{array}{cccc} P(\overline{A} \cup B) &=& 0.85 & [1] \\ \\[-3mm] P(\overline{A} \cap B) &=& 0.41 & [2]\\ \\[-3mm] P(\overline{A \cup B}) &=& 0.21 & [3]\end{array}\)

$\displaystyle \text{Find: }\; P(A\,|\,\overline{B})$

$\displaystyle \text{The "opposite" of }P(A\,|\,\overline{B})\,\text{ is }\,P(\overline{A}\,|\,\overline{B})$

$\displaystyle \text{Bayes' Theorem: }\;P(\overline{A}\,|\,\overline{B}) \;=\;\frac{P(\overline{A} \cap \overline{B})}{P(\overline{B})}\;\;[4]$


$\displaystyle \text{DeMorgan's Law: }\:\overline{A \cup B} \;=\;\overline{A} \cap \overline{B}$

$\displaystyle \text{So [3] becomes: }\;P(\overline{A} \cap \overline{B}) \:=\:0.21\;\;[5]$


$\displaystyle \text{Consider the "opposite" of [1]: }\;P(\overline{\overline{A} \cup B}) \;=\;0.15$

$\displaystyle \text{DeMorgan's Law: }\;\overline{(\overline{A} \cup B)} \;=\;A \cap \overline{B}$

$\displaystyle \text{So, the opposite of [1] is: }\;P(A \cap \overline{B}) \:=\:0.15\;\;[6]$


$\displaystyle \text{Combining [5] and [6], we have: }\;P(\overline{B}) \;=\;P(\overline{A} \;\cap \;\overline{B}) \;+\; P(A \;\cap\; \overline{B}) \;=\;0.15 \;+\; 0.25 \;=\;0.36\;\;[7]$


$\displaystyle \text{Substitute [5] and [7] into [4]: }\;P(\overline{A}\,|\overline{B}) \;=\;\frac{0.21}{0.36} \;=\;\frac{7}{12}$


$\displaystyle \text{Therefore: }\;P(A\,|\,\overline{B}) \;=\;1-\frac{7}{12} \;=\;\frac{5}{12}$