1. ## Conditional probability

Suppose we have 2 events: $s, e$.

We know:

$P(e)=.7$
$P(e|\neg s)=.15$
$P(e|s)=1$

What is $P(s)$?

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I know that $P(s) = 1 - P(\neg s) \Rightarrow P(\neg s)=1-P(s)$ and that

$P(e|s) = \frac{P(s|e)P(e)}{P(s)}=\frac{P(s|e)P(e)}{1-P(\neg s)}=\frac{.7P(s|e)}{1-P(\neg s)}=1$

$\Rightarrow .7P(s|e)=1-P(\neg s)$

$\Rightarrow P(\neg s)=1-.7P(s|e)$

$\Rightarrow P(s) = .7P(s|e)$

but I sort of go around in circles...

2. Originally Posted by scorpion007
Suppose we have 2 events: $s, e$.

We know:

$P(e)=.7$
$P(e|\neg s)=.15$
$P(e|s)=1$

What is $P(s)$?

----------

I know that $P(s) = 1 - P(\neg s) \Rightarrow P(\neg s)=1-P(s)$ and that

$P(e|s) = \frac{P(s|e)P(e)}{P(s)}=\frac{P(s|e)P(e)}{1-P(\neg s)}=\frac{.7P(s|e)}{1-P(\neg s)}=1$

$\Rightarrow .7P(s|e)=1-P(\neg s)$

$\Rightarrow P(\neg s)=1-.7P(s|e)$

$\Rightarrow P(s) = .7P(s|e)$

but I sort of go around in circles...
hi

P(e|s)=P(S n e)/p(s)

P(S n e)= P(s) ---1

P(e|s')=P(s' n e)/P(s')

0.15=[P(e)-P(S n e)]/[1-P(s)]

0.15[1-P(s)]=0.7-P(s) from 1

3. Hello, scorpion007!

$\text{Suppose we have two events: }\:e,\, s.$

We know: . $\begin{Bmatrix}P(e) &=& 0.7 & [1]\\
P(e\,|\sim\!s) &=& 0.15 & [2]\\
P(e \,|\, s) &=& 1 & [3] \end{Bmatrix}$

What is $P(s)$ ?

I filled in a chart . . .

From [1]: . $P(e) \:=\:0.7 \quad\Rightarrow\quad P(\sim\!e) \:=\:0.3$

. . $\begin{array}{c||c|c||c|}
& P(s) & P(\sim\!s) & \text{Total} \\ \hline\hline
P(e) & & & {\color{red}0.7} \\ \hline
P(\sim\!e) & & & {\color{red}0.3} \\ \hline\hline
\text{Total} & {\color{red}P(s)} & {\color{red}P(\sim\!s)} & 1.0 \end{array}$

From [3]: . $P(e\,|\,s) \:=\:\frac{P(e\,\wedge\,s)}{P(s)} \:=\:1$

Hence: . $P(e\wedge s) \:=\:P(s) \quad\Rightarrow\quad P(\sim\!e\wedge s) \:=\:0$

. . $\begin{array}{c||c|c||c|}
& P(s) & P(\sim\!s) & \text{Total} \\ \hline\hline
P(e) & {\color{red}P(s)} & & 0.7 \\ \hline
P(\sim\!e) & {\color{red}0} & & 0.3 \\ \hline\hline
\text{Total} & P(s) & P(\sim\!s) & 1.0 \end{array}$

From [2]: . $P(e\,|\sim\!s) \:=\:\frac{P(e \wedge\sim\!s)}{P(\sim\!s)} \:=\:0.15$

Hence: . $P(e\,\wedge \sim\!s) \:=\:0.15P(\sim\!s) \quad\Rightarrow\quad P(\sim\!e \wedge \sim\!s) \:=\:0.85P(\sim\!s)$

. . $\begin{array}{c||c|c||c|}
& P(s) & P(\sim\!s) & \text{Total} \\ \hline\hline
P(e) & P(s) & {\color{red}0.15P(\sim\!s)} & 0.7 \\ \hline
P(\sim\!e) & 0 & {\color{red}0.85P(\sim\!s)} & 0.3 \\ \hline\hline
\text{Total} & P(s) & P(\sim\!s) & 1.0 \end{array}$

Reading across, we have: . $\begin{array}{ccccc}
P(s) + 0.15P(\sim\!s) &=& 0.7 & [1] \\
0 + 0.85P(\sim\!s) &=& 0.3 & [2] \end{array}$

From [2], we have: . $0.85P(\sim\!s) \:=\:0.3 \quad\Rightarrow\quad P(\sim\!s) \:=\:\frac{0.3}{0.85} \:=\:\frac{6}{17}$

Therefore: . $P(s) \;=\;1-\frac{6}{17} \;=\;\frac{11}{17}$

4. thank you both very much!

0.15=[P(e)-P(S n e)]/[1-P(s)]

is it true in general that

$P(A\cap B) = P(A) - P(A\cap \neg B)$ ?

6. Originally Posted by scorpion007
is it true in general that

$P(A\cap B) = P(A) - P(A\cap \neg B)$ ?
in general P(A n B')=P(A)-P(A n B)

It should be clear when you draw the sets