probability of three event A questionnaire survey on the use of SNS was conducted for students at A University.  As a result,
 we got the following: 
$55\%$ using Twitter ,
$53\%$ using Facebook ,
$20\%$ using Twitter and facebook both,
$19\%$ use both Facebook and
 Instagram.
$76\%$ use at least  Twitter
 and / or Instagram
$72\%$ use at least one of Facebook
 and Instagram 
$49\%$ use only one of twitter ,facebook, or instagram
At this time, find the next
 ratio respectively. 
$1$. Percentage of using both
 Twitter and Instagram 
$2$. Percentage of using all
 of Twitter, Facebook and Instagram 
$3$. Percentage of not using
 either Twitter, Facebook or Instagram 
I was confused, for $P(T\cap F)=20\%$ does this also include $P(T\cap F \cap I )$ ?
for $P(F) $ only = $P(F\cap T)$, is it correct only facebook is $14\%$?
$P(F)=53\% - P(F\cap T) - P(F\cap I)=53\%-19\%-20\% =14\%$?

i could find $P(T \cup F) = P(T)+P(F)-P(T \cap  F)=55\% +53\%-20\%=88% $ is this right?
$P(T) =35$ but i dont know $P(T \cap I)$

P(F)=14
i manage to find :
$P(T \cap I) = 17 $ and $P(I)=38$ 
but when i count $P(T \cup F \cup I) = P(T)+P(F)+P(I)-P(T \cap F) - P(T \cap I) - Ｐ(I \cap F) + P(T \cap F \cap I) = $
$49=55+53+38-20-19-17 +P(T \cap F \cap I) $ 
$P(T \cap F \cap I)=-41$
 A: Hint :


*

*$P(T\cup F)+P(T\cap F)=P(T)+P(F)$

*$P(T\cup F \cup I)=P(T)+P(F)+P(I)-P(T\cap F)-P(T\cap I)-P(I\cap F) + P(T\cap F \cap I)$

*$P(\overline A) = 1 - P(A)$
A: We may draw the Venn Diagram.

$ \begin{cases}
c+e+g+h=0.55 \\
b+e+f+h=0.53 \\
e+h=0.2 \\
f+h=0.19 \\
c+d+e+f+g+h=0.76 \\
b+d+e+f+g+h=0.72 \\
b+c+d=0.49 \\
a+b+c+d+e+f+g+h=1
\end{cases}$
and we want to find


*

*$g+h=?$

*$h=?$

*$a=?$
We have $e=0.2-h$ and $f=0.19-h$. The system is equivalent to  
$ \begin{cases}
e=0.2-h \\
f=0.19-h \\
c+g=0.35 \\
b=h+0.14 \\
c+d+g=0.37+h \\
b+d+g=0.33+h \\
b+c+d=0.49 \\
a+b+c+d+g=0.61+h
\end{cases}$
$ \begin{cases}
e=0.2-h \\
f=0.19-h \\
b=h+0.14 \\
c=0.35-g \\
d=0.02+h \\
d=0.19-g \\
c+d=0.35-h \\
a+d=0.12
\end{cases}$
So, $0.02+h=d=0.19-g $ and hence $g=0.17-h$
$0.35-h=c+d=0.35-(0.17-h)+0.02+h$
$h=0.05$.
$a=0.05$, $b=0.19$, $c=0.23$, $d=0.07$, $e=0.15$, $f=0.14$, $g=0.12$.

Your symbols are a bit confusing. $P(T\cup F\cup I)$ is not $0.49$
What we have is $P((T\setminus(F \cup I))\cup( F\setminus(T \cup I))\cup( I\setminus (F\cup T)))=0.49$.
Note that $P((T\setminus(F \cup I))\cup( F\setminus(T \cup I))\cup( I\setminus (F\cup T)))=P(T\setminus(F \cup I))+P(F\setminus(T \cup I))+P( I\setminus (F\cup T))$.
$P(T\setminus(F \cup I))=P(T)-P(F\cap T)-P(I \cap T)+P(T\cap F\cap I)$
We can also expand $P(F\setminus(T \cup I))$ and $P( I\setminus (F\cup T))$ and yield
\begin{align*}
&\;P((T\setminus(F \cup I))\cup( F\setminus(T \cup I))\cup( I\setminus (F\cup T)))\\
=&\;P(T)+P(F)+P(I)-2P(F\cap T)-2P(I \cap T)-2P(F \cap I)+3P(T\cap F\cap I)
\end{align*}
$\displaystyle 0.49=0.55+0.53+0.38-2(0.17)-2(0.19)-2(0.2)+3P(T\cap F\cap I)$
$P(T\cap F\cap I)=0.05$
