probability of three event

Question

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) $
$P(T \cap F \cap I)=-41$

Answer 1

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

1. $g+h=?$

2. $h=?$

3. $a=?$

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 $g=0.17-h$

$0.35-h=c+d=0.35-(0.17-h)+0.02+h$

$h=0.05$.

$a=0.05$, $b=0.19$, $c=0.23$, $d=0.07$, $e=0.15$, $f=0.14$, $g=0.12$.

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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*}

$\displaystyle 0.49=0.55+0.53+0.38-2(0.17)-2(0.19)-2(0.2)+3P(T\cap F\cap I)$

$P(T\cap F\cap I)=0.05$

Answer 2

Hint :

1. $P(T\cup F)+P(T\cap F)=P(T)+P(F)$

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

3. $P(\overline A) = 1 - P(A)$