Answer:
Option 3.
Step-by-step explanation:
Let T=Total, B=Brother and S=Sister. By using given diagram, we get
[tex]n(B)=28+68=96[/tex]
[tex]n(S)=82+68=150[/tex]
[tex]n(T)=28+68+82+52=230[/tex]
[tex]n(B\cap S)=68[/tex]
Using this information, we get
[tex]P(B)=\dfrac{n(B)}{n(T)}=\dfrac{96}{230}\approx 0.42[/tex]
[tex]P(S)=\dfrac{n(S)}{n(T)}=\dfrac{150}{230}\approx 0.65[/tex]
[tex]P(S|B)=\dfrac{P(B\cap S)}{P(B)}=\dfrac{n(B\cap S)}{n(B)}=\dfrac{68}{96}\approx 0.71[/tex]
[tex]P(B|S)=\dfrac{P(B\cap S)}{P(S)}=\dfrac{n(B\cap S)}{n(S)}=\dfrac{68}{150}\approx 0.45[/tex]
The probability P(S|B) has the largest value.
Therefore, the correct option is 3.
