Conditional Probabilities
The table gives us information about the favorite leisure activities of students at a school
We are required to find the probability of some events, given the occurrence of another event.
a) P(Sports | Female). It's the probability that a student likes sport if it's a female.
There are 334 female students, from which 39 like sports, thus:
[tex]P(S|F)=\frac{39}{334}=11.68\text{ \%}[/tex]b) P(Female | Sports). It's the probability that a student is known to like sports and it's also a female.
There are 106 students that like sports, from which 39 are female, thus:
[tex]P(F|S)=\frac{39}{106}=36.79\text{ \%}[/tex]c) P(Reading | Male). It's the probability that a male student likes reading.
There are 366 male students from which 76 like reading, thus:
[tex]P(R|M)=\frac{76}{366}=20.77\text{ \%}[/tex]d) P(Male | Reading). It's the probability that a student is known to like reading and it's also a male. There are 161 students who like reading out of which 76 are male, thus:also a male. d)
[tex]P(M|R)=\frac{76}{161}=47.20\text{ \%}[/tex]e) P(Hiking | Male). There are 366 male students out of which 58 like hiking, thus:H
[tex]P(H|M)=\frac{58}{366}=15.85\text{ \%}[/tex]f) P(Hiking | Female). There are 334 female students out of which 48 like hiking, thus:
[tex]P(H|F)=\frac{48}{334}=14.37\text{ \%}[/tex]g) P(Male | Shopping). There are 139 students who like shopping and from them, 68 are male, thus:39 students who like shopping and fr
[tex]P(M|S)=\frac{68}{139}=48.92\text{ \%}[/tex]h) P(Shopping | Female). There are 334 female students from which 71 like shopping, thus:
[tex]P(S|F)=\frac{71}{334}=21.26\text{ \%}[/tex]I) P(Phoning | Male). There are 366 male students from which 54 like phoning, thus:
[tex]P(P|M)=\frac{54}{366}=14.75\text{ \%}[/tex]ich 39 are female, thus:
S
