Answer:
[tex]\frac{5}{29}[/tex]
Step-by-step explanation:
Let [tex]n(A)[/tex] represent students playing basketball, [tex]n(B)[/tex] represent students playing baseball.
Then, [tex]n(A)=7[/tex], [tex]n(B)=24[/tex]
Let [tex]n(S)[/tex] be the total number of students. So, [tex]n(S)=29[/tex].
Now,
[tex]P(A)=\frac{n(A)}{n(S)}=\frac{7}{29}[/tex]
[tex]P(B)=\frac{n(B)}{n(S)}=\frac{24}{29}[/tex]
3 students play neither of the sport. So, students playing either of the two sports is given as:
[tex]n(A\cup B)=n(S)-3\\n(A\cup B)=29-3=26[/tex]
∴ [tex]P(A\cup B)=\frac{n(A\cup B)}{n(S)}=\frac{26}{29}[/tex]
From the probability addition theorem,
[tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]
Where, [tex]P(A\cap B)[/tex] is the probability that a student chosen randomly from the class plays both basketball and baseball.
Plug in all the values and solve for [tex]P(A\cap B)[/tex] . This gives,
[tex]\frac{26}{29}=\frac{7}{29}+\frac{24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{7+24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{31}{29}+P(A\cap B)\\\\P(A\cap B=\frac{31}{29}-\frac{26}{29}\\\\P(A\cap B=\frac{31-26}{29}=\frac{5}{29}[/tex]
Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is [tex]\frac{5}{29}[/tex]
