Question

In a certain Algebra 2 class of 29 students, 5 of them play basketball and 16 of them play baseball. There are 10 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

Answer 1

ANSWER
$P(both\:sports)= \frac{2}{29}$

EXPLANATION
Let the number who plays both games be
$x$
Then those who play only basketball
$= 5 - x$

and those who play only baseball
$= 16 - x$

We were given that 10 students play neither sports.

We can then write the following equation,

$(5 - x) + x + (16 - x) + 10 = 29$

This implies that,

$- x + x - x = 29 - 5 - 16 - 10$

This simplifies to,

$- x = - 2$

This gives us,

$x = 2$

Therefore the number of students play both basketball and baseball is 2.

The probability that a student chosen from the class plays both basketball and baseball ball

$= \frac{2}{29}$