Question

At a certain high school, 15 students play stringed instruments and

20 students play brass instruments. Five students play neither
stringed nor brass instruments. What is the probability that a
randomly selected student plays both a stringed and a brass
instrument?

Answer 1

0 is the probability that a randomly selected student plays both a stringed and a brass instrument.

Explanation:

Given that:

Number of students who play stringed instruments, N(A) = 15

Number of students who play brass instruments, N(B) = 20

Number of students who play neither, N(
`A \cup B`)' = 5

To find:

The probability that a randomly selected students plays both = ?

Solution:

Total Number of students = N(A)+N(B)+N(
`A \cup B`)' =15 + 20 + 5 = 40

(As there is no student common in both the instruments we can simply add the three values to find the total number of students)

As per the venn diagram, no student plays both the instruments i.e.

`N(A\cap B) =0`

Formula for probability of an event E can be observed as:

`P(E) = \frac{\text{Number of favorable cases}}{\text {Total number of cases}}`

`P(A\cap B) = (N(A\cap B))/(N(U))\\\Rightarrow (0)/(40)\\\Rightarrow P(A\cap B) = 0`

So, 0 is the probability that a randomly selected student plays both a stringed and a brass instrument.