Question

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student tested will get version A of the test? Express your answer as a percent, and round to the nearest tenth

Answer 1

1.2%

Explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A =
`(1)/(4)`.

Thus, the probability that the student does not receive version A =
`1-(1)/(4)` =
`(3)/(4)`.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

`(1)/(4)* (1)/(4)* (1)/(4)* (3)/(4)* (3)/(4)`+
`(1)/(4)* (1)/(4)* (1)/(4)* (1)/(4)* (3)/(4)`+
`(1)/(4)* (1)/(4)* (1)/(4)* (1)/(4)* (1)/(4)`

=
`((1)/(4))^3* ((3)/(4))^2+((1)/(4))^4* ((3)/(4))+((1)/(4))^5`

=
`((1)/(4))^3* ((3)/(4))[(3)/(4)+(1)/(4)+((1)/(4))^2]`

=
`((3)/(4^4))[1+(1)/(16)]`

=
`((3)/(256))[(17)/(16)]`

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.