Answer:
1.2%
Step-by-step 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 = [tex]\frac{1}{4}[/tex].
Thus, the probability that the student does not receive version A = [tex]1-\frac{1}{4}[/tex] = [tex]\frac{3}{4}[/tex].
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,
[tex]\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}[/tex]+[tex]\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}[/tex]+[tex]\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}[/tex]
= [tex](\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5[/tex]
= [tex](\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2][/tex]
= [tex](\frac{3}{4^4})[1+\frac{1}{16}][/tex]
= [tex](\frac{3}{256})[\frac{17}{16}][/tex]
= 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%.
