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 =
.
Thus, the probability that the student does not receive version A =
=
.
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,
+
+![\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7B1%7D%7B4%7D)
= ![(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B4%7D%29%5E3%5Ctimes%20%28%5Cfrac%7B3%7D%7B4%7D%29%5E2%2B%28%5Cfrac%7B1%7D%7B4%7D%29%5E4%5Ctimes%20%28%5Cfrac%7B3%7D%7B4%7D%29%2B%28%5Cfrac%7B1%7D%7B4%7D%29%5E5)
= ![(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B4%7D%29%5E3%5Ctimes%20%28%5Cfrac%7B3%7D%7B4%7D%29%5B%5Cfrac%7B3%7D%7B4%7D%2B%5Cfrac%7B1%7D%7B4%7D%2B%28%5Cfrac%7B1%7D%7B4%7D%29%5E2%5D)
= ![(\frac{3}{4^4})[1+\frac{1}{16}]](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%7D%7B4%5E4%7D%29%5B1%2B%5Cfrac%7B1%7D%7B16%7D%5D)
= ![(\frac{3}{256})[\frac{17}{16}]](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%7D%7B256%7D%29%5B%5Cfrac%7B17%7D%7B16%7D%5D)
= 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%.