I'm a tad bit unsure,but still my approach would be as below:- :)
The experiment is conducted 10 times,and we need to consider the cases and add up those cases in which he gets less than (50*10=500,if he gives the exam fairly),and maximum before 500 is when he cheats and remains uncaught for 6 times out of 10,which makes his score (80*6=480).
Now,let's assume p be the probability he cheats and succeeds in that,(p=1-0.3=0.7) ,and probability of failure be (q=0.3),now it's just adding up the binomial terms for which number of success<=6,I'll create a table below:-
| Success |
Marks |
Probability |
| 6 |
80*6=480 |
$\binom{10}{6}(0.7)^{6}0.3^{4}$ |
| 5 |
80*5=400 |
$\binom{10}{5}(0.7)^{5}0.3^{5}$ |
| 4 |
80*4=320 |
$\binom{10}{4}(0.7)^{4}0.3^{6}$ |
| 3 |
80*3=240 |
$\binom{10}{3}(0.7)^{3}0.3^{7}$ |
| 2 |
80*2=160 |
$\binom{10}{2}(0.7)^{2}0.3^{8}$ |
| 1 |
80*1=80 |
$\binom{10}{1}(0.7)^{1}0.3^{9}$ |
| 0 |
80*0=0 |
$\binom{10}{0}(0.7)^{0}0.3^{10}$ |
Summing up the above binomial probability terms will yield the ans
