Hello community, I solved this question and got it correct. However, I am a bit unsure about my approach towards the question. And the solution is not available. Hence, would be really grateful if someone could share a detailed solution for me to cross-check my approach.
Thank you
Correct answer - 20
My approach to the question: 1 out of 10 times, the thief is caught. If min. amount is then 1810 = 180 < 181 (fine amount), 18 is not possible.
If min. amount is 20, then 2010 = 200 > 181, still left with some returns even after getting caught. Hence, min. is 20.
If you want an algebraic approach, here’s my solution:
I’m using the expected value formula, E(X)=x_1 p_1+x_2 p_2+...
And I plug some of the numbers in: E(X)=(-181)(\frac{1}{10}) +x_2 (\frac{9}{10})
Since we only want a minimum x_2 to cover the thief’s losses, we’ll modify our equation to: 0<(-181)(\frac{1}{10}) +x_2 (\frac{9}{10})
Now we can solve for x_2: 0<-18.1 +x_2 (\frac{9}{10}) 18.1<x_2 (\frac{9}{10}) 18.1(\frac{10}{9})<x_2 20.\bar{1}<x_2
From this, we know that the x_2 (or the minimum combined value of the items they steal) has to be above 20.\bar{1} for the thief to cover their losses. Rounding 20.\bar{1} to an integer gives us 20, so the correct answer choice must be 20.
