Why can't i use the "direct" way to get correct answer?

[aimhier]

Manhattan GMAT Challenge Problem of the Week - 13 Feb 2012
Each of three investments has a 20% of becoming worthless within a year of purchase, independently of what happens to the other two investments. If Simone invests an equal sum in each of these three investments on January 1, the approximate chance that by the end of the year, she loses no more than 1/3 of her original investment is

A. 90%
B. 80%
C. 70%
D. 60%
E. 40%
Why can't i use the "direct" method to get the correct answer A? no more than 1/3rd" being lost means that she got all three win:(0.8)(0.8)(0.8) = 0.064, or exactly 1 loss : (0.2)(0.8)(0.8) = 0.128 they don't add up to 90% ; WHY?

[jnelson0612]

Manhattan GMAT Challenge Problem of the Week - 13 Feb 2012
Each of three investments has a 20% of becoming worthless within a year of purchase, independently of what happens to the other two investments. If Simone invests an equal sum in each of these three investments on January 1, the approximate chance that by the end of the year, she loses no more than 1/3 of her original investment is

A. 90%
B. 80%
C. 70%
D. 60%
E. 40%
Why can't i use the "direct" method to get the correct answer A? no more than 1/3rd" being lost means that she got all three win:(0.8)(0.8)(0.8) = 0.064, or exactly 1 loss : (0.2)(0.8)(0.8) = 0.128 they don't add up to 90% ; WHY?

Okay, let's think this through very systematically:
Our desired outcome is for her to lose no more than 1/3 of her original investment. So I want her to either have all retain their value OR have two remain their value while only one becomes worthless.

Let's call them investments A, B, and C. Here are the chances for the various outcomes:
1) A, B, and C all retain value: .8 * .8 * .8 = about 51%
2) A becomes worthless; B and C retain value: .2 * .8 * .8 = about 13%
3) B becomes worthless; A and C retain value: 13% (notice that this is the exact same calculation that we see in #2
4) C becomes worthless; A and B retain value: 13% (again, same calculation)

So what I want is to have outcome 1 OR outcome 2 OR outcome 3 OR outcome 4. In probability when something OR something else is acceptable I add the probabilities. 51% + 13% + 13% + 13% = about 90%.

Does this help make things more clear? I can't just count the probability of the failure of one stock one time; I have to count that probability for each individual stock, as shown above.