Each of three investments has a 20% of becoming worthless within a yea

Simone invests in three investments A, B, C
Each of three investments has a 20% of becoming worthless,
so each of three investments has a 80% of becoming NOT worthless (staying the same value of becoming profitable)

Approximate chance of losing not greater 1/3 of original investment
= [A, B, C becoming NOT worthless + A or B or C becoming worthless and other two NOT worthless]
= [\(\frac{80}{100}*\frac{80}{100}*\frac{80}{100}\) + \(C^1_3\)\(\frac{20}{100}*\frac{80}{100}*\frac{80}{100}\)]
= \(8*8[\frac{8}{1000} + \frac{6}{1000}\)] \(* 100\%\)
= \(89.6\%\)

Answer A

She can afford to loose 1/3rd of he original investment it means she can only afford to loose one of her investment.
Probability of investment becoming worthless = 20% = 1/5
P(she looses not more than 1/3 of her original investment) = 1 - p(she looses more than 1 investment = looses 2 of her investment and looses all three of her investments)
P (required) = 1-3*(4/5*1/5*1/5) - 1/5*1/5*/15 = 1-13/125 = 112/125 ~ 88. 4% ~ 90%

MANHATTAN GMAT OFFICIAL SOLUTION:

The problem asks for the approximate chance that no more than 1/3 of the original investment is lost. We can apply the “1 – x” technique: what’s the chance that more than 1/3 of the original investment is lost? There are two outcomes we have to separately measure:

(a) All 3 investments become worthless.
(b) 2 of the 3 investments become worthless, while 1 doesn’t.

Outcome (a): The probability is (0.2)(0.2)(0.2) = 0.008, or a little less than 1%.

Outcome (b): Call the investments X, Y, and Z. The probability that X retains value, while Y and Z become worthless, is (0.8)(0.2)(0.2) = 0.032. Now, we have to do the same thing for the specific scenarios in which Y retains value (while X and Z don’t) and in which Z retains value (while X and Y don’t). Each of those scenarios results in the same math: 0.032. Thus, we can simply multiply 0.032 by 3 to get 0.096, or a little less than 10%.

The sum of these two probabilities is 0.008 + 0.096 = 0.104, or a little more than 10%. Finally, subtracting from 100% and rounding, we find that the probability we were looking for is approximately 90%.

The correct answer is A.

This problem illustrates the power of diversification in financial investments. All else being equal, it’s less risky to hold a third of your money in three uncorrelated (independent) but otherwise equivalent investments than to put all your eggs in one of the baskets. That said, be wary of historical correlations! Housing price changes in different US cities were not so correlated—and then they became highly correlated during the recent housing crisis (they all fell together), fatally undermining spreadsheet models that assumed that these price changes were independent.

To lose no more than 1/3 of her original investments means either she loses none of her 3 investments, or she loses exactly one of her 3 investments. If she doesn’t lose an investment, we will call it a “win,” otherwise, we will call it a “lose.” Thus, we have two scenarios to consider:

Scenario 1: win-win-win

P(win-win-win) = 0.8 x 0.8 x 0.8 = 0.512

Scenario 2: win-win-lose

P(win-win-lose) = 0.8 x 0.8 x 0.2 = 0.128

We also see that win-win-lose (2 wins and 1 loss) can occur in 3!/2! = 3 ways.

So the total probability of win-win-lose in any combo is 0.128 x 3 = 0.384

So the overall probability is 0.512 + 0.384 = 0.896 = 89.6% ≈ 90%.

Answer: A

I solved it but I have my doubts on the way the question is posed:

What if two investments go to 0, and one grows so extreme that it makes up for more than what is lost? She still retains more than or equal to 2/3 of her initial investment?
Or, what if one goes to 0, and the other two lose 50% over the same time?...

I find that irritating

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