Confusion on 2 solutions of a Probability ques

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Question: There are 5 items <A,B,C,D and E>. There are 3 trials given to me in which I can pick any item out of the five with replacement. What is the probability of choosing at least one time A?

I arrived at two solutions:
Solution 1: Choose a trial out of the 3 trials, in which I choose A (3C1). In rest two trials I can have any of the 5 items (5*5). The sample space is (5*5*5)
So the answer becomes: (3*5*5)/(5*5*5) = 3/5

Solution 2: Probability of not choosing A = (4*4*4)/(5*5*5). So, the probability of choosing A becomes 1- (4*4*4)/(5*5*5) = 61/125.

The 2nd solution seems to me as the correct one but can someone help me in letting me know what have I done wrong in the 1st solution. Thanks!

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@ntncoldfire,

The probability of getting an A on any trial involves ADDING the probabilities for independent events. (giving 3/5)
The probability of NOT getting an A on any is a combined event requiring MULTIPLYING probabilities.

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@fresco,

I don't think the first calculation represents anything meaningful. Do it for six trials...

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@markr,

Yes I see what you mean. Are you suggesting we need to calculate the number of "cubes" containing A in the 125 cube sample space ? That does indeed look like 61.

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@fresco,

The additive approach would go like this:
[C(3,1)*1*(5-1)*(5-1) + C(3,2)*1*1*(5-1) + C(3,3)*1*1*1)] / 5^3
= (48+12+1) / 125
= 61/125

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@markr,

Yes. Looking at 3 mutually perpendicular faces of a 5- cube meeting at the AAA corner, you get 25+20+16 =61

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@fresco,

Can you give a bit of a description of this? It sounds interesting, but I can't visualize it. How about a 3-dimensional example:

3 samplings with replacement from the set (1, 2, 3).

I assume you'll get to 19 by 3x3 + 3x2 + 2*2, but why (geometrically)?

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@markr,

A three sampling situation of {1,2,3} would give you a sample space like a Rubiks cube.
1,2,3 would make a number line on each of the x,y,z axes (origin left hand corner). So to count the triples (conjoint events) involving item 1, start with all nine cubes in the x-y plane (z=1) then add the six remaining cubes in the y-z plane (x=1) and finally the four remaining cubes in the x-z plane (y=1). 9+6+4=19

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@fresco,

Ah, so the 5-cube is a 5x5x5 cube, not a 5-dimensional hypercube.

Thanks!

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Confusion on 2 solutions of a Probability ques

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