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5 spheres (S), 5 pyramids (Y), 5 cubes (C) in container. Randomly select two of the shapes at the same time. Probability that one shape is a cube or one is a pyramid?

I work this problem as: $P(C) + P(Y)-P(C \cup Y)$ where $P(C \cup Y)= P(C) \cdot P(Y|C) = 1/3 \cdot 5/14=5/42$. I get $5/14$ by saying the $P(Y|C)$ means that the given C part of this indicates a change in sample space to $14$ items vs the original $15$.

My math works as: $1/5+1/5-5/42$ which does NOT equal the books answer of $4/7$. I'm off by $1/42$ but don't see the error and have no hair left to pull.

Can you assist?

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    If you're randomly selecting two shapes at the same time, is this equivalent to selecting a random shape with replacement two times? – 2017-02-25

3 Answers 3

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The most likely interpretation of the problem statement is that we want the probability of not selecting two spheres. That is, two cubes, or two pyramids, or a cube and a pyramid are also successful outcomes.

Since simultaneously selecting two shapes is like sequentially selecting them without replacement, the probability of selecting two identical shapes is

$$ \frac{5}{15} \cdot \frac{4}{14} = \frac{2}{21} \enspace. $$

The probability of drawing a cube or a pyramid is then $1 - \frac{2}{21} = \frac{19}{21}$.

This is a very straightforward approach, but not the only one to get the desired result. Let's compute the probability of drawing at least one cube:

$$ \frac{5}{15} + \frac{10}{15}\cdot \frac{5}{14} = \frac{4}{7}\enspace. $$

The probability of drawing both a cube and a pyramid is

$$ 2 \cdot \frac{5}{15} \cdot \frac{5}{14} = \frac{5}{21} \enspace. $$

Finally, the probability of drawing at least a cube or a pyramid is

$$ \frac{4}{7} + \frac{4}{7} - \frac{5}{21} = \frac{19}{21} \enspace, $$

as expected. So, either the $4/7$ answer is wrong, or the "most likely interpretation" is wrong.

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    so you feel the answer is $19/21$ – 2017-02-25
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    Your first approach makes sense to me completely. I guess I was not considering the other shape. – 2017-02-25
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    If I interpreted the question correctly, it's more than just a feeling. It may be helpful if you quoted the question *verbatim*, in case you haven't already done so. – 2017-02-25
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    "Sandra randomly chooses two of the shapes from the set at the same time. What is the probability that one shape is a cube or a pyramid?" Then the problem shows a box with 5 cubes, 5 pyramids, and 5 cylinders.. then the answers are given as (a) 4/7 (b) 5/9 (c) 29/42 (d) 5/42. book says (a) is answer. – 2017-02-25
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    The probability that one selected shape is a cube is $4/7$ and so is the probability that one selected shape is a pyramid. The probability that Sandra doesn't select two spheres is not $4/7$. – 2017-02-25
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    Your first approach makes sense although I was hoping for an equation to use. – 2017-02-25
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    Something like $1 - P(SS)$? – 2017-02-25
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Other than Fabio's interpretation (which is perfectly reasonable), I can see one other possible interpretation of this problem, which is that you want exactly one cube or exactly one pyramid (or both) ... In which case getting 2 pyramids or getting two cubes is also not what you want.

The chance of getting 2 pyramids is, like the chance of getting 2 spheres, $\frac{2}{21}$. Same for 2 cubes.

So, with this interpretation, the probability of getting 'one cube or one pyramid' is $1-3*\frac{2}{21}=\frac{15}{21}=\frac{5}{7}$ .... Which is yet again not $\frac{4}{7}$ ...

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Hint -

Mistake in your method is that you are skipping a few cases. Cases included in this question are 1 shape is cube other can be sphere or 1 shape can be pyramid other can be sphere or both are cube or both are pyramid or one is pyramid and other cube.

Further notice that shape 1 can be sphere and other is cube or pyramid.

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    so how would I write the equation for this? Would this produce 4/7? – 2017-02-25