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There are three buttons which are painted red on one side and white on the other. If we tosses the buttons into the air, calculate the probability that all three come up the same color.

Remarks: A wrong way of thinking about this problem is to say that there are four ways they can fall. All red showing, all white showing, two reds and a white or two whites and a red. Hence, it seem that out of four possibilities, there are two favorable cases and hence the probability is 1/2.

Who knows it well, please answers my question. Thank you very much. :)

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    $\begingroup$ The issue is assuming that all of those $4$ events are of equal probability. $\endgroup$ – Cameron Buie Sep 1 '15 at 4:08
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You have to distinguish between the three buttons. There are the following possible outcomes:

$\color{red}{rrr}$

$rrw$

$rwr$

$wrr$

$rww$

$wrw$

$wwr$

$\color{red}{www}$

The favorable outcomes are marked red. Now you can divide the number of favorable outcomes by the number of possible outcomes.

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While there are four distinct events as identified, they are not equiprobable.   We must include weighting by counting what equiprobable outcomes form each event.

Assuming that there is no bias for any particular button showing white or red, then each of the three buttons has an equiprobable choice of two colours. Thus there are $8$ equiprobable outcomes.

Now there is only one such way the three buttons could all show white. Similarly for the all red even.

However, when one button is of a different colour than the other two, there are three buttons that could be that different colour.   Hence the weight of each of these events is thrice that of each of the all-same-colour events.

Thus we have that: $\quad$ $\Pr(\text{all white}) = \frac 1 8 \\ \Pr(\text{2 white, 1 red}) = \frac 3 8 \\ \Pr(\text{2 red, 1 white}) = \frac 3 8 \\ \Pr(\text{all red}) = \frac 1 8$

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