Question 1195890
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A bag contains 8 red marbles, 3 blue marbles and 5 green marbles. 
If three marbles are drawn out of the bag, what is the exact probability 
that all three marbles drawn will be green?
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<pre>
At the start, there are  8 + 3 + 5 = 16  marbles in the bag.

Of them, 5 marbles are green.


The probability that the first drawn marble is green equals  {{{5/16}}}.

After that, we have 15 marbles in the bag with 4 green marbles.
So, the probability that the first two drawn marbles are green equals  {{{(5/16)*(4/15)}}}.


Analyzing further in the same way, we find that the probability
that the first three drawn marbles are green equals 

    {{{(5/16)*(4/15)*(3/14)}}} = reduce the fractions = {{{(1/4)*(1/3)*(3/14)}}} = {{{(1/4)*(1/1)*(1/14)}}} = {{{1/56}}}.   <U>ANSWER</U>
</pre>

Solved.


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Ignore the solution by &nbsp;@MathLover1, &nbsp;since it is totally wrong.


It is a good/classic example on how this problem &nbsp;SHOULD &nbsp;NOT &nbsp;be solved.



Avoid such errors in the future &nbsp;(&nbsp;!&nbsp;)



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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;After reading the post by Alan, I find it necessary to explain 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;on how to read the problem and what does it really mean.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;When such problem goes without explicit pointing on &nbsp;" with or without replacement ",

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;then by &nbsp;DEFAULT &nbsp;and by the context, &nbsp;it &nbsp;ALWAYS &nbsp;means &nbsp;" without replacement ".