OK so I have a bag of marbles containing: 3 Catseyes, 4 Agates and 7 Oxbloods. If all have an equal chance of being picked and two are drawn, what are the chances of at least 1 of the chosen being an Agate.
According to my limited learning I worked it out as follows:
Is this correct? If not then how do I work it out?
Hello, TommyBoy22!
Sorry, your work is way off . . .
What formulas (if any) are you using?
There are 14 marbles: 3 CEs, 4 AGs, 7 OBs.I have a bag of marbles containing: 3 Catseyes, 4 Agates and 7 Oxbloods.
If all have an equal chance of being picked and two are drawn,
what is the probability of at least one Agate being chosen?
Two marbles are drawn.
. . There are: . possible outcomes.
The opposite of "at least one AG" is "no AGs."
There are: 4 AGs and 10 Others.
There are: . ways to draw two Others (no AGs).
. . Hence, there are: . ways to draw at least one AG.
Therefore: .
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From your work, I assume you are working with a sequence of draws.
In that case, we must consider all the cases.
We want at least one AG.
. . This means: (1 AG and 1 Other), or (2 AGs).
There are two ways to get one AG and one Other:
. . AG, then Other: .
. . Other, then AG: .
There is one way to get two AGs:
. . AG, then AG: .
Therefore: .