# Thread: Losing my marbles.... Probability

1. ## Losing my marbles.... Probability

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:

$P(at least 1 agate)= \frac{4}{14} + \frac{4}{13} = \frac{54}{91}$
Is this correct? If not then how do I work it out?

2. Originally Posted by TommyBoy22
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:

$P(at least 1 agate)= \frac{4}{14} + \frac{4}{13} = \frac{54}{91}$
Is this correct? If not then how do I work it out?
Hi TommyBoy,

$P(\text{at least 1 agate})=1-(\frac{10}{14})(\frac{9}{13})$

3. Hello, TommyBoy22!

Sorry, your work is way off . . .
What formulas (if any) are you using?

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?
There are 14 marbles: 3 CEs, 4 AGs, 7 OBs.

Two marbles are drawn.
. . There are: . $_{14}C_2 \:=\:91$ possible outcomes.

The opposite of "at least one AG" is "no AGs."

There are: 4 AGs and 10 Others.
There are: . $_{10}C_2 \:=\:45$ ways to draw two Others (no AGs).
. . Hence, there are: . $91 - 45 \:=\:46$ ways to draw at least one AG.

Therefore: . $P(\text{at least one AG}) \:=\:\frac{46}{91}$

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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: . $\frac{4}{14}\cdot\frac{10}{13} \:=\:\frac{20}{91}$

. . Other, then AG: . $\frac{10}{14}\cdot\frac{4}{13} \:=\:\frac{20}{91}$

There is one way to get two AGs:

. . AG, then AG: . $\frac{4}{14}\cdot\frac{3}{13} \:=\:\frac{6}{91}$

Therefore: . $P(\text{one AG or two AGs}) \;=\;\frac{20}{91} + \frac{20}{91} + \frac{6}{91} \;=\;\frac{46}{91}$

4. Hi thanks guys.. yeah no theory used.. just made sense in my head