Probability Question: 4 tiles have P, 3 tiles have E and 2 tiles have T.

I need help with this..

A bag contains 9 letter tiles. 4 tiles have P, 3 tiles have E and 2 tiles have T. 3 tiles are selected without replacement. Calculate the probability that the word PET can be spelt with the 3 tiles.

Mary123 said:

I need help with this..

A bag contains 9 letter tiles. 4 tiles have P, 3 tiles have E and 2 tiles have T. 3 tiles are selected without replacement. Calculate the probability that the word PET can be spelt with the 3 tiles.

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Mary123 said:

A bag contains 9 letter tiles. 4 tiles have P, 3 tiles have E and 2 tiles have T. 3 tiles are selected without replacement. Calculate the probability that the word PET can be spelled with the 3 tiles.

Click to expand...

Please note that this is sampling without replacement.
Without my going into details, look at this calculation.
See the term \(\displaystyle 9x^3\). That tells us that there are nine different samples of three selections.
Here are four of them:\(\displaystyle <P,P,P>,~<P,T,T>,~<E,E,T>,~\&~<T,E,P>\)
There is only one way to draw the first sample. So the probability is \(\displaystyle \dfrac{4}{9}\cdot\dfrac{3}{8}\cdot\dfrac{2}{7}\).

The second sample can be drawn in three ways: \(\displaystyle PTT,~TPT,~TTP\). So the probability is \(\displaystyle 3\cdot\dfrac{4}{9}\cdot\dfrac{2}{8}\cdot\dfrac{1}{7}\).

But only the last one can be rearranged to spell \(\displaystyle PET\)
So what is the probability of drawing the last one?

Answer show some work of your own.