Probabilities

[jsbeckton]

An experimenter is studying the effects of temperature, pressure and type of cataylst for a reaction. 3 different pressures, 4 different temperatures and 5 different catalysts are used. Suppose 5 different experimental runs are to be made on the same day. If the 5 are ramdomally selected from among all possibilities, so that any group of 5 has the same probability of selection, what is the probability that a different catalyst will be used in each run?

I was thinking that the anwser woud be given by this equation:

\(\displaystyle \begin{array}{l}
P( A\nolimits_1 { \cap A}\nolimits_2 \cap \nolimits_{} A\nolimits_3 { \cap A}\nolimits_4 { \cap A}\nolimits_5 ) = P\left( { A\nolimits_1 } \right) \times P\left( { A\nolimits_2 } \right) \times P\left( { A\nolimits_3 } \right) \times P\left( { A\nolimits_4 } \right) \times P\left( { A\nolimits_5 } \right) \\
\\
= \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} \\
\\
= \frac{1}{{3125}} \\
\\
\approx .00032 \\
\end{array}\)

Which is not the anwser in the back (.0456)
Can anyone tell me where I'm going wrong here? Thanks.

 

[pka]

Because (3)(4)(5)=60 so there are 60 different possible experiments.
If the five catalysis A, B, C, D & E, then there are 12 of the 60 that contain say A.
Thus we can select the first experiment is any way from the 60.
Select the second in 48 ways from 59, so that the first two have different catalysis.
To continue until we select five different experiments.
To get the answer \(\displaystyle \prod\limits_{k = 0}^4 {\frac{{60 - 12k}}{{60 - k}}} = 0.0456\)

 

[jsbeckton]

Thanks, that makes more sence than my book.