Question

My Notes Ask Your Teacher 1 2/3 points | Previous Answers An experimenter is studying the effects of temperature, pressure, and type of catalyst on yield from a certain chemical reaction. She considers 7 different temperatures, 5 different pressures, and 8 different catalysts. (a) If any particular experimental run involves the use of a single temperature, a single pressure and, a single catatyst, how many experimental runs are possible? 280 ways (b) How many experimental runs are there that involve the use of the lowest temperature and two lowest pressures? (Hint: There is only one choice for temperature and two choices for pressure.) 16 ways (c) Suppose that two different experimental runs are to be made on the first day of experimentation. If the two are randomly selected from all possibilities, so that any group of two runs has the same probability of selection, what is the probability that a different catalyst is used on each run? (Hint: Count the number of ways you can have a sequence of two runs with different catalysts, then divide that by the total number of sequences of ANY two different run.) it Answer Save Progress Practice Another Version

Answer 1

c) Number experimental runs were possible using 7 different temperatures, 5 different pressures and 8 different catalysts = 7*5*8 = 280

Let A = {two randomly selected experimental runs have different catalysts}

Since each outcome in A is equally likely,

The probability of event A i.e.

  $\fn_cm P(A) = \frac{N(A)}{N}$

where, N(A) = the number of outcomes in A

N = the total number of outcomes possible

Now, to determine number of combinations of size 8 formed from 280 experimental runs,

$\fn_cm N = \binom{280}{8} = 8.47\times 10^{14}$

Next, to determine the number of experimental runs that can be made for each individual catalyst.For each catalyst there are 7 different temperatures and 8 different pressures. So for one catalyst there is

7*5*1 = 35 different experimental runs

Since each catalyst has 35 experimental runs associated with it, by the product rule

$\fn_cm N(A)= 35^{8} = 2.25\times 10^{12}$

So,

$\fn_cm P(A) = \frac{N(A)}{N}$

$\fn_cm P(A) = \frac{2.25\times 10^{12}}{8.47\times 10^{14}}$

$\fn_cm =2.6564\times 10^{-3}$

= 0.0026564