Question 1 : We would like to test 3 different treatments on a particu-
lar type of plant. A worker at the local greenhouse will allow us to use 3
among 10 plants. Note: Each treatment will be used on one and only one
plant.
a) In how many different ways can we assign the plants to the treatments?
b) Suppose that 2 of the ten plants are older than the other 8 plants. Assuming that we are choosing the plants randomly, what is the probability that at least one of the two older plants will be chosen among the three selected
plants?
There are 10C3 10 choose 3 ways to select a plant, and 3 solutions we could use on a plant. You should be able to calculate this.
The easier way to calculate b) is to observe that there are
8C3 ways to select only younger plants out of 10C3 total ways to select plants. Thus the prob of only younger plants is
(8C3)/(10C3) and the prob that at least on plant is an older one is
1 - (8C3)/(10C3)
To solve question 1a, we need to find the number of different ways we can assign the plants to the treatments.
First, we need to choose 3 plants out of the 10 available. This can be done in "10 choose 3" ways, which can be denoted as 10C3 or written as "10! / (3! * (10-3)!)". This calculation gives us the number of ways we can select 3 plants from a pool of 10.
Once we have chosen the 3 plants, we need to assign each of them to one of the 3 treatments. Since each treatment can be used on one and only one plant, we have 3 options for assigning the first plant, 2 options for assigning the second plant, and only 1 option for assigning the third plant. Therefore, the total number of ways to assign the plants to the treatments is 3 x 2 x 1 = 6.
So the answer to question 1a is that there are 6 different ways we can assign the plants to the treatments.
Moving on to question 1b, we need to find the probability that at least one of the two older plants will be chosen among the three selected plants.
Out of the 10 plants, 2 are older and 8 are younger. We want to calculate the probability that at least one of the two older plants will be chosen, which means we want to find the probability of selecting 1 or 2 older plants out of the 3 selected plants.
To do this, we can calculate the probability of selecting only younger plants and subtract it from 1 to get the probability of selecting at least one older plant.
The number of ways to select only younger plants can be calculated by choosing 3 plants out of the 8 younger plants. This is represented as "8 choose 3" or 8C3, which is equal to "8! / (3! * (8-3)!)".
The total number of ways to select 3 plants out of the 10 available is "10 choose 3" or 10C3, which we calculated earlier.
So, the probability of selecting only younger plants is (8C3) / (10C3).
Finally, the probability of selecting at least one older plant is 1 minus the probability of selecting only younger plants. Therefore, the probability can be calculated as:
1 - (8C3) / (10C3)
This will give us the probability that at least one of the two older plants will be chosen among the three selected plants.