Explanation:
In problem 16 in section 4.2, we have a box of 11 markers, with 6 containing ink and 5 not containing ink. We randomly select a sample of 8 markers and define the random variable Y as the number of markers selected that do not have ink.
To determine how many different values are possible for the random variable Y, we need to consider the different combinations of markers that can be selected.
The maximum number of markers without ink that can be selected is 5, since there are only 5 markers without ink in the box. The minimum number of markers without ink that can be selected is 0, as it is possible to select all 8 markers with ink.
To find the number of different values for Y, we need to consider the number of ways we can choose k markers without ink from the 5 available, where k ranges from 0 to 5. This can be calculated using the combination formula:
C(n, k) = n! / (k! * (n-k)!)
where n is the total number of markers without ink and k is the number of markers without ink selected.
Using this formula, we can calculate the number of different values for Y:
C(5, 0) = 5! / (0! * (5-0)!) = 1
C(5, 1) = 5! / (1! * (5-1)!) = 5
C(5, 2) = 5! / (2! * (5-2)!) = 10
C(5, 3) = 5! / (3! * (5-3)!) = 10
C(5, 4) = 5! / (4! * (5-4)!) = 5
C(5, 5) = 5! / (5! * (5-5)!) = 1
Therefore, the random variable Y can take on 6 different values: 0, 1, 2, 3, 4, and 5.
To complete the probability density function (PDF) table, we need to calculate the probability of each value of Y occurring.
Let's denote P(Y = y) as the probability of Y taking on the value y.
P(Y = 0) = C(5, 0) * C(6, 8-0) / C(11, 8)
P(Y = 1) = C(5, 1) * C(6, 8-1) / C(11, 8)
P(Y = 2) = C(5, 2) * C(6, 8-2) / C(11, 8)
P(Y = 3) = C(5, 3) * C(6, 8-3) / C(11, 8)
P(Y = 4) = C(5, 4) * C(6, 8-4) / C(11, 8)
P(Y = 5) = C(5, 5) * C(6, 8-5) / C(11, 8)
These probabilities can be calculated using the combination formula mentioned earlier.