A set of 60 different days is selected from a given year. Assume that all sets of cardinality 60 are equally likely. Also, for simplicity, assume that the year has only 360 days, divided into twelve 30-day months. Evaluate the probabilities of the following events.
1.Exactly 5 days are selected from each of the 12 months.
2.None of the selected days is from January.
3.There exist 3 different months such that exactly 20 days are selected from each one of these months.
To evaluate the probabilities of the given events:
1. Exactly 5 days are selected from each of the 12 months:
In a set of 60 different days, if exactly 5 days are selected from each of the 12 months, then the total number of ways to select these days is given by selecting 5 days from each month, which can be calculated as:
C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5)
Now, the total number of ways to select any 60 different days from 360 days is:
C(360, 60)
Therefore, the probability of exactly 5 days selected from each of the 12 months is given by:
[C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5) * C(30, 5)] / C(360, 60)
2. None of the selected days is from January:
Since there are 30 days in January, to not select any day from January, we need to select all the 60 days from the remaining 330 days (360 - 30).
Therefore, the probability of none of the selected days being from January is given by:
C(330, 60) / C(360, 60)
3. There exist 3 different months such that exactly 20 days are selected from each one of these months:
To have exactly 20 days selected from each of the 3 different months, we need to select 20 days from each of these months. The total number of ways to do this is:
C(30, 20) * C(30, 20) * C(30, 20)
Now, we need to choose these 3 months from the 12 available months. The total number of ways to do this is given by:
C(12, 3)
Therefore, the probability of there existing 3 different months such that exactly 20 days are selected from each one of them is given by:
[C(30, 20) * C(30, 20) * C(30, 20)] * C(12, 3) / C(360, 60)
To evaluate the probabilities of the given events, we need to calculate the total number of possible sets of 60 different days that can be selected from a year with 360 days.
1. Exactly 5 days are selected from each of the 12 months:
To calculate the number of sets that satisfy this condition, we start by choosing 5 days from the first month (30 days), then 5 days from the second month, and so on, until we choose 5 days from the twelfth month.
The number of possible sets can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!)
Where n is the total number of items to choose from, and r is the number of items to select.
So, for the first month, there are C(30, 5) ways to choose 5 days, for the second month there are C(30, 5) ways, and so on.
Therefore, the total number of sets that have exactly 5 days from each of the 12 months is:
N1 = C(30, 5) * C(30, 5) * ... * C(30, 5) (12 times)
2. None of the selected days is from January:
Since we have already selected exactly 5 days from each of the other 11 months, we need to choose the remaining 60 - 5 * 11 = 5 days from the month of January (which has 30 days).
Again, we can use the combination formula to find the number of sets that satisfy this condition:
N2 = C(30, 5)
3. There exist 3 different months such that exactly 20 days are selected from each one of these months:
First, we need to choose 20 days from each of the three selected months. Following the same logic as in the first event, the number of sets that satisfy this condition is:
N3 = C(30, 20) * C(30, 20) * C(30, 20)
Finally, to calculate the probabilities, we divide the number of sets that satisfy each event by the total number of possible sets (which is the same for all events):
P1 = N1 / C(360, 60)
P2 = N2 / C(360, 60)
P3 = N3 / C(360, 60)