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 probability of the following events.
1) Exactly 5 days are selected from each of the 12 months.
a) (30 C 5)^2 / 360 C 60
b) (30 C 5) / 360 C 60
c) 30^12 / 360^60
d) 360 / (30 C 5)^12
2) None of the selected days is from January.
a) 1 - (30 C 12)/ (360 C 60)
b) 1 - (330 C 60)/(360 C 60)
c) (330 C 60) / (360 C 60)
d) 1/12
3) There exist 3 different months such that exactly 20 days are selected from each one of these months.
a) (30 C 20)^3 / (360 C 60)
b) (90 C 60) / (360 C 60)
c) ((12 C 3) * (30 C 20)^3) / ( 360 C 60)
d) 1/4
NOTE: n C k = n! / k! (n - k)!
1) Exactly 5 days are selected from each of the 12 months.
a) (30 C 5)^2 / 360 C 60 - This is the probability of selecting exactly 5 days from one month squared, divided by the total number of possible sets of 60 days. Incorrect.
b) (30 C 5) / 360 C 60 - This is the probability of selecting exactly 5 days from one month, divided by the total number of possible sets of 60 days. Incorrect.
c) 30^12 / 360^60 - This is the probability of selecting any 60 days out of the year, without any restrictions. Incorrect.
d) 360 / (30 C 5)^12 - This is the probability of selecting any 60 days out of the year, divided by the total number of possible sets of 5 days from each month, raised to the power of 12 (since it happens in all 12 months). Incorrect.
Answer: None of the options are correct.
2) None of the selected days is from January.
a) 1 - (30 C 12)/ (360 C 60) - This is the probability of selecting any 60 days out of the year, minus the probability of selecting 12 days from January and any 48 days from the rest of the year, divided by the total number of all possible sets of 60 days. Incorrect.
b) 1 - (330 C 60)/(360 C 60) - This is the probability of selecting any 60 days out of the year, minus the probability of selecting 60 days from January and any 300 days from the rest of the year, divided by the total number of all possible sets of 60 days. Incorrect.
c) (330 C 60) / (360 C 60) - This is the probability of selecting 60 days from the rest of the year without selecting any day from January, divided by the total number of all possible sets of 60 days. Correct.
d) 1/12 - This is the probability of selecting any day from the year that is not from January. Incorrect.
Answer: c) (330 C 60) / (360 C 60)
3) There exist 3 different months such that exactly 20 days are selected from each one of these months.
a) (30 C 20)^3 / (360 C 60) - This is the probability of selecting exactly 20 days from one month cubed, divided by the total number of all possible sets of 60 days. Incorrect.
b) (90 C 60) / (360 C 60) - This is the probability of selecting any 60 days from three months, divided by the total number of all possible sets of 60 days. Incorrect.
c) ((12 C 3) * (30 C 20)^3) / ( 360 C 60) - This is the probability of selecting exactly 20 days from three different months (which can be chosen in 12 C 3 ways) cubed, divided by the total number of all possible sets of 60 days. Correct.
d) 1/4 - This is an arbitrary probability and does not consider the specific requirements given. Incorrect.
Answer: c) ((12 C 3) * (30 C 20)^3) / ( 360 C 60)
To evaluate the probability of the given events, we need to understand the problem and apply the appropriate formulas.
1) Exactly 5 days are selected from each of the 12 months:
The total number of ways to select 60 days from the 360-day year is given by (360 C 60) because we are choosing a subset of 60 days from a total of 360 days.
The number of ways to choose 5 days from a 30-day month is given by (30 C 5), and since we have 12 months, the number of ways to choose 5 days from each month is (30 C 5)^12.
Therefore, the probability is (30 C 5)^12 / (360 C 60).
Answer: a) (30 C 5)^2 / (360 C 60)
2) None of the selected days is from January:
The number of ways to choose 60 days from the remaining 330 days (excluding January) is given by (330 C 60).
The total number of ways to choose 60 days from the 360-day year is (360 C 60).
Therefore, the probability is 1 - (330 C 60) / (360 C 60).
Answer: b) 1 - (330 C 60) / (360 C 60)
3) There exist 3 different months such that exactly 20 days are selected from each one of these months:
The number of ways to choose 20 days from a 30-day month is given by (30 C 20), and since we have 3 different months, the number of ways to choose 20 days from each of these months is (30 C 20)^3.
To choose the remaining 60 - (20 * 3) = 60 - 60 = 0 days from the remaining 9 months, we have only one option.
The total number of ways to choose 60 days from the 360-day year is (360 C 60).
Therefore, the probability is (30 C 20)^3 / (360 C 60).
Answer: a) (30 C 20)^3 / (360 C 60)
Note: The formulas mentioned in the options are not correct.