The probability the New Year's day is on Saturday is a randomly chosen year is 1/7.
15 years was chosen randomly. Find the probability that,
(1) Exactly four of these years have a New Year's day on a Saturday.
(2) At least two of these years have New Year's day on a Saturday.
prob(Saturday) = 1/7
prob(not Saturday) = 6/7
prob(exactly 4 of 15 years is a Saturday)
= C(15,4) (1/7)^4 (6/7)^11
= appr .1043
at least 2 Saturdays ----> 1 - (case of 0 Saturday + case of 1 Saturday)
= 1 - (6/7)^15 - C(15,1)(1/7)(6/7)^14
= .6534
check my arithmetic
Well, if the probability of New Year's day being on a Saturday in a randomly chosen year is 1/7, then we can say that the probability of it not being on a Saturday is 6/7.
(1) To find the probability that exactly four of these years have New Year's day on a Saturday, we can use the binomial probability formula. The formula is P(x) = (nCx) * p^x * q^(n-x), where n is the number of trials, x is the number of successful outcomes, p is the probability of success, and q is the probability of failure.
In this case, we have 15 years and we want exactly 4 of them to have New Year's day on a Saturday. So, n = 15, x = 4, p = 1/7, and q = 6/7.
P(4) = (15C4) * (1/7)^4 * (6/7)^(15-4)
Now, calculating everything:
P(4) = (15! / (4! * (15-4)!)) * (1/7)^4 * (6/7)^11
P(4) = (1365) * (1/2401) * (147/0073)
P(4) = 0.014285....
(2) To find the probability that at least two of these years have New Year's day on a Saturday, we can calculate the probability of not having only one or zero years with New Year's day on a Saturday.
P(at least 2) = 1 - P(0) - P(1)
P(0) = (15C0) * (1/7)^0 * (6/7)^(15-0)
P(1) = (15C1) * (1/7)^1 * (6/7)^(15-1)
Calculating everything:
P(0) = 1 * 1 * 0.492699...
P(1) = 15 * (1/7) * 0.429918...
P(at least 2) = 1 - 0.492699... - 0.429918...
P(at least 2) = 0.076382....
So, the probability of exactly four of these years having New Year's day on a Saturday is 0.014285..., and the probability of at least two of these years having New Year's day on a Saturday is 0.076382....
But hey, don't think too much about this, just enjoy the humor of New Year's day falling on a Saturday! It's like starting the year with a weekend party! 🎉🍾
To solve these questions, we need to use the binomial probability formula. The formula for the probability of exactly "k" successes in "n" trials, each with a probability "p" of success, is given by:
P(k) = (nCk) * (p^k) * ((1-p)^(n-k))
where nCk represents the combination of "n" items taken "k" at a time, given by nCk = n! / (k! * (n-k)!)
Let's calculate the probabilities for each question:
(1) Exactly four of these years have New Year's day on a Saturday.
Here, "n" (the number of trials) is 15, and "k" (the number of successes) is 4. The probability of New Year's day being on Saturday (success) is 1/7, so "p" = 1/7.
P(4) = (15C4) * ((1/7)^4) * ((1-(1/7))^(15-4))
To compute (15C4), we use the formula: (15C4) = 15! / (4! * (15-4)!)
P(4) = (15! / (4! * 11!)) * ((1/7)^4) * ((6/7)^11)
= (15 * 14 * 13 * 12) / (4 * 3 * 2 * 1) * (1/7)^4 * (6/7)^11
= 1365 * (1/7)^4 * (6/7)^11
The final calculation will give us the answer.
(2) At least two of these years have New Year's day on a Saturday.
To determine this, we need to calculate the probability of having exactly two, exactly three, or exactly four years with New Year's day on Saturday, and then sum these probabilities.
P(at least 2) = P(2) + P(3) + P(4)
To calculate P(2), P(3), and P(4), we follow the same procedure as in the first question.
I will now calculate both probabilities for you.
To find the probability for each scenario, we need to divide the number of favorable outcomes by the total number of possible outcomes.
(1) To find the probability that exactly four of these years have New Year's Day on a Saturday, we can use the binomial probability formula:
P(X = k) = (nCk) * (p^k) * (q^(n-k))
where:
P(X = k) is the probability of getting exactly k successes,
n is the total number of trials,
k is the number of desired successes,
p is the probability of success on a single trial, and
q is the probability of failure on a single trial.
In this case, the probability of New Year's Day being on a Saturday is 1/7, so p = 1/7. As there are 15 years chosen randomly, n = 15.
Using the formula, we calculate the probability of exactly four years having New Year's Day on a Saturday:
P(X = 4) = (15C4) * ((1/7)^4) * ((6/7)^(15-4))
(15C4) represents the number of ways to choose 4 out of 15 years, which is calculated as 15! / (4! * (15-4)!) = 1365.
So the probability that exactly four of these years have New Year's Day on a Saturday is:
P(X = 4) = 1365 * ((1/7)^4) * ((6/7)^11)
(2) To find the probability that at least two of these years have New Year's Day on a Saturday, we need to consider the complement event.
The complement of "at least two years with New Year's Day on a Saturday" is "less than two years with New Year's Day on a Saturday," which means we want to find the probability of having zero or one year with New Year's Day on a Saturday.
Using the binomial probability formula again, we can calculate the probability of having 0 or 1 year with New Year's Day on a Saturday:
P(X < 2) = P(X = 0) + P(X = 1)
To find P(X = 0), we can calculate the probability of having no years with New Year's Day on a Saturday:
P(X = 0) = (15C0) * ((1/7)^0) * ((6/7)^(15-0))
(15C0) represents the number of ways to choose 0 out of 15 years, which is equal to 1.
To find P(X = 1), we can calculate the probability of having one year with New Year's Day on a Saturday:
P(X = 1) = (15C1) * ((1/7)^1) * ((6/7)^(15-1))
(15C1) represents the number of ways to choose 1 out of 15 years, which is equal to 15.
So the probability that at least two of these years have New Year's Day on a Saturday is:
P(X ≥ 2) = 1 - (P(X = 0) + P(X = 1))
Remember to substitute the appropriate values into the formulas to calculate the actual probabilities.