Question 1
Positive integer \(n\), then the following statements are true:
1. \(n\) is greater than zero: A positive integer \(n\) is, by definition, greater than zero. It is a whole number that is not zero or negative.
2. \(n\) is not a fraction or decimal: Positive integers are discrete values that are not expressed as fractions or decimals. They are whole numbers.
3. \(n\) can be incremented or decremented by 1: Since \(n\) is a positive integer, it can be increased or decreased by a value of 1. For example, if \(n = 5\), then \(n + 1 = 6\) and \(n - 1 = 4\).
4. \(n\) is a count or a label: Positive integers are often used as counts or labels in various contexts. For example, they can represent the number of items in a set, the position in a sequence, or the order of a term in a series.
It's important to note that these statements are true only within the context of positive integers. If we were to extend our discussion to include other number systems or real numbers, these statements might not hold true.
Question 2
To determine the convergence of the sequence given by \(a_n = \frac{3^n}{n!}\), we can use the ratio test or the root test.
1. Ratio Test:
The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series converges.
Let's apply the ratio test to the sequence \(a_n = \frac{3^n}{n!}\):
\[
\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{{n \to \infty}} \left| \frac{\frac{3^{n+1}}{(n+1)!}}{\frac{3^n}{n!}} \right|
= \lim_{{n \to \infty}} \left| \frac{3^{n+1} \cdot n!}{3^n \cdot (n+1)!} \right|
= \lim_{{n \to \infty}} \left| \frac{3}{n+1} \right|
= 0
\]
Since the limit is 0, which is less than 1, the series converges by the ratio test.
2. Root Test:
The root test states that if the limit of the \(n\)th root of the absolute value of each term is less than 1, then the series converges.
Applying the root test to our sequence:
\[
\lim_{{n \to \infty}} \sqrt[n]{\left| \frac{3^n}{n!} \right|}
= \lim_{{n \to \infty}} \frac{3}{\sqrt[n]{n!}}
\]
To simplify further, we can use Stirling's approximation for \(n!\):
\[
n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n
\]
Substituting Stirling's approximation into the root test expression:
\[
\lim_{{n \to \infty}} \frac{3}{\sqrt[n]{n!}}
= \lim_{{n \to \infty}} \frac{3}{\sqrt[n]{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}}
= \lim_{{n \to \infty}} \frac{3}{\left(\sqrt{2\pi n}\right)^{1/n} \left(\frac{n}{e}\right)}
= \frac{3}{\frac{1}{e}}
= 3e
\]
Since the limit is greater than 1 (approximately 8.1548), the series diverges by the root test.
Therefore, based on the ratio test, we can conclude that the series given by \(a_n = \frac{3^n}{n!}\) converges.
Question 3
A magician has 20 coins in his pocket. Twelve of these coins are normal fair coins (with one head and onetail) and eight are defective coins with heads on both sides. The magician randomly draws a coin from hispocket and dips it. Given that the dipped coin shows a head, what is the probability that it is defective?
To solve this problem, we can use Bayes' theorem. Let's denote the events as follows: A: The coin drawn is defective. B: The flipped coin shows a head. We want to find the conditional probability that the coin is defective given that it shows a head, P(A|B). According to Bayes' theorem: P(A|B) = (P(B|A) * P(A)) / P(B) P(B|A) is the probability that the coin shows a head given that it is defective. Since all the defective coins have heads on both sides, P(B|A) = 1. P(A) is the probability of drawing a defective coin. Out of the 20 coins in the magician's pocket, 8 are defective. Therefore, P(A) = 8/20 = 2/5. P(B) is the probability that the coin shows a head, regardless of whether it is defective or not. There are two ways this can happen: either the coin drawn is defective and shows a head, or the coin drawn is normal and shows a head. So: P(B) = P(B|A) * P(A) + P(B|A') * P(A') P(B|A') is the probability that the coin shows a head given that it is not defective. For a normal coin, the probability of showing a head is 1/2. P(A') is the probability of drawing a normal coin, which is 12/20 = 3/5. Therefore: P(B) = 1 * (2/5) + (1/2) * (3/5) = 2/5 + 3/10 = 7/10. Now we can substitute these values into Bayes' theorem: P(A|B) = (1 * (2/5)) / (7/10) = (2/5) * (10/7) = 4/7. Therefore, the probability that the coin is defective given that it shows a head is 4/7.
Question 4
A class has 30 girls and 22 boys. A teacher selects 15 students at random from the class to participate in acompetition. What is the probability that out of the 15 selected students, 6 of them are boys
To find the probability that out of the 15 selected students, exactly 6 of them are boys, we can use the concept of combinations. The total number of ways to select 15 students out of the 52 students in the class (30 girls + 22 boys) is given by the combination formula: C(52, 15) = 52! / (15! * (52 - 15)!) = 52! / (15! * 37!) Now, let's calculate the number of ways to select 6 boys from the 22 available boys and 9 girls from the 30 available girls: C(22, 6) = 22! / (6! * (22 - 6)!) = 22! / (6! * 16!) C(30, 9) = 30! / (9! * (30 - 9)!) = 30! / (9! * 21!) To calculate the desired probability, we divide the number of favorable outcomes (selecting 6 boys and 9 girls) by the total number of possible outcomes (selecting any 15 students): P(6 boys) = (C(22, 6) * C(30, 9)) / C(52, 15) P(6 boys) = (22! / (6! * 16!)) * (30! / (9! * 21!)) / (52! / (15! * 37!)) Calculating this expression will give you the probability that exactly 6 out of the 15 selected students are boys.
Question 5
An island consists of four kinds of people: Tetas, Jekas, Frekas and Hekas. The following information isknown:Every Heka is either a Teta or a Jeka, but not bothAll Frekas are JekasNo Frekas are TetasConsider the following statements:[i.] No Tetas are Jekas[ii.] Some Hekas are FrekasWhich of these statements are necessarily true based only on the information above
Based on the given information, we can determine which of the statements are necessarily true: [i.] No Tetas are Jekas: This statement is necessarily true. It is mentioned that every Heka is either a Teta or a Jeka but not both. Since there are no Tetas that are Hekas, and all Hekas are either Tetas or Jekas, it implies that there are no Tetas that are Jekas. [ii.] Some Hekas are Frekas: This statement cannot be necessarily true based on the given information. It is possible that all Hekas are Tetas, and since no Tetas are Jekas, it means no Hekas are Jekas. In this case, there would be no Hekas that are Frekas. To summarize, statement [i.] "No Tetas are Jekas" is necessarily true based on the given information, but statement [ii.] "Some Hekas are Frekas" is not necessarily true.
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