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 #6
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Solution:

(a) The number of ordered triples \( (a,b,c),\) where \( 1 \le a, b,c \le n,\) is \(n^3\) We can classify these triples into three categories.

Case 1:  \(a,b,c\) are all equal.

There are \(n\) ordered triples in this case.

Case 2: Two of \(a,b,c\) are equal, and the third is different.

There are \(n\) ways to choose the two equal values, then \(n-1\) ways to choose the third value that is different. There are then \(3\) ways to arrange the numbers within the ordered triple, so the number of ordered triples in this case is \(3n(n-1).\)

Case 3: \(a,b,c\) are all different.

There are \(\dbinom{n}{3}\) ways to choose three different values. There are then \(3!=6\) ways to arrange them, so the number of ordered triples in this case is \(6\dbinom{n}{3}\)

By counting the number of triples \((a,b,c)\) in two different ways, we arrive at the conclusion 

\(n^3=n+3n(n-1)+6\dbinom{n}{3}. \)

(b) The number of subsets of \(\{1,2,3,...,n+2\}\) containing three different numbers is \(\dbinom{n+2}{3}\)

We can also count the number of subsets as follows. Let the subset be \(\{a,b,c\},\) where \(a  Then the middle element  \(b\) is at least \(2\) and at most  \(n+1\) so we can let  \(b=k+1\) where  \(1\le k \le n.\)

Once the middle element  \(b=k+1\) is chosen, the smallest element  \(a\) must be between \(1\) and  \(k\) inclusive, so there are  \(k\) ways of choosing the smallest element  \(a\). The largest element  \(c\) must be between  \(k+2\) and  \(n+2\) inclusive, so there are  \((n+2)-(k-2)+1\) ways of choosing the largest element  \(c\). So the number of subsets where the middle element is  \(k+1\) is  \(k(n-k+1)\) Summing over  \(1 \le k \le n\) we get

\(\dbinom{n+2}{3}=(1)(n)+(2)(n-1)+(3)(n-2)+\cdots+(k)(n-k+1)+\cdots+(n)(1).\)
 

Nov 6, 2022
 #9
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Let us call the circle's center \(O\). We first note that if \(A\) and \(B\) are points on the circle, then triangle \(AOB\) is isosceles with \(AO=BO\). Therefore, if \(AOB\) is an obtuse triangle, then the obtuse angle must be at \(O\). So \(AOB\) is an obtuse triangle if and only if minor arc \(AB\) has measure of more than  \(\dfrac{\pi}{2}\)\((90^{\circ})\).

Now, let the three randomly chosen points be \(A_0\),\(A_1\), and \(A_2\). Let \(\theta\) be the measure of minor arc \(A_0A_1\). Since \(\theta\) is equally likely to be any value from \(0\) to \(\pi\), the probability that it is less than \(\dfrac{\pi}{2}\) is \(\dfrac{1}{2}\).

Now suppose that \(\theta<\dfrac{\pi}{2}\). For the problem's condition to hold, it is necessary and sufficient for point \(A_2\) to lie within \(\dfrac{\pi}{2}\) of both \(A_0\) and \(A_0\) along the circumference. As the diagram below shows, this is the same as saying that \(A_2\) must lie along a particular arc of measure \(\pi-\theta\).



The probability of this occurrence is \(\dfrac{\pi-\theta}{2\pi}=\dfrac{1}{2}-\dfrac{\theta}{2\pi}\), since \(A_2\) is equally likely to go anywhere on the circle. Since the average value of \(\theta\) between \(0\) and \(\dfrac{\pi}{2}\) is \(\dfrac{\pi}{4}\), it follows that the overall probability for \(\theta<\dfrac{\pi}{2}\) is \(\dfrac{1}{2}-\dfrac{\pi/4}{2\pi}=\dfrac{3}{8}\).

Since the probability that \(\theta<\dfrac{\pi}{2}\) is 1/2, our final probability is \(\dfrac{1}{2}\cdot\dfrac{3}{8}=\dfrac{3}{16}\).

Oct 22, 2022
 #2
 #2
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I think you mean that the question is:

 

In the SuperLottery, three balls are drawn (at random, without replacement) from ten white balls numbered from 1 to 10, and one SuperBall is drawn (at random) from ten red balls numbered from 11 to 20. When you buy a ticket, you choose three numbers from 1 to 10 and one number from 11 to 20.

If the numbers on your ticket match at least two of the white balls or match the red SuperBall, then you win a super prize. What is the probability that you win a super prize?

 

Answer:

 

We compute the complement: we'll count the number of losing tickets.

 

To have a losing ticket, you must have at most one correct white ball, and miss the SuperBall.

You miss all 3 white balls if your ticket contains 3 of the 7 white numbers that were not drawn, so there are \(\dbinom{7}{3}=\dfrac{7\cdot6\cdot5}{6}=35\) possibilities.

You hit 1 white ball and miss the others if your ticket contains 1 of the 3 white numbers that were drawn and 2 of the 7 white numbers that were not drawn, so there are \(3\dbinom{7}{3}=\dfrac{3\cdot7\cdot6}{2}=63\) possibilities.

You miss the SuperBall if you have one of the 9 red numbers that were not drawn.

Therefore, there are \((35+63)\cdot9=882\) losing tickets.

Hence, there are \(1200-882=318\) winning tickets, and your probability of winning a super prize is

 

\(\dfrac{318}{1200}=\boxed{\dfrac{53}{200}}.\)
 

 

Note: We can approach this problem using direct counting, but there are a number of cases:

  1.  Matching 3 white balls with any super ball.
  2. Matching 2 white balls with any super ball.
  3. Matching 1 white ball and matching the super ball.
  4. Matching 0 white balls and matching the super ball.

Given the numerous cases here, a complementary counting approach is a faster approach.

Oct 17, 2022
 #6
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+1

Solution:

(a) There are four attributes, and there are three choices for each attribute. In a Set deck, there is one card for each combination of choices, so there are \(3^4=\boxed{81}\) cards.

(b) Note that if we choose any two cards, there is exactly one card that will complete a set. For example, consider the following two cards:

 





Since the first card has three shapes and the second card has one shape, the third card must have two shapes. Since both cards are red, the third card must also be red, and so on. Each attribute of the third card is determined uniquely by the first two cards, which gives us the following set:

 





There are \(\dbinom{81}{2}\) ways to choose two cards, and two cards uniquely generate a set. However, for each set, there are 3 ways that we could have chosen the two cards that generate the set. Therefore, the total number of possible sets is \(\dfrac{1}{3}\dbinom{81}{2}=\boxed{1080}.\)

(c) We want all three cards to be different with respect to every attribute.

There are \(81 \) ways to choose the first card in the set. Then for the second card, there are 2 ways to choose each attribute (since every attribute must be different from the first card), so there are \(2^4=16\) ways to choose the second card. Then the third card is uniquely determined (since every attribute must be different from the first two cards).

This gives us a count of \(81\cdot16=1296.\) But the order of the cards does not matter, so there are \(\dfrac{1296}{3!}=\boxed{216}\) sets in this case.

(d) There is one attribute where all three cards are the same, and they are different for every other attribute.

There are 4 ways to choose the attribute for which the cards are the same, then 3 options for this attribute. There are \(3^3=27\) ways to choose the first card (one attribute has already been determined), then \(2^3=8\) ways to choose the second card, and then the third card is uniquely determined.

So there are \(\dfrac{4\cdot3\cdot27\cdot8}{3!}=\boxed{432}\) sets in this case.

(e) There are two attributes where all three cards are the same, and they are different for every other attribute.

There are \(\dbinom{4}{2}=6\) ways to choose the two attributes for which they are the same, then 3 options for these attributes. There are \(3^2=9\) ways to choose the first card, then \(2^2=4\) ways to choose the second card, and then the third card is uniquely determined.

So there are \(\dfrac{6\cdot3^2\cdot9\cdot4}{3!}=\boxed{324}\) sets in this case.

(f) There are three attributes where all three cards are the same, and they are different for the remaining attribute.

There are \(\dbinom{4}{3}=4\) ways to choose the three attributes for which they are the same, then 3 options for these attributes. There are exactly three cards that meet these conditions, which form a set.

So there are \(4\cdot3^3=\boxed{108}\) sets in this case.

As a check, the answers for parts (c)-(f) add up to \(216+432+324+108=1080\) which is the answer for part (b).

Oct 8, 2022