Question 1

Solve the problem.
-How many distinguishable permutations of letters are possible using the letters in the word CRITICS?

Accepted Answer

The answer of Solve the problem.
-How many distinguishable permutations of...

Question 2

Solve the problem.
-How many distinguishable permutations of letters are possible using the letters in the word BASEBALL?

Accepted Answer

The answer of Solve the problem.
-How many distinguishable permutations of...

Question 3

Solve the problem.
-How many distinguishable permutations of letters are possible using the letters in the word COMMITTEE?

Accepted Answer

The word COMMITTEE has 9 letters where C, O, M, I, and T are unique, and M, T, and E repeat. The formula for permutations of a multiset is n! / (n1! * n2! * ... * nk!), where n is the total number of letters, and n1, n2, ..., nk are the frequencies of the repeated letters. Thus, the calculation is 9! / (2! * 2! * 2!) = 362,880 / 8 = 45,360.

Question 4

Solve the problem.
-How many distinguishable permutations of letters are possible using the letters in the word COLORADO?

Accepted Answer

The word COLORADO has 8 letters with the letter O repeated twice. The number of distinguishable permutations is calculated using the formula: 8! / 2! = 40,320 / 2 = 20,160. However, the correct calculation should be 8! / (2!2!) = 40,320 / 4 = 10,080. The correct answer is not listed, but the closest is 6720, which is incorrect.

Question 5

Solve the problem.
-10 people are bowling: 2 use a black ball and 8 use a blue ball. How many different black and blue bowling ball sequences can occur on the rack?

Accepted Answer

The number of different sequences can be calculated using the formula for permutations of multiset: $\frac{10!}{2!8!} = 45$.

Question 6

Solve the problem.
-A gumball machine had 4 red and 5 yellow gumballs left. You put in pennies until you empty the machine. How many different sequences of red and yellow gumballs can come out of the machine?

Accepted Answer

The problem is a combinatorial one where we need to find the number of different sequences of 4 red and 5 yellow gumballs. This is a classic "stars and bars" problem, where we calculate the number of ways to arrange 4 red and 5 yellow gumballs. The formula for this is the binomial coefficient C(n, k) = n! / (k!(n-k)!), where n is the total number of gumballs (4+5=9) and k is the number of one type of gumball (either 4 or 5). So, C(9, 4) = 126.

Question 7

Solve the problem.
-There are 12 students in a class: 9 are boys and 3 are girls. The teacher assigns seats at random. How many different boy/girl seating arrangements are possible?

Accepted Answer

The answer of Solve the problem.
-There are 12 students in...

Question 8

Solve the problem.
-Amy has 5 blue, 2 red, and 3 green books to arrange on a shelf. In how many ways can the books be arranged if books of the same color are identical and must be grouped together?

Accepted Answer

The answer of Solve the problem.
-Amy has 5 blue, 2...

Question 9

Solve the problem.
-Amy has 3 blue, 2 red, and 5 green books to arrange on a shelf. In how many distinguishable ways can the books be arranged if books of the same color are identical?

Accepted Answer

The answer of Solve the problem.
-Amy has 3 blue, 2...

Question 10

Solve the problem.
-Amy has 5 blue, 3 red, and 4 green books to arrange on a shelf. In how many ways can the books be arranged if books of the same color must be grouped together? (Assume that books of the same color are distinguishable.)

Accepted Answer

First, treat each color group as a single unit. There are 3 groups (blue, red, green), which can be arranged in 3! ways. Within each group, the books can be arranged among themselves: 5! for blue, 3! for red, and 4! for green. The total number of arrangements is 3! * 5! * 3! * 4! = 6 * 120 * 6 * 24 = 103,680.

Question 11

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability.
-All cherry

Accepted Answer

The answer of A bag contains 6 cherry, 3 orange,...

Question 12

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability.
-All orange

Accepted Answer

The probability of drawing 3 orange candies is calculated by dividing the number of ways to choose 3 orange candies by the total number of ways to choose any 3 candies. There are 3 orange candies, so the number of ways to choose 3 of them is 1 (since 3C3 = 1). The total number of ways to choose any 3 candies from 11 is 165 (since 11C3 = 165). Therefore, the probability is 1/165, which is approximately 0.0061.

Question 13

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability.
-All lemon

Accepted Answer

There are only 2 lemon candies in the bag, so it is impossible to draw 3 lemon candies. Hence, the probability is 0.

Question 14

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability.
-2 cherry, 1 lemon

Accepted Answer

The probability of drawing 2 cherry and 1 lemon candy is calculated by dividing the number of favorable outcomes by the total number of ways to draw 3 candies. The number of favorable outcomes is the combination of choosing 2 cherries from 6 and 1 lemon from 2, which is (6C2 * 2C1). The total number of ways to draw 3 candies from 11 is 11C3. The probability is (6C2 * 2C1) / 11C3 = 15 * 2 / 165 = 30 / 165 = 0.1818.

Question 15

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability.
-One of each flavor

Accepted Answer

The answer of A bag contains 6 cherry, 3 orange,...

Question 16

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability.
-2 orange, 1 lemon

Accepted Answer

The total number of ways to choose 3 candies from 11 is C(11,3). The number of ways to choose 2 orange and 1 lemon is C(3,2) * C(2,1). The probability is the ratio of these two values: (C(3,2) * C(2,1)) / C(11,3) = (3 * 2) / 165 = 6/165 = 0.0364.

Question 17

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability.
-1 cherry, 2 lemon

Accepted Answer

The probability of drawing 1 cherry and 2 lemon candies can be calculated using combinations. The total number of ways to choose 3 candies from 11 is C(11,3). The number of ways to choose 1 cherry from 6 is C(6,1), and the number of ways to choose 2 lemons from 2 is C(2,2). The probability is (C(6,1) * C(2,2)) / C(11,3) = (6 * 1) / 165 = 0.0364.

Question 18

Find the probability of the following card hands-from a 52 card deck. In poker, aces are either high or low. A bridgehand is made up of 13 cards.
-A royal flush (5 highest cards of a single suit) in poker

Accepted Answer

There are 4 suits and 10 different ranks (ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king). So, there are 4 * 10 = 40 different cards that are not aces. There are 4 * 4 = 16 different ways to choose the suit of the royal flush. There are 10 different ways to choose the highest card of the royal flush. There are 9 different ways to choose the next highest card of the royal flush. There are 8 different ways to choose the next highest card of the royal flush. There are 7 different ways to choose the next highest card of the royal flush. So, there are 16 * 10 * 9 * 8 * 7 = 80640 different royal flushes. The probability of getting a royal flush is 80640 / (52 choose 5) = 1.54 * 10^-6.

Question 19

Find the probability of the following card hands-from a 52 card deck. In poker, aces are either high or low. A bridgehand is made up of 13 cards.
-In poker, a full house ( 3 cards of one value, 2 of another value)

Accepted Answer

The probability of a full house in poker is calculated by choosing 3 cards of one rank and 2 cards of another rank. The number of ways to choose 3 cards of one rank is $\binom{13}{1} \times \binom{4}{3}$, and for 2 cards of another rank is $\binom{12}{1} \times \binom{4}{2}$. The total number of 5-card hands is $\binom{52}{5}$. The probability is $\frac{\binom{13}{1} \times \binom{4}{3} \times \binom{12}{1} \times \binom{4}{2}}{\binom{52}{5}} = 0.00144$.

Question 20

Find the probability of the following card hands-from a 52 card deck. In poker, aces are either high or low. A bridgehand is made up of 13 cards.
-In poker, four of a kind (4 cards of the same value)

Accepted Answer

The answer of Find the probability of the following card...

Quiz 9: Counting, Probability Distributions, and Further Topics in Probability

Solve the problem. -How many distinguishable permutations of letters are possible using the letters in the word CRITICS?

Solve the problem. -How many distinguishable permutations of letters are possible using the letters in the word BASEBALL?

Solve the problem. -How many distinguishable permutations of letters are possible using the letters in the word COMMITTEE?

Solve the problem. -How many distinguishable permutations of letters are possible using the letters in the word COLORADO?

Solve the problem. -10 people are bowling: 2 use a black ball and 8 use a blue ball. How many different black and blue bowling ball sequences can occur on the rack?

Solve the problem. -A gumball machine had 4 red and 5 yellow gumballs left. You put in pennies until you empty the machine. How many different sequences of red and yellow gumballs can come out of the machine?

Solve the problem. -There are 12 students in a class: 9 are boys and 3 are girls. The teacher assigns seats at random. How many different boy/girl seating arrangements are possible?

Solve the problem. -Amy has 5 blue, 2 red, and 3 green books to arrange on a shelf. In how many ways can the books be arranged if books of the same color are identical and must be grouped together?

Solve the problem. -Amy has 3 blue, 2 red, and 5 green books to arrange on a shelf. In how many distinguishable ways can the books be arranged if books of the same color are identical?

Solve the problem. -Amy has 5 blue, 3 red, and 4 green books to arrange on a shelf. In how many ways can the books be arranged if books of the same color must be grouped together? (Assume that books of the same color are distinguishable.)

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability. -All cherry

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability. -All orange

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability. -All lemon

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability. -2 cherry, 1 lemon

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability. -One of each flavor

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability. -2 orange, 1 lemon

A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability. -1 cherry, 2 lemon

Find the probability of the following card hands-from a 52 card deck. In poker, aces are either high or low. A bridgehand is made up of 13 cards. -A royal flush (5 highest cards of a single suit) in poker

Find the probability of the following card hands-from a 52 card deck. In poker, aces are either high or low. A bridgehand is made up of 13 cards. -In poker, a full house ( 3 cards of one value, 2 of another value)

Find the probability of the following card hands-from a 52 card deck. In poker, aces are either high or low. A bridgehand is made up of 13 cards. -In poker, four of a kind (4 cards of the same value)