A bag contains 5 green and 7 red jellybeans. How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?
To determine the number of ways 5 jellybeans can be withdrawn from the bag so that the number of green jellybeans withdrawn is less than 2, we consider the scenarios for the number of green jellybeans that can be withdrawn: either 0 or 1 green jellybean.
Total number of ways = C(7,5)+(C(5,1)×C(7,4))=21+175=196.
A certain congressional committee consists of 14 senators and 13 representatives. How many ways can a subcommittee of 5 be formed if at least 4 of the members must be representatives?
We need to consider two scenarios: one with 4 representatives and 1 senator, and another with all 5 members being representatives.
Total number of ways = C(13,4)×C(14,1)+C(13,5)=10010+1287=11297.
If a coin is tossed 5 times, a six-sided die is rolled 2 times, and three cards are drawn from a deck of 52 cards without replacement, how many different outcomes are possible?
Consider each event independently and then multiply the number of outcomes for each event:
Total outcomes = 25×62×C(52,3).
Calculating the values gives 32×36×22100=25,459,200 different outcomes.
3 cards are drawn from a standard deck without replacement. What is the probability that at least one of the cards drawn is a 3?
Calculating the probability gives approximately 0.2174 or 2174/10000 when rounded to four decimal places.
Lorenzo is picking out some movies to rent, and he is primarily interested in documentaries and mysteries. He has narrowed down his selections to 17 documentaries and 14 mysteries.
Step 1: Total Combinations of 5 Movies
Lorenzo has 17 documentaries and 14 mysteries, making a total of 31 movies. The number of different combinations of 5 movies he can rent from these 31 movies is given by the combination formula
C(n,k)=n!k!(n−k)!
where n is the total number of items and k is the number of items to choose.
Here, n=31 (total movies) and k=5 (movies to choose).
Step 2: Combinations of 5 Movies with at Least One Mystery
To calculate this, we can subtract the number of combinations with no mysteries from the total combinations of 5 movies.
Let’s calculate these values.
Answers
Step 1: The total number of different combinations of 5 movies Lorenzo can rent from his selection is 169,911.
Step 2: The number of different combinations of 5 movies he can rent if he wants at least one mystery is 163,723.
In choosing music to play at a charity fundraising event, Cory needs to have an equal number of symphonies from Brahms, Haydn, and Mendelssohn. He has a selection of 4 Brahms, 104 Haydn, and 17 Mendelssohn symphonies and needs to set up a schedule of 9 symphonies to be played.
Since Cory needs to select an equal number of symphonies from each composer, he will choose 3 from each. The number of different schedules possible is the product of the combinations of choosing 3 symphonies from each composer’s offerings.
The number of ways to choose 3 symphonies from 4 by Brahms is:
\binom{4}{3}
For Haydn’s 104 symphonies, the number of ways to choose 3 is:
\binom{104}{3}
And for Mendelssohn’s 17 symphonies, the number of ways to choose 3 is:
\binom{17}{3}
Therefore, the total number of different schedules that Cory can create is given by:
\binom{4}{3} \times \binom{104}{3} \times \binom{17}{3}
Calculating this product yields a total of approximately 4.95 \times 10^8 different schedules possible, expressed in scientific notation rounded to the hundredths place.
Answer: There are approximately 4.95 \times 10^8 different schedules possible.
An English teacher needs to pick 13 books to put on his reading list for the next school year, and he needs to plan the order in which they should be read. He has narrowed down his choices to 6 novels, 6 plays, 7 poetry books, and 5 nonfiction books.
Step 1. If he wants to include no more than 4 nonfiction books, how many different reading schedules are possible?
To solve this, we need to consider all possible combinations of choosing nonfiction books and then choosing the remaining books from the novels, plays, and poetry books to make up the total of 13. The teacher can choose 0, 1, 2, 3, or 4 nonfiction books.
The total number of reading schedules is given by:
\sum_{k=0}^{4} \binom{5}{k} \binom{19}{13-k}
where \binom{n}{k} is the binomial coefficient representing the number of ways to choose k elements from a set of n elements.
Calculating the sum of these combinations gives us the total number of possible reading schedules, which is approximately 2.42 \times 10^6 in scientific notation rounded to the hundredths place.
Answer for Step 1: 2.42 \times 10^6
Step 2. If he wants to include all 6 plays, how many different reading schedules are possible?
Since the teacher wants to include all 6 plays, he now needs to choose the remaining 7 books from the novels, poetry books, and up to 4 nonfiction books.
The total number of reading schedules in this case is given by:
\sum_{k=0}^{4} \binom{5}{k} \binom{18}{7-k}
Calculating the sum of these combinations gives us the total number of possible reading schedules, which is approximately 3.17 \times 10^4 in scientific notation rounded to the hundredths place.
Answer for Step 2: 3.17 \times 10^4
Zane is planting trees along his driveway, and he has 5 sycamores and 5 cypress trees to plant in one row. We need to calculate the probability that he randomly plants the trees so that all 5 sycamores are next to each other and all 5 cypress trees are next to each other.
Since the sycamores must be planted together and the cypress trees must also be planted together, we can consider each group of trees as a single unit. Therefore, we are essentially arranging two units (one unit of sycamores and one unit of cypress trees).
The number of ways to arrange these two units is 2! (2 factorial). Within each unit, the trees can be arranged among themselves. So for the sycamores, there are 5! (5 factorial) arrangements, and for the cypress trees, there are also 5! arrangements.
The total number of ways to arrange the trees without any restriction would be 10! (10 factorial), since there are 10 trees in total.
Therefore, the probability P is given by:
P = \frac{2! \times 5! \times 5!}{10!}
Calculating this, we find that the probability is approximately 0.0079 when rounded to four decimal places.
Answer: The probability is 0.0079.
If you draw a queen or lower from a standard deck of cards, I will pay you $4. If not, you pay me $16. (Aces are considered the highest card in the deck.)
Step 1. Find the expected value of the proposition.
The deck has 4 suits, and each suit has 13 cards. Since Aces are the highest, the Queen or lower cards include one Queen, one Jack, one 10, one 9, one 8, one 7, one 6, one 5, one 4, one 3, and one 2. This totals to 11 cards per suit that are Queen or lower.
Therefore, there are 4 \times 11 = 44 such cards in the entire deck.
The probability of drawing a Queen or lower card from the deck is then \frac{44}{52}. If this event occurs, you win $4.
The probability of not drawing a Queen or lower card is \frac{52 - 44}{52} = \frac{8}{52}. If this event occurs, you lose $16.
The expected value (EV) of the game can be calculated as follows:
EV = \left(\frac{44}{52} \times 4\right) + \left(\frac{8}{52} \times -16\right)
Calculating the EV gives:
EV = \left(\frac{44}{52} \times 4\right) + \left(\frac{8}{52} \times -16\right) = -\$0.62
Step 2. If you played this game 833 times, how much would you expect to win or lose?
The total expected value over 833 games is:
833 \times EV = 833 \times -\$0.62 = -\$516.46
Therefore, if you played the game 833 times, you would expect to lose approximately $516.46.
Answer for Step 1: $-0.62
Answer for Step 2: $-516.46