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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?
There are \({7 \choose 5}=\frac{7!}{2!5!}=21\) ways of drawing only red jellybeans, and \({5 \choose 1}{7 \choose 4}=\frac{5!}{4!1!}\frac{7!}{3!4!}=(5)(35)=175\) ways of drawing 4 red and 1 green jellybean, for a total of 196 possible ways of drawing 5 jellybeans such that less than 2 green jellybeans are drawn.
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?
If all of the members are representatives, then there are \({13 \choose 5}=\frac{13!}{8!5!}=1,287\) ways of selecting the committee.
If only four of the members are representatives, then there are \({14 \choose 1}=\frac{14!}{13!1!}=14\) ways of selecting the senator and \({13 \choose 4}=\frac{13!}{9!4!}=715\) ways of selecting the representatives, so there are \(14 \times 715=10,010\) ways of selecting the committee.
This is a combined total of \(1,287+10,010=11,297\) ways of selecting the committee.
If a coin is tossed 5 times, and then a standard six-sided die is rolled 2 times, and finally a group of three cards are drawn from a standard deck of 52 cards without replacement, how many different outcomes are possible?
There are \(2^5\times6^2\times52\times51\times50=152,755,200\) 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? Express your answer as a fraction or a decimal number rounded to four decimal places.
\(P(\text{At least one 3})=1-P(\text{No threes})=1-(\frac{48}{52}\times\frac{47}{51}\times\frac{46}{50})=1-\frac{103776}{132600}=\frac{28824}{132600}=\frac{1201}{5525}\)
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. How many different combinations of 5 movies can he rent?
There are \({31 \choose 5}=\frac{31!}{26!5!}=169,911\) different combinations of movies that he can rent.
How many different combinations of 5 movies can he rent if he wants at least one mystery?
We know how many possible combinations there are (169,911), so the way to determine how many combinations involve at least one mystery is to find out how many combinations DON’T include at least one mystery (in other words, are all documentaries) and subtract this from the total number of possible combinations.
There are \({14 \choose 5}=\frac{14!}{9!5!}=2,002\) different combinations of only documentaries that he can rent.
Therefore, there are \(169,911-2,002=167,909\) different combinations that include at least one mystery.
In choosing what music to play at a charity fund raising event, Cory needs to have an equal number of symphonies from Brahms, Haydn, and Mendelssohn. If he is setting up a schedule of the 9 symphonies to be played, and he has 4 Brahms, 104 Haydn, and 17 Mendelssohn symphonies from which to choose, how many different schedules are possible? Express your answer in scientific notation rounding to the hundredths place.
He needs to choose 3 symphonies from each composer, so…
There are \({4 \choose 3}=\frac{4!}{3!1!}=4\) ways of choosing the Brahms symphonies;
There are \({104 \choose 3}=\frac{104!}{101!3!}=182,104\) ways of choosing the Haydn symphonies; and
There are \({17 \choose 3}=\frac{17!}{14!3!}=680\) ways of choosing the Mendelssohn symphonies.
This means that overall there are \(4\times 182104\times 680=495,322,880\) different combinations of symphonies that can be performed.
However, to make a schedule he also needs to decide on the order in which they will be performed. Since there are 9 symphonies being performed, each of the 495,322,880 combinations can be performed in \(9!=362,880\) different orders, resulting in 179,742,766,694,400 possible schedules, or \(1.80\times 10^{14}\) in scientific notation.
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. If he wants to include no more than 4 nonfiction books, how many different reading schedules are possible? Express your answer in scientific notation rounding to the hundredths place.
If the maximum number of allowed nonfiction books is 4, then we can determine how many different combinations of books are possible by finding the total number of combinations of books and subtracting the number of combinations that inclues all 5 nonfiction books.
There are \({24 \choose 13}=\frac{24!}{11!13!}=2,496,144\) total combinations of books
There are \({19 \choose 8}=\frac{19!}{11!8!}=75,582\) combinations of books that include all 5 nonfiction books (and that he therefore will not choose).
Therefore, there are \(2,496,144-75,582=2,420,562\) combinations of books that he might select.
Each of these combinations has \(13!=6,227,020,800\) different orders in which it can be arranged, so there are \(6,227,020,800\times 2,420,562=15,072,889,921,689,600\) or approximately \(1.51\times 10^{16}\) possible reading schedules.
If he wants to include all 6 plays, how many different reading schedules are possible? Express your answer in scientific notation rounding to the hundredths place.
If all 6 plays are included, there are 18 remaining books and 7 remaining selections, so there are \({18 \choose 7}=\frac{18!}{11!7!}=31,824\) possible combinations of books.
Each of these combinations has \(13!=6,227,020,800\) different orders in which it can be arranged, so there are \(6,227,020,800\times 31,824=198,168,709,939,200\) or approximately \(1.98\times 10^{14}\) possible reading schedules.
Zane is planting trees along his driveway, and he has 5 sycamores and 5 cypress trees to plant in one row. What is 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? Express your answer as a fraction or a decimal number rounded to four decimal places.
There are \({10 \choose 5}=\frac{10!}{5!5!}=252\) possible ways of planting the trees. There are only two ways of planting them that satisfy the described conditions: SSSSSCCCCC or CCCCCSSSSS. Therefore, the probability that they are planted in this way is \(\frac{2}{252}=\frac{1}{126}\)
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.) Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values.
There are 44 cards that are a queen or lower and 8 cards that are not (the four kings and four aces). Therefore, \(E=\frac{44}{52}(4)+\frac{8}{52}(-16)=\frac{176}{52}-\frac{128}{52}=\frac{48}{52}=\frac{12}{13} \approx 0.92\)
If you played this game 833 times how much would you expect to win or lose? Round your answer to two decimal places. Losses must be expressed as negative values.
The expected value of each round of the game is \(\frac{12}{13}\), so the expected value of 833 rounds is \(\frac{12}{13}(833)=\frac{9996}{13} \approx768.92\). I would expect to make about $768.92