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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?
In this situation, we can have either 5 reds or 4 reda and one green. There are two different scenarios to look at. The number of choices in the scenario with one green and four red can be represented as follows: \(\binom{5}{1}*\binom{7}{4}\) The number of choices in the scenario with five red and zero green can be represented as follows: \(\binom{7}{5}\). The total number of choices then is: \(\binom{5}{1}*\binom{7}{4} + \binom{7}{5} = (5)*(35) + (21) = 196\)
Thus, there are 196 possible ways to withdraw 5 jellybeans so that the number of green jelly beans is less than 2.
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?
For at least 4 members to be representatives, there are two possible cases. There is the case with 4 representatives and there is the case with 5 representatives. The possible arrangements for the first case are: \(\binom{14}{1}*\binom{13}{4}\) The possible arrangements for the second case are: \(\binom{13}{5}\).
The total number of choices then is: \(\binom{14}{1}*\binom{13}{4} + \binom{13}{5} = (14)*(715) + (1287) = 11,297\)
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?
The different possible outcomes are represented by \((2)^5*(6)^2 * \frac {52!}{(52 - 5)!} = 359280230400\)
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.
The probability that at least one is a three can be found by taking the complement of the probability of no card being a three.
\(P = 1 - \frac{48}{52} * \frac{47}{51} * \frac{46}{50} \approx 0.2173756\)
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.
He could rent \(\binom{31}{5} = 169911\) different combinations.
Another way to compute this involves a bit more work.
The available options are: $ 5D 0M, 4D 1M, 3D 2M, 2D 3M, 1D 4M, 0D 5M$, which gives us six different cases to consider.
In this case, the number of different combinations becomes: \(\binom{17}{5} + \binom{17}{4}*\binom{14}{1} + \binom{17}{3}*\binom{14}{2} + \binom{17}{2}*\binom{14}{3} + \binom{17}{1}*\binom{14}{4} + \binom{14}{5}= 169911\), which is the same as in the simpler case above.
To find the probability of at least one mystery, it is best to find the complement of the probability of no mysteries.
The probability can be found by using the deck of cards approach:
\(P = 1 - \frac{17}{31} * \frac{16}{30} * \frac{15}{29} * \frac{14}{28} * \frac{13}{27} \approx 0.9635809\)
The number of combinations can be found using the total possible combinations minus the combinations that include no mysteries.
Number of combinations = \(169911 - \binom{17}{5} = 163723\). This can also be found by multiplying the probability by the total number of possible combinations.
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.
The number of different possible schedules can be given by \(\binom{4}{3} * \binom{104}{3} * \binom{17}{3}= 4.95 \times 10^8\)
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 choose no more than four nonfiction books, then we need to either evaluate four different cases or take the total possible outcomes and subtract the one outcome with five nonfiction books.
The total possible outcomes are: \(\binom{24}{13} = 2496144\)
The number of outcomes with five nonfiction books are: \(\binom{5}{5}*\binom{19}{8} = 75582\)
Thus, the number of possible outcomes with no more than 4 nonfiction books is \(2496144 - 75582 = 2.42 \times 10^6\)
If he wants to include all six plays, then he has \(\binom{6}{6}*\binom{18}{7} = 3.18 \times 10^5\). Restricting the selection to include all six plays reduces the available pool to 18 with 7 additional slots to fill.
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.
This problem could be given by \(P = 2 *\frac{5}{10} * \frac{4}{9} * \frac{3}{8} * \frac{2}{7} * \frac{1}{6} \approx 7.94 \times 10^{-3}\). This is multiplied by two because you could have the cypress trees first in order or the sycamores first in order.
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.)
The expected value is \(4 \times \frac{44}{52} - 16 \times \frac{8}{52} = 0.92\)
If the game were played 833 times, the expectation from one game should be multiplied by 833 to find the total expectation. Expectation = 768.92