For five jellybeans to be less than 2 green jellybeans we should break up the subsets into the number of combinations where there are only 1 green jellybean and 0 jellybeans and find their sum. If there are no green jelly beans, then the number of combinations of all red jellybeans is given by \(\binom{7}{5}\). If there is one green jelly bean, then the combinations are \(\binom{7}{4}\) \(\binom{5}{1}\).
\(\binom{7}{5}\) + \(\binom{7}{4}\)\(\cdot\)\(\binom{5}{1}\) = 196
A subcommittee of 5 of 14 senators and 13 representatives where 4 of the members must be representatives can also be represented by the sum of 2 possibilities. One of which is if there is 4 representatives and 1 senator and the other is if they are all representatives. If one senator is chosen it can be be written as \(\binom{13}{4}\) \(\cdot\) \(\binom{14}{1}\) = 10010. If all 5 are representatives, that is written as \(\binom{13}{5}\) = 1287. The sum would be 11297.
The number of different outcomes will be the product of each
event.
A coin being tossed 5 times = \(2^5\).
A standard die being rolled twice = 36.
A group of three cards drawn without replacement = 52 \(\cdot\) 51 \(\cdot\) 50 = 132600.
The product is 152,755,200.
The probability that at least one is a three can be calculated as the sum of the number of outcomes drawing one 3, two 3’s and three 3’s divided by the total number of outcomes.
One three is drawn = \(\binom{4}{1}\) \(\cdot\) $ = 4512
Two threes are drawn = \(\binom{4}{2}\)
\(\cdot\) \(\binom{48}{1}\) = 288
Three threes are drawn = $ = 4
Total number of outcomes = $ = 22100
Probability at least one three = \(\frac{4804}{22100}\) = 21.74%
How many different combinations of 5 movies to rent is given by \(\binom{31}{5}\) = 169911
The number of different outcomes if there is at least one mystery is
given by the sum of outcomes from 1 mystery, 2 mysteries… five
mystery’s.
One mystery = \(\binom{14}{1}\) \(\cdot\) \(\binom{17}{4}\) = 14 \(\cdot\) 2380 = 33320
Two mystery’s = \(\binom{14}{2}\) \(\cdot\) \(\binom{17}{3}\) = 91 \(\cdot\) 680 = 61880
Three mystery’s = \(\binom{14}{3}\)
\(\cdot\) \(\binom{17}{2}\) = 364 \(\cdot\) 136 = 49504
Four mystery’s = \(\binom{14}{4}\)
\(\cdot\) \(\binom{17}{1}\) = 1001 \(\cdot\) 17 = 17017
Five mystery’s = \(\binom{14}{5}\) =
2002
Sum = 163723
The number of different schedules is given by the product of outcomes
of subsets of 3 elements per composer.
Brahms = \(\binom{4}{3}\) = 4
Haydn = \(\binom{104}{3}\) =
182104
Mendelssohn = \(\binom{17}{3}\) =
680
#of different schedules is 495,322,880
The total number of permutations would be \(\frac{24!}{11!}\) = 1.55 x \(10^{16}\). If we remove all permutations where 5 non fiction books are being included which is
the number of combinations of 8 books that are not nonfiction which
is \(\binom{19}{8}\)
Each one of those combinations combined with the five nonfiction books
have 13! combinations each.
(1.55 x \(10^16)-\)$ \(\cdot\) 13! =
(1.55 x $10^16)-(4.71 x \(10^{14}\))
The total number of possibilities would be \(\frac{10!}{5!5!}\) = 252. There are only 2 permutations which are desired so the answer would be \(\frac{2}{252}\) or \(\frac{1}{126}\)
The expected value is the probability multiplied by the cost.
The probability that it is a queen or lower is \(\binom{44}{52}\) = .846 The probability
that it is a King or Ace is \(\binom{8}{52}\) = .154 Expected Value =
(.154)(16) + (.846)(4) = $.92
The expected value after 833 games would be 833 \(\cdot\) .92 or $766.36