Mathematics, 18.05.2021 19:20 shawdyylinda

An English teacher needs to pick 9 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 13 novels, 17 plays, and 13 nonfiction books. If he wants to include an equal number of novels, plays, and nonfiction books, how many different reading schedules are possible

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Answer from: jgunt8277

Different schedule Possible = 55,621,280

Step-by-step explanation:

Given - An English teacher needs to pick 9 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 13 novels, 17 plays, and 13 nonfiction books.

To find - If he wants to include an equal number of novels, plays, and nonfiction books, how many different reading schedules are possible ?

Formula used -

Number of combinations of n things taking r at a time : $^{n}C_{r}$ = $\frac{n!}{r! (n-r)!}$

Proof -

Given that, An English teacher needs to pick 9 books and wants to include equal number of novels, plays, and nonfiction books

i.e. he will Choose

3 books from Novels

3 books from Plays

3 books from Non-fiction books

Now,

Given that,

Total number of novel = 13

Total number of plays = 17

Total number of non-fiction books = 13

So,

Different schedule Possible = $^{13}C_{3}* ^{17}C_{3}*^{13}C_{3}$

= $\frac{13!}{3! (13-3)!}* \frac{17!}{3! (17-3)!}* \frac{13!}{3! (13-3)!}$

= $\frac{13!}{3! (10)!}* \frac{17!}{3! (14)!}* \frac{13!}{3! (10)!}$

= $\frac{13(12)(11)10!}{3! (10)!}* \frac{17(16)(15)14!}{3! (14)!}* \frac{13(12)(11)10!}{3! (10)!}$

= $\frac{13(12)(11)}{3(2)(1)}* \frac{17(16)(15)}{3(2)(1)}* \frac{13(12)(11)}{3(2)(1)}$

= {13(2)(11)}*{17(8)(5)}* {13(2)(11)}

= 55,621,280

⇒Different schedule Possible = 55,621,280

Answer from: Quest

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64. therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560

Answer from: Quest

60: 45 =12: 9 =4: 3 so the answer is 4: 3

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An English teacher needs to pick 9 books to put on his reading list for the next school year, and he...

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