Chapter 3.1 Exercise 10

A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?

At first I though that this was a conditional probability because what if 1 or 2 or 3 aces had been dealt in the first 12 cards? Then I realized that because we are told nothing about the first twelve cards it is a simple question of what is the probability that any card drawn is an ace. \(P(\text{Ace}) = \frac{4}{52} = \frac{1}{13}\). Therefore:

\[P(\text{Ace on 13}^{\text{th}} \text{ draw}) = \frac{1}{13}\]

This is similar to the gambler’s fallacy. There’s an equal chance of any card in the deck being an ace. Knowing nothing about the first 12 cards drawn we cannot infer anything about the 13th card and it still has a \(\frac{1}{13}\) chance of being an ace.

Since this was too easy, let’s do another…

Chapter 3.1 Exercise 12

A symphony orchestra has in its repertoire 30 Haydn symphonies, 15 modern works, and 9 Beethoven symphonies. Its program always consists of a Haydn symphony followed by a modern work, and then a Beethoven symphony.
(a) How many different programs can it play?

This is a simple counting problem. Each of the 30 Haydn symphonies can be followed by any one of the 15 modern works, and then each of those combinations can be followed by any one of the 9 Beethoven symphonies, so the number of programs equals \(30 \times 15 \times 9\).

prog <- 30 * 15 * 9
prog
## [1] 4050
There are 4050 possible programs.
(b) How many different programs are there if the three pieces can be played in any order?

Since there are three types of music, we can order those three types in 3! or 6 different ways:

first second third
Haydn modern Beethoven
Haydn Beethoven modern
modern Haydn Beethoven
modern Beethoven Haydn
Beethoven Haydn modern
Beethoven modern Haydn

Since the commutative property of multiplication, states that \(a \times b = b \times a\), this means that no matter what order you multiply the three numbers: 30, 15, and 9, you will still get the same product we found in part (a), 4,050.

So, for each of the 6 dfferent orders there are 4,050 different combinations of songs. So the total number of possible programs would be 6 times 4,050.

prog2 <- 6 * 4050
prog2
## [1] 24300
There are 24300 possible programs.
(c) How many different three-piece programs are there if more than one piece from the same category can be played and they can be played in any order?

If any 3 of the 30 + 15 + 9 (or 54) pieces of music can be played in any order then the number or possible programs is equal to the number of permutations of 54 songs taken 3 at a time, which we denote by \(_{54}P_3\) or \((54)_3\):

\[ (54)_3 = \frac{54!}{(54-3)!} \]

prog3 <- factorial(54)/factorial(54-3)
prog3
## [1] 148824
There are 148824 possible programs.