Question

5-38. a better chance of winning
each of the problems below describes two different games you can play with a random number generator. state
the theoretical probability of game i and game ii for each round. which gives you a better chance of winning?

a.

b.

round	game i	probability for game i	game ii	probability for game ii	winning game
b.	picking a multiple of 5 from the integers between 1 and 20		picking a multiple of 5 from the integers between 1 and 40
c	picking a multiple of 7 from the integers between 1 and 40		picking a multiple of 6 form the integers between 1 and 25

Explanation:

Part a

Step1: Count primes 1-20

Primes 1-20: 2,3,5,7,11,13,17,19 (8 total).
Probability: $\frac{8}{20} = \frac{2}{5}$

Step2: Count primes 21-40

Primes 21-40: 23,29,31,37 (4 total).
Probability: $\frac{4}{20} = \frac{1}{5}$

Step3: Compare probabilities

$\frac{2}{5} > \frac{1}{5}$, so Game I wins.

Part b

Step1: Count multiples 5 (1-20)

Multiples of 5 1-20: 5,10,15,20 (4 total).
Probability: $\frac{4}{20} = \frac{1}{5}$

Step2: Count multiples 5 (1-40)

Multiples of 5 1-40: 8 total.
Probability: $\frac{8}{40} = \frac{1}{5}$

Step3: Compare probabilities

$\frac{1}{5} = \frac{1}{5}$, so no better game.

Part c

Step1: Count multiples 7 (1-40)

Multiples of 7 1-40: 7,14,21,28,35 (5 total).
Probability: $\frac{5}{40} = \frac{1}{8}$

Step2: Count multiples 6 (1-25)

Multiples of 6 1-25: 6,12,18,24 (4 total).
Probability: $\frac{4}{25}$

Step3: Compare probabilities

$\frac{1}{8}=0.125$, $\frac{4}{25}=0.16$, so $\frac{4}{25} > \frac{1}{8}$, Game II wins.

Answer:

Round	Probability for Game I	Probability for Game II	Winning Game
b.	$\frac{1}{5}$	$\frac{1}{5}$	Neither (equal chance)
c.	$\frac{1}{8}$	$\frac{4}{25}$	Game II