Question

A sixteen-sided number cube has the numbers 1 through 16 on each face. each face is equally likely to show after a roll. what is the probability that you will roll an even number or an odd prime number? round to the nearest thousandth.
a. 0.063
b. 0.813
c. 0.219
d. 0.875

Answer 1

P(even number) = 8/16 = 1/2...sample space is 16, there are 8 even numbers (2,4,6,8,10,12,14,16)
P (odd prime number) = 5/16...sample space is 16, there are 5 odd primes (3,5,7,11,13)

P (both) = 1/2 + 5/16 = 8/16 + 5/16 = 13/16 = 0.8125 rounds to 0.813

Answer 2

Answer:

B. 0.813

Step-by-step explanation:

A sixteen-sided number cube has the numbers 1 through 16 on each face.

So, $|\ S\ |=16$

Let us assume that, A be the event that the number will be an even number. So,

$A=\left \{ 2,4,6,8,10,12,14,16 \right \}$ and $|\ A\ |=8$

Then,

$P(A)=\dfrac{|\ A\ |}{|\ S\ |}=\dfrac{8}{16}$

Let us assume that, B be the event that the number will be an odd prime number.

$B=\left \{3,5,7,11,13 \right \}$ and $|\ B\ |=5$

Then,

$P(B)=\dfrac{|\ B\ |}{|\ S\ |}=\dfrac{5}{16}$

So the probability that you will roll an even number or an odd prime number will be,

$P(A\cup B)=P(A)+P(B)-P(A\cup B)$

$=\dfrac{8}{16}+\dfrac{5}{16}-0$ ( as independent events)

$=\dfrac{13}{16}$

$=0.813$