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Dimas [21]
3 years ago
6

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
Mathematics
2 answers:
faltersainse [42]3 years ago
4 0

Answer:

B. <em>0.813</em>

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


il63 [147K]3 years ago
3 0
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
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<em><u>Solution:</u></em>

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