Question

A game of chance consists of spinning an arrow that comes to rest at one of the
numbers 10, 16, 21, 29, 31, 36, 44 and 51.
Find the probability that it will point at
a. A prime number.
b. An odd number.
c. A perfect square number.
d. A number less than 10.

Answer 1

Answer:

A - [10,16,21,29,31,36,44,51]

n(A) - 8

a) E - a prime number [29,31]

n(E) - 2

P = $\frac{n(E)}{n(A)}= \frac{2}{8}=\frac{1}{4}$

b) E - an odd number [21,29,31,51]

n(E)- 4

P = $\frac{n(E)}{n(A)} =\frac{4}{8} =\frac{1}{2}$

c) E - a perfect square number [36]

n(E) - 1

P=$\frac{n(E)}{n(A)}= \frac{1}{8}$

d) E - a number less than 10

n(E) - 0

P = $\frac{n(E)}{n(A)}=\frac{0}{8}=0$