Question

A prize wheel is divided into 12 equally likely spaces labeled 1 to 12.

Answer 1

Answer:

Events E and F are independent.

Step-by-step explanation:

E = {multiple of 3}  = {3, 6, 9, 12}

P(E) = 4/12

F = {even number}  = {2, 4, 6, 8. 10, 12}

P(F) = 6/12

E and F = {even and multiple of 3} = {6, 12}

P(E∩F) = 2/12

In order for two events to be independent the following relationship must be true:

$P(E)*P(F) = P(E\cap F)$

Testing this property:

$P(E)*P(F) = \frac{4}{12}*\frac{6}{12}=\frac{24}{144}=\frac{1}{6} \\P(E\cap F) = \frac{2}{12}=\frac{1}{6} \\P(E)*P(F) = P(E\cap F)$

The relationship holds true, thus events E and F are independent.