Question

A spinner is divided into 10 equal sized sections. Seven sections are colored red and three sections are colored blue. Half of the sections are labeled with an even number. Three sections of the spinner are both colored red and labeled with an even number.
What is the probability that the arrow of the spinner will stop on a section colored red given that it stops on a section labeled with an even number

Answer 1

The probability is 3/5. Since there are 5 out of 10 sections labeled with an even number, the denominator of the desired probability is 5. Since three sections are both colored red and labeled with an even number, the numerator is 3. The probability is 3/5. We can also use conditional probability: P(A|B)=P(AUB)/P(B). Let A=section colored red, B=section labeled with even number. P(AUB)=3/10, while P(B) is 1/2, so P(A|B)=3/5.