Question

The arrows on two spinners are spun one time. the first spinner is divided equally into the colors red, blue, yellow, and green. the second spinner is divided equally into the numbers 1,2,3 and 4. All possible outcomes are shown in the table. what is the probability the first spinner lands on red or the second spinner land on an even number?

Answer 1

Given data:

The total number in sample space is n(S)= 16.

The events in which first spinner lands on red and second on an event number is,

$E=\mleft\lbrace(R,\text{ 1), (R, 2), (R, 3), (R, 4), (}B,\text{ 2), (B, 4), (G, 2), (G, 4), (Y, 2), (Y, 4)}\mright\rbrace$

The total number of favorable events is,

$n(E)=10$

The expression for the probability that the first spinner lands on red and second on an event number is,

$\begin{gathered} P(E)=\frac{n(E_{})}{n(S)} \\ =\frac{10}{16} \\ =\frac{5}{8} \end{gathered}$

Thus, the probability that the first spinner lands on red and second on an event number is 5/8, so (C) option is correct.