Question

in a single experiment, a die is tossed and a spinner with the letters a, b, and c is spun. each letter is equally likely. draw your sample space and then find the probability of getting a 2 or a b.
a. the sample space is 9. the probability of getting a b is \\( \frac { 1 } { 9 } \\).
b. the sample space is 18. the probability of getting a 2 or a b is \\( \frac { 18 } { 8 } = \frac { 9 } { 4 } \\).
c. the sample space is 18. the probability of getting a 2 or a b is \\( \frac { 8 } { 18 } = \frac { 4 } { 9 } \\).
d. the sample space is 9. the probability of getting a 2 or a b is \\( \frac { 8 } { 18 } = \frac { 4 } { 9 } \\).

Explanation:

Step1: Calculate sample space

A die has 6 outcomes (\(1,2,3,4,5,6\)) and a spinner has 3 outcomes (\(A,B,C\)). By the fundamental counting principle, the sample space \(n(S)=6\times3 = 18\).

Step2: Use the formula for probability of the union of two events

Let \(A\) be the event of getting a \(2\) and \(B\) be the event of getting a \(B\).
\(n(A)=3\) (i.e., \((2,A),(2,B),(2,C)\)), \(n(B)=6\) (i.e., \((1,B),(2,B),(3,B),(4,B),(5,B),(6,B)\)) and \(n(A\cap B) = 1\) (i.e., \((2,B)\)).
By the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), \(P(A)=\frac{3}{18}\), \(P(B)=\frac{6}{18}\), \(P(A\cap B)=\frac{1}{18}\).
\(P(A\cup B)=\frac{3 + 6-1}{18}=\frac{8}{18}=\frac{4}{9}\)

Answer:

C. The sample space is 18. The probability of getting a 2 or a B is \(\frac{8}{18}=\frac{4}{9}\)