#### Answer

1/36

#### Work Step by Step

The problem asks the probability of the pointer landing on 1 and then 3 if spun twice.
These two spins are considered independent events because the occurrence of each spin has no effect on the probability of the other.
With that being said,
P(A and B)= P(A) • P(B)
P(A and B)= 1/6 • 1/6
P(A and B)= 1/36
The probability of the spinner landing on 1 and then 3 when spun twice is 1/36.