Spinner Probability

[Zelda22]

Hello,

I'm not sure if I'm doing this exercise correctly.

a) Event -Second sping is different from the first spin.

A={ (0,1), (1,0)
p(A)= 3/5 * 2/5 + 2/5 * 3/5
= 6/25 + 6/25
= 12/25

Could I just subtract the probability of both spins being the same from 1?
I got the same answer.
p(A)= 1-p(A')
= 1 - 13/25
= 12/25

b) Event- both spins are 1

p(A)= (1,1)
= 2/5 * 2/5
= 4/25

Thank you

 

[BigBeachBanana]

Yes, those look correct.
For event A, yes you can use either method.

 

[Zelda22]

Thank you.