Finding probability of P(B' and C') : Do I multiply or just add?

Question

For the following I am supposed to find P(B' and C'). So, the players are not forward and not guard. That leaves only 2 out of 10 players. Is P(B' and C') then .2 or do we have to multiply PB' by PC' to get .36?

A basketball team is composed of 10 players. The team is made up of players who play either guard, forward, or center position. Four of the ten are guards; four of the ten are forwards; and two of the ten are centers. The numbers of the players are 1, 2, 3, 4, for the guards; 5, 6, 7, 8 for the forwards; and 9 and 10 for the centers. The starting five are numbered 1, 3, 5, 7, and 9. Let a player be selected at random from the ten. Define the following events:

A = player selected has a number from 1 to 8.

B = player selected is a guard.

C = player selected is a forward.

D = player selected is a starter.

E = player selected is a center.

Answer 1

A = player selected has a number from 1 to 8

B = player selected is a guard

C = player selected is a forward

D = player selected is a starter

E = player selected is a center

P(A) = 8/10 = 0.8

P(B) = 4/10 = 0.4

P(C) = 4/10 = 0.4

P(D) = 5/10 = 0.5

P(E) = 2/10 = 0.2

We need to find the complement for the two events: not guard nor forward.

P(B') = 1 - P(B) = 1 - 0.4 = 0.6 ⇒ {5, 6, 7, 8, 9, 10}

P(C') = 1 - P(C) = 1 - 0.4 = 0.6 ⇒ {1, 2, 3, 4, 9, 10}

P(B' and C') = P(B') × P(C') = 0.6*0.6 = 0.36

The probability of a player who is not a guard nor forward is about 0.36. This one you multiply because we are looking at events, not necessarily players themselves.