Let's analyse options one by one.
(a) Independence of P and Q implies that probability (P ∩ Q) = 0
No. This is the case for mutually exclusive events that if one occurs then the chance of another one occuring is ruled out.
Infact Independence of P and Q means if P happens, then outcome of Q won't be affected by that.
Means P(Q|P) =Q because it has not effect whether P happens or not
which makes P(P $\cap$ Q) =P(P).P(Q) because again P(P|Q)= P.
(b) Probability (P ∪ Q) ≥ Probability (P) + Probability (Q)
This can hold true for mutually exclusive events P and Q but if P and Q can happen together(implies independence)
then P(P $\cap$ Q) $\neq$ 0.
So P( P U Q )= P(P) + P(Q) - P(P $\cap$ Q)
So, in this case P(PUQ) $\leq$ P(P) +P(Q)
(c) If P and Q are mutually exclusive, then they must be independent.
Absolutely false. When P and Q are mutually exclusive, then either P occurs or Q occurs but not both simultaneously. So if P happens, chance of Q happening gets ruled out and vice-versa. Hence, mutually exclusive events are dependent.
(d) Probability (P ∩ Q) ≤ Probability (P)
Consider 2 cases
Case 1 : P and Q are mutually exclusive.
Let S be sample space of outcomes of a single toss of a coin.
Let P : Outcome is head
Q : Outcome is tails.
Obviously, P(P $\cap$ Q) =0 and P(P)= $\frac{1}{2}$
Equality satisfied.
Case 2: P and Q are independent.
P : Outcome of getting head on a single toss of a coin
Q : Outcome of getting 4 on dice.
P(P $\cap$ Q) = $\frac{1}{12}$
P(P)=$\frac{1}{2}$
Equality satisfied.
Option (d) TRUE.