Final answer:
To show that P[E∩F]≥0.4 given P[E]=0.8 and P[F]=0.6, we use a rearranged probability principle which states P[E∩F] is at least the sum of the individual probabilities minus 1, giving us P[E∩F]≥0.4.
Step-by-step explanation:
To demonstrate that P[E∩F]≥0.4 given that P[E]=0.8 and P[F]=0.6, we can use the principle that the probability of the intersection of two events is at least as large as the sum of their probabilities minus 1. This is because:
P[E] + P[F] - P[E∩F] ≤ 1
This rearranges to:
P[E∩F] ≥ P[E] + P[F] - 1
Substituting the given probabilities into this equation yields:
P[E∩F] ≥ 0.8 + 0.6 - 1 = 0.4
Therefore, P[E∩F] must be greater than or equal to 0.4 to satisfy the probability rules.