If P(A) = 0·3, P(B) = 0·4, P(C) = 0·8 P(AB) = 0·08, P(AC) = 0·28 P(BC) = P, P(ABC) = 0·09 P (A + B + C) ≥ 0·75 then—Select one:a. 0·23 ≤ P ≤ 0·48b. 0·32 ≤ P ≤ 0·72c. 0·25 ≤ P ≤ 0·71d. None of these

The question is asking for the range of values that P, the probability of event B and C occurring together, can take given the other probabilities provided.

The formula for the probability of the union of three events A, B, and C is:

P(A + B + C) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC)

We know that P(A + B + C) ≥ 0.75. Substituting the given values into the formula, we get:

0.75 ≤ 0.3 + 0.4 + 0.8 - 0.08 - 0.28 - P + 0.09

Solving for P, we get:

P ≤ 0.48

We also know that the probability of any event is between 0 and 1. Since P(BC) is the intersection of events B and C, it cannot be greater than the probability of either B or C. Therefore, the lower limit of P is 0.

So, the range of P is 0 ≤ P ≤ 0.48.

Therefore, the correct answer is not listed among the options provided. The correct range should be 0 ≤ P ≤ 0.48.