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If events A and B are independent, P(A) = 0.4, and P(B) = 0.5, what is P(A^c intersect B^c)?

Mathematics

ScanSolve AI Answer

Answer

B\text{B}

Step-by-step solution

5 steps
  1. Identify the given information.

    We are given that events AA and BB are independent.
    We are given the probabilities:
    P(A)=0.4P(A) = 0.4
    P(B)=0.5P(B) = 0.5

  2. Calculate the probabilities of the complements of events AA and BB.

    The probability of the complement of an event EE, denoted P(Ec)P(E^c), is 1−P(E)1 - P(E).
    For event AA:

    P(Ac)=1−P(A)=1−0.4=0.6P(A^c) = 1 - P(A) = 1 - 0.4 = 0.6

    For event BB:

    P(Bc)=1−P(B)=1−0.5=0.5P(B^c) = 1 - P(B) = 1 - 0.5 = 0.5
  3. Use the property of independent events.

    If events AA and BB are independent, then their complements AcA^c and BcB^c are also independent.
    For independent events E1E_1 and E2E_2, the probability of their intersection is P(E1∩E2)=P(E1)P(E2)P(E_1 \cap E_2) = P(E_1)P(E_2).
    Therefore, for AcA^c and BcB^c:

    P(Ac∩Bc)=P(Ac)P(Bc)P(A^c \cap B^c) = P(A^c)P(B^c)
  4. Substitute the calculated values and find the result.

    P(Ac∩Bc)=(0.6)(0.5)P(A^c \cap B^c) = (0.6)(0.5) P(Ac∩Bc)=0.3P(A^c \cap B^c) = 0.3
  5. Compare the result with the given options.

    The calculated value 0.3 matches option B.

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