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Given P(A) = 0.8 and P(A intersect B) = 0.7, find P(B') under different conditions for A and B.

Mathematics

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Answers

2 parts
b

0.125\text{0.125}

c

0.15\text{0.15}

Step-by-step solution

9 steps

Given:
P(A)=0.8P(A) = 0.8
P(A∩B)=0.7P(A \cap B) = 0.7

Part a· 3 steps

Find P(B′)P(B') if A and B are mutually exclusive.

  1. Understand the condition for mutually exclusive events.

    If events A and B are mutually exclusive, then P(A∩B)=0P(A \cap B) = 0.

  2. Compare with the given information.

    The problem states that P(A∩B)=0.7P(A \cap B) = 0.7.

  3. Conclude.

    Since P(A∩B)=0.7≠0P(A \cap B) = 0.7 \neq 0, events A and B cannot be mutually exclusive. Therefore, it is impossible for A and B to be mutually exclusive under the given conditions.

Part b

0.125\text{0.125}

Find P(B′)P(B') if A and B are independent.

  1. Use the condition for independent events.

    If events A and B are independent, then P(A∩B)=P(A)⋅P(B)P(A \cap B) = P(A) \cdot P(B).

  2. Substitute the given values to find P(B)P(B).

    0.7=0.8⋅P(B)0.7 = 0.8 \cdot P(B) P(B)=0.70.8=78P(B) = \frac{0.7}{0.8} = \frac{7}{8} P(B)=0.875P(B) = 0.875

  3. Calculate P(B′)P(B').

    P(B′)=1−P(B)P(B') = 1 - P(B) P(B′)=1−0.875P(B') = 1 - 0.875 0.125\boxed{\text{0.125}}

Part c

0.15\text{0.15}

Find P(B′)P(B') if P(A∪B)=0.95P(A \cup B) = 0.95.

  1. Use the formula for the probability of the union of two events.

    P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

  2. Substitute the given values to find P(B)P(B).

    0.95=0.8+P(B)−0.70.95 = 0.8 + P(B) - 0.7 0.95=0.1+P(B)0.95 = 0.1 + P(B) P(B)=0.95−0.1P(B) = 0.95 - 0.1 P(B)=0.85P(B) = 0.85

  3. Calculate P(B′)P(B').

    P(B′)=1−P(B)P(B') = 1 - P(B) P(B′)=1−0.85P(B') = 1 - 0.85 0.15\boxed{\text{0.15}}
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