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Math Help - probability problem

  1. #1
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    probability problem

    An airline has 3 flights from Chicago to New York, Atlanta, and Los Angeles. Let  A denote the event that the New York flight is full and define events  B and  C analogously.

    And  P(A) = 0.6, \ P(B) = 0.5, \ P(C) = 0.4 and the three events are independent. Compute the following probabilities:

    (a) all 3 flights are full. at least one flight is not full.

    So  P(ABC) = P(A)P(B)P(C) = 0.12 and  1-P(ABC) = 0.88 .

    (b) only the New York flight is full. exactly one of the three flights is full.

    So  P(\text{only NY flight is full}) = P(A) - P(AB) - P(AC) + P(ABC) ?

    And the probability that exactly one flight is full is:   P(A) - P(AB) - P(AC) + P(ABC)  + P(B) - P(AB) - P(BC) + P(ABC) + P(C) - P(AC) - P(BC) + P(ABC)

    are these correct?
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  2. #2
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    Hello, shilz222!

    An airline has 3 flights from Chicago to NY, Atlanta, and LA.
    Let A denote the event that the New York flight is full
    and define events B and C analogously.

    And: . P(A) = 0.6,\;P(B) = 0.5,\;P(C) = 0.4
    and the three events are independent.

    Compute the following probabilities:

    (a) (i) all 3 flights are full. (ii) at least one flight is not full.

    \text{(i) }\;P(A \cap B \cap C) \;= \;P(A)\cdot P(B)\cdot P(C) \;= \;0.12 . . . . Yes!

    \text{(ii) }\;1 - P(A\cap B\cap C) \;= \;0.88 . . . . Right!

    (b) (i) only the New York flight is full. (ii) exactly one of the three flights is full.
    \text{(i) }\;P(A \cap \overline{B} \cap \overline{C}) \;=\;(0.6)(0.5)(0.6) \;=\;0.18


    \text{(ii)}\;\begin{array}{ccccc}P(A \cap \overline{B} \cap \overline{C}) & = & (0.6)(0.5)(0.6) & = & 0.18 \\<br />
P(\overline{A} \cap B \cap \overline{C}) &=& (0.4)(0.5)(0.6) &=& 0.12 \\<br />
P(\overline{A} \cap \overline{B} \cap C) &=& (0.4)(0.5)(0.4) &=& 0.08\end{array}

    P(\text{exactly one full}) \;=\;0.18 + 0.12 + 0.08 \;=\;0.38

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