1. ## 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?

2. 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.
$\text{(i) }\;P(A \cap \overline{B} \cap \overline{C}) \;=\;(0.6)(0.5)(0.6) \;=\;0.18$

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

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