To find the extramental probability that at least 5 flights arrive, we can use the binomial distribution formula:
P(X ≥ 5) = 1 - P(X < 5)
where X is the number of flights that arrive on time out of 10 flights.
The probability that a flight arrives on time is 0.6, and the probability that a flight arrives late is 0.3.
Using the binomial distribution formula and the complement rule, we have:
P(X < 5) = ΣP(X = x) for x = 0 to 4
= C(10,0)(0.6)^0(0.4)^10 + C(10,1)(0.6)^1(0.4)^9 + C(10,2)(0.6)^2(0.4)^8 + C(10,3)(0.6)^3(0.4)^7 + C(10,4)(0.6)^4(0.4)^6
= 0.000105
Therefore,
P(X ≥ 5) = 1 - P(X < 5)
= 1 - 0.000105
= 0.999895
This means that the extramental probability that at least 5 flights arrive is 0.999895.
To find the extramental probability that at least 2 flights arrive late, we can use a similar approach:
P(Y ≥ 2) = 1 - P(Y < 2)
where Y is the number of flights that arrive late out of 10 flights.
Using the complement rule and the binomial distribution formula, we have:
P(Y < 2) = ΣP(Y = y) for y = 0 to 1
= C(10,0)(0.3)^0(0.7)^10 + C(10,1)(0.3)^1(0.7)^9
= 0.0282485
Therefore,
P(Y ≥ 2) = 1 - P(Y < 2)
= 1 - 0.0282485
= 0.9717515
This means that the extramental probability that at least 2 flights arrive late is 0.9717515.
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