Question

From experience, an airline knows that only of the passengers booked on a flight from New York to Los Angeles actually board their flight. A random sample of booked passengers from New York to Los Angeles is chosen. Find the probability that of them board their flight.

Do not round your intermediate computations, and round your answer to three decimal places.

Answer 1

Explanation:

6 or 7.

that means either exactly 6 or exactly 7 out of the 10 board the flight.

these are non-depending (or non-overlapping) scenarios. so, for this "or" remained we can simply add the 2 individual probabilities P(6) and P(7).

P(6) is the probability of exactly 6 passengers out of 10 will board the flight.

because of the 75% certainty overall we can conclude that the probabilty for each booked passenger to board is 0.75 (75% means 75 out of 100 = 0.75).

and the probabilty to not board is

1 - 0.75 = 0.25

so, for a specific group of exactly 6 booked passengers out of 10 their probabilty to board is

0.75⁶ × 0.25⁴ = 0.000695229...

6 passengers board, 4 do not board.

but this is for only one possible grouping of 6 passengers out of 10.

since the sequence does not matter, and we have no repetitions (every passenger counts as 1), we have combinations :

C(10, 6) = 10! / (6! × (10-6)!) = 10!/(6!×4!) = 10×9×8×7/4! =

= 5×3×2×7 = 210

therefore,

P(6) = 210 × 0.75⁶ × 0.25⁴ = 0.145998001...

and for a specific group of exactly 7 booked passengers out of 10 their probabilty to board is

0.75⁷ × 0.25³ = 0.002085686...

7 passengers board, 3 do not board.

but this is for only one possible grouping of 7 passengers out of 10.

since the sequence does not matter, and we have no repetitions (every passenger counts as 1), we have combinations :

C(10, 7) = 10! / (7! × (10-7)!) = 10!/(7!×3!) = 10×9×8/3! =

= 5×3×8 = 120

therefore,

P(7) = 120 × 0.75⁷ × 0.25³ = 0.250282288...

so, the probability that either 6 or 7 booked passengers are actually boarding is

P(6) + P(7) = 0.396280289... ≈ 0.396