Question

From experience, an airline knows that only 80% of the passengers booked for a certain flight actually show up. If 6 passengers are randomly selected, find the

probability that at least 5 of them show up.
Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

Answer 1

Answer: about 66% probability

Explanation:

p = 0.8 is the success of a show up

q= 0.2 is the fail rate probability

Pr(at least 5 successful showups)

= (6 choose 5 )* 0.8^5 * 0.2 + (6 choose 6)* 0.8^6

= 6 * 0.8^5 *0.2 + 0.8^6

= 6* 0.32768*0.2 + 0.262144

= 0.393216 + 0.262144

= 0.65536 = 0.66

From MysticAlanCheng