Question

It is reported that 73% of men use computers. Choose 7 men at random. a. Find the probability that all 7 use a computer. a. b. Find the probability that at least 1 doesn't use a computer. b. 10. 10. The Washington airport ranked 8 in major U.S. airports in on-time arrivals in the last quarter of 2008 with an 81.6% on-time rate. Choose 5 arrivals at random and find the probability that at least 1 was not on time.

Answer 1

a) 0.11047398 b) 0.889526 10.0.63821480241 Explanation: If in any problem, if probability of success is given (p) and probability of failure is (1-p). Then to find probability of k successes out of n chosen samples is given by "Ck (p)k(1-P)n-k. (Using binomial distribution) 9. This follows binomial distribution. probability of success (p) = 0.73 So, a) P (out of 7 randomly chosen samples all 7 use computer) = 7C7 (p)?(1-0)° = 0.11047398 b)P (out of 7 randomly chosen samples at least one does not use computer) = 1 - Plout of 7 randomly chosen samples at less than one does not uses computer) = 1 - 7Co (p) (1-P) = 0.889526 10. Here also binomial distribution is being followed. probability of success (p) = 0.816 Pl out of 5 randomly chosen at least 1 was not on time) =1- Pl out of 5 randomly chosen less than 1 was not on time) =1-5Co (p)5(1-p)º =0.63821480241