Delta Airlines' flights from Boston to Seattle are on time 70 % of the time. Suppose 6 flights are randomly selected, and the number on-time flights is recorded. 1.The probability that at least 4 flights are on time is

P(at least 4)  = 0.74431

Step-by-step explanation:

Here

Probability of the flights on time is = 70% = 70/100= 0.7

q= 0.3

n= 6

As the probability of success and failure is constant and the n is also fixed then the binomail probability distribution can be used to solve this.

P (x= 4) = 6c4 *(0.7)^4 (0.3)^2= 15*0.2401* 0.09= 0.324135

P (x= 5) = 6c5 *(0.7)^5(0.3)^1= 6*0.16807* 0.3=0.302526

P (x= 6) = 6c6 *(0.7)^6 (0.3)^0= 0.117649P

P (x= 0) = 6c0 *(0.7)^0 (0.3)^6= 0.000729

P (x= 1) = 6c1 *(0.7)^1 (0.3)^5=0.010206

P (x= 2) = 6c 2*(0.7)^2 (0.3)^4=0.059535

P (x= 3) = 6c3 *(0.7)^3 (0.3)^3=0.18522

Probability of at least 4 means that probability of greater than 4 inclusive of 4.

This can be obtained in two ways.

First is by subtracting the probabilities of less than 4.

Second is by adding probabilities greater than 4 including 4.

We get the same answer with any of the two methods used.

P(at least 4) = 1- P(x<4)

                     = 1- [ 0.000729+ 0.010206 +0.059535+ 0.18522]

                     = 1- 0.25569

                       = 0.74431

P (at least 4) = P(x=4) +P(X=5) + P(x=6)

                        =0.324135+ 0.302526+ 0.117649

                        =0.74431

8^3.

Step-by-step explanation:

= [ (6^3)^6*1/2] /  [9^9]^1/2

= 6^9 / 9^9/2

= 6^9 / (3^2)^9/2

=  6^9 /3^9

=  512

= 8^3.