The probability of a customer arrival at a grocery service counter in any one second is equal to 0.4. Assume that customers arrive in a random stream, so an arrival in any one second is independent of all others. (Round your answers to four decimal places.) (a) Find the probability that the first arrival will occur during the seventh one-second interval. 0.0187 Correct: Your answer is correct. (b) Find the probability that the first arrival will not occur until at least the seventh one-second interval.

a. approximately 0.0187

b. 0.047

Step-by-step explanation:

q = 1-p

= 1-0.4

q = 0.6

a. the probability that the first arrival will occur during seventh one-second interval

probability(7) = 0.6⁷⁻¹ x 0.4

= 0.6⁶ x 0.4

= 0.046656 x 0.4

= 0.0186624

approximately 0.0187

b. probability that the first arrival will not occur until at least the seventh one second interval

p(y≥7) = 1-p(x<7)

= 1-[(0.4)(0.6)⁰ + (0.4)(0.6)¹ +(0.4)(0.6)²+(0.4)(0.6)³+(0.4)(0.6)⁴+(0.4)(0.6)⁵]

= 1-(0.4+0.24+0.144+0.0864+0.05184+0.031104

= 1-0.95334

= 0.04667

= 0.047

there are no short passages to look at

6 and 24

step-by-step explanation:

6 x 4 = 24