Answer:
a. approximately 0.0187
b. 0.047
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