Suppose that the probability that a person books an airline ticket using an online travel website is 0.72. Consider a sample of ten randomly selected people who recently booked an airline ticket. What is the probability that at least nine out of ten people used an online travel website when they booked their airline ticket

0.183

Step-by-step explanation:

This is a question on Binomial Probability

Formula =nCx × p^x × q^n - x

p = 0.72

q = 1 - p

= 1 - 0.72

= 0.28

x = number of successes = 9

n = 10

The probability that at least nine out of ten people used an online travel website when they booked their airline ticket

At least 9 out of 10 means

x ≥ 9 = x = 9 and x = 10

Hence,

P(x ≥ 9) = 10C9 × (0.72^9 × 0.28^10 - 9) + 10C10 × ( 0.72^10 × 0.28^10 - 10)

P(x ≥ 9) = 0.14559635388 + 0.0374390623

= 0.18303541631

≈ 0.183

a=24; b=12

step-by-step explanation:

a/b=2: 1

a=2b

36/(1+2)=12

b=12

a=24