Ameeting is scheduled involving 12 people. suppose that each person has an 87 percent chance of arriving on time, independently of the other people involved. what is the probability that one or more of the 12 people arrives late?

We solve this using binomial probability equation.. The formula is given as:

P = [n! / (n – r)! r!] p^r * q^(n – r)

where n is total number of people = 12, r number of people arriving on time = 0, p is chance of arriving on time = 0.87, q is 1 – p = 0.13

P = [12! / (12 – 0)! 0!] 0.87^0 * 0.13^(12 – 0)

P = 2.33 x 10^-11 = 2.33 x 10^-9 %

So the chance that 1 or more arrives late is very low ~2.33 x 10^-11  or 2.33 x 10^-9 %

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

73.0769230769% of the games were won