Question

Three couples and two individuals (eight total people) have been invited to an investment seminar and have agreed to attend. Suppose the probability that any particular couple or individual arrives late is .4 (a couple will travel together in the same vehicle, so either both people will be on time or else both people will arrive late). Assume that different couples and individuals are on time or late independently of one another. Calculate the probability that 6 or more people arrive to the seminar on time. Hint: consider the disjoint cases: exactly 6 are on time, exactly 7 are on time, exactly 8 are on time, then think of all the combinations of couples and individuals that yield that amount.

Answer 1

P( 6 or more people arrive to the seminar on time )=P(6 are on time)+P(7 on time)+P(8 on time)

=P(2 individual are late or a couple is late)+P(1 individual is late)+P(none of them is late)

=P(no couple late and 2 individual late)+P(1 couple and no individual) +P(no couple late and 1 individual late)+P(no couple and no individual late)

=(₃C₃)*(0.6)³(0.4)⁰)*(₂C₀)*(0.6)⁰(0.4)²+(₃C₂)*(0.6)²(0.4)¹)*(₂C₂)*(0.6)²(0.4)⁰+(₃C₃)*(0.6)³(0.4)⁰)*(₂C₁)*(0.6)¹(0.4)¹+(₃C₃)*(0.6)³(0.4)⁰)*(₂C₀)*(0.6)²(0.4)⁰

=0.03456+0.15552+0.10368+0.07776 =0.37152