Among 10 people traveling in a group 2 have outdated passpor

Among 10 people traveling in a group, 2 have outdated passports. It is known that inspectors will
check the passports of 20% of the people in any group passing their desks. The group can go as a
whole (all 10) to one desk or can split into two groups of 5 and use two different desks. How should
members of the group arrange themselves to maximize the probability of getting by the inspectors
without having the outdated passports detected? I do not know how to figure out each probability... The section is on counting rules - i am guessing it is without replacement and order does not matter. Thank you for your help

Solution

Inspectors would check 20% of people in the group

Hence if 10 people, 2 would be checked or if 5 one person would be checked

If 10 people go to one desk, p for having outdated passport = 0.2

X - no of people having correct passports in 10 is binomial with p = 0.8

They would be caught only if Inspector selects any one person from the two.

Y - no of people having outdated passports and p for Y =0.2

P(Y>=1) = 1-P(Y=0)

= 1-(0.8)¹⁰

= 0.8926

---------------------------------------------

If goes into two groups then either each group can have one outdated or one group no outdated and other groups 2 outdated.

For each group one outdated prob of being caught

= P(y=1) in I group + P(y=1 in second group)-P( P(y=1) in I group *P(y=1 in second group)

as these two events are independent

= {1-(0.8)⁵}+{1-(0.8)⁵}-{1-(0.8)⁵}{1-(0.8)⁵}

=2(0,67232)-0.67232²

= 0.8926 (same as the first case)

----------------------------------------------

If one group has both outdated and other has none then prob of being not caught

= 1(Prob of atleast one in the group having outdated 2)

Here p for good passport =0.6

= P(Y>=1)

= 1-P(y=0)

= 1-(0.6)⁵

=0.92224

Hence this is the best method to go.

Divide into two groups with 2 outdated in one group