Among 10 people traveling in a group, 2 have outdated passports.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members of the group arrange themselves to maximize the probability of getting by the inspectors without having the outdated passports detected?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Hence if 10 people, 2 would be checked or if 5 one person would be checked If 10 people go to one desk, p for having outdated passport = 0.2 X - no of people having correct passports in 10 is binomial with p = 0.8 They would be caught only if Inspector selects any one person from the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 Y - no of people having outdated passports and p for Y =0.2 P(Y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= 0.8926 --------------------------------------------- If goes into two groups then either each group can have one outdated or one group no outdated and other groups 2 outdated. For each group one outdated prob of being caught = P(y=1) in I group + P(y=1 in second group)-P( P(y=1) in I group *P(y=1 in second group) as these two events are independent =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= 0.8926 (same as the first case) ---------------------------------------------- If one group has both outdated and other has none then prob of being not caught = 1(Prob of atleast one in the group having outdated 2) Here p for good passport =0.6 = P(Y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 Divide into two groups with 2 outdated in one group.

1. Among 10 people traveling in a group, 2 have outdated passports.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without having the outdated passports detected?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

2. = 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