Among 10 people traveling in a group 2 have outdated passpor

Among 10 people traveling in a group, 2 have outdated passports.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members of the group arrange themselves to maximize the probability of getting by the inspectors without having the outdated passports detected?

Solution

Out of 10 people 2 have outdated passports

Hence p = prob of a person having outdated passport = 0.2

If all 10 go to one desk

inspectors will check only 2.

Let X = no of people not having passport checked by inspector

X is binomial with (2,0.2)

P(x is atleast 1)=1-P(0)

=1-0.8²

= 0.36

Hence prob of not being caught = 0.64

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If 5 people go to one desk, then there are 3 possibilities for G1 and G2.

(5,0) (3,2)

(4,1) (4,1)

(3,2) (5,0)

In the above cases only one person will be checked as 20% of 5 is 1

For case I, being not caught

= P(X=0 in 1 trial) = 0.6

For case II, being not caught = P(X=0)*P(X=0)

= 0.8*0.8 = 0.64

For case III being not caught is

3/5= 0.6

Hence either all to one desk or split into two groups with one outdated person in one group is the best

For case II,