Joe, Ken and Ben, they live in a shared house. Each Sunday each of the tenants...

Joe, Ken and Ben, they live in a shared house. Each Sunday each of the tenants choose uniformly at random and independence of other tenant one part of the house: kitchen(K), living room(L),bathroom(B),garage(G). and cleans it during the week that follows( here week means a period of 7 days starting on Sunday)

(a) what is the probability that the garage is not cleaned during one week?

(b) what is the probability that the kitchen is cleaned exactly once during one week?

Solutions

Expert Solution

So here each of the tenants choose uniformly at random and independence of other tenant the probability that

kitchen(K), living room(L),bathroom(B),garage(G) gets cleaned is 1/4 (uniformly at random)

a) Hence P(Garage no cleaned) = P(Garage not cleaned by Joe)* P(Garage not cleaned by Ken)* P(Garage not cleaned by Ben)

=3/4*3/4*3/4 (Since the events of picking a part is mutually exclusive the prob that garage is not picked by any one is 3/4)

=27/64

=0.421875

b)P(Kitchen cleaned by once)=

P(Kitchen selected by Joe)*P(Kitchen not selected by Ken)*P(Kitchen not selected by Ben)

+P(Kitchen not selected by Joe)*P(Kitchen selected by Ken)*P(Kitchen not selected by Ben)

+P(Kitchen not selected by Joe)*P(Kitchen not selected by Ken)*P(Kitchen selected by Ben)

=1/4*3/4*3/4 + 3/4*1/4*3/4+3/4*3/4*1/4

=0.421875

Accidentally both the answers from a and b are coming the same. There is no relation however between the answers being the same.

Hope the above answer has helped you in understanding the problem. Please upvote the ans if it has really helped you. Good Luck!!

milcah answered 2 hours ago

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