The old lady who lives in a shoe has 7 boys and 5 girls Each

The old lady who lives in a shoe has 7 boys and 5 girls. Each week she assigns 3 different jobs (sweeping,mopping,dusting) to her children by picking names out of a sack. The old lady places the name back in the sack each time, which means there is a chance that one child will get all three jobs!!!

Tammy and Billy are two of her children

a)For each selection there is a 10% chance she will pick Tammy, a 10% Chance she will pick Billy and an 8% chance she will select one of her other 10 children. What is the probability that Tammy gets exactly one job and Bobby will get none?

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Solution

We have following probabilites:

P(Timmy)=0.10

P(Billy)=0.10

P(other child)=0.08

(a)

Since selection are independent from each other so the probability that Tammy gets exactly one job and Bobby will get none is

P(Tammy)*P(other child)*P(other child)=0.10*0.08*0.08=0.00064

Hence, the probability that Tammy gets exactly one job and Bobby will get none is 0.00064.