Q:A list of the names of the people of the entire 1990 foreign-born population in neighborhood V was generated, with each person’s name appearing once. The names of 2 different people will be randomly selected from the list. Which of the following is closest to the probability that both names selected will be names of people whose region of origin was “Other”?
A 0.01
B 0.11
C 0.25
D 0.39
E 0.49
my approach:(11/100)*(10/99)*2
Hey,
Your logic is right, where you are making the mistake is that the question is asking for ‘selection’ i.e combination. So you don’t need to *2 to arrange the selections.
So the question can be simplified to probability of selecting 2 individuals from Others = 11C2/100C2 = (11x10)/(100x99)
Hey,
Thanks I understood your solution, but didn’t understand what is wrong with mine. Because I thought order doesn’t matter, I multiplied by 2.
Let A,B belong to Others & you choose A&B from the pool.
Because the order doesn’t matter AB (i.e first you chose A then you chose B) and BA will be the same. By multiplying it with *2 you are contradicting the fact that order doesn’t matter.
Hope this helps.
If we do (11/100)(10/99), doesn’t it mean 11/100 is first and 10/99 is second, so don’t we do (11/100)(10/99) + (11/100)*(10/99)
Hey,
So when it is a selection order doesn’t matter. So whether 11/100)10/99 case Object 1 is picked in spot 1 or spot 2, it’s one in the same, like the example I gave in my previous reply.
If you still aren’t clear don’t worry you can revise Combinatorics again and come back.
Hey,
Thanks will do.