Question 1105926
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You are selecting 3 of the 15 states in the list, so the denominator of the probability fraction for all three cases is "15 choose 3": {{{C(15,3) = (15*14*13)/(3*2*1) = 455}}}.<br>
In all three cases you are looking for states with populations of more than 9 million.  10 of the 15 have populations of more than 9 million; 5 do not.<br>
a) To have none of the three state you choose having populations over 9 million, you need to choose 3 of the 5 states with lower populations (5 choose 3) and none of the 10 with higher populations (10 choose 0).<br>
The probability for this case to 3 decimal places is {{{(C(5,3)*C(10,0))/C(15,3) = (10*1)/455 = .022}}}<br>
b) To have exactly one of the three state you choose having populations over 9 million, you need to choose 2 of the 5 states with lower populations (5 choose 2) and 1 of the 10 with higher populations (10 choose 1).<br>
The probability for this case to 3 decimal places is {{{(C(5,2)*(C(10,1)))/C(15,3) = (10*10)/455 = .220}}}<br>
c) This case is just the "opposite" or complement of the first case: "at least one" is the complement of "none".  So the probability for this case is 1-.022 = .978.