<li>A coin is tossed 3 times. Given that at least one head appears, what is the probability that exactly 2 heads appear?</li>
</ol>
<p>A) 3/8
B) 3/7
C) 3/4
D) 5/8
E) 7/8</p>
<li>If f(x) = x/(x-1) and f^2(x) = f(f(x)), f^3(x) = f(f^2(x))…, f^n(x) = f(f^n-1(x)), where n is a positive integer greater than 1, what is the smallest value of n such that f^n(x) = f(x)?</li>
</ol>
<p>A) 2
B) 3
C) 4
D) 6
E) No value of n works</p>
<p>thanks for the help</p>
<p>1) 3/8
2) 3 (f^2(x)=x)</p>
<p>explanations?</p>
<p>1 more..these kinds of questions always gets me:</p>
<p>A purse contains 5 different coins (penny, nickel, dime, quarter, half-dollar). How many different sums of money can be made using one or more coins?</p>
<p>A) 5
B) 10
C) 32
D) 120
E) none of the above</p>
<p>I read the explanations for these problems in the Barron's book, but still didn't understand them.</p>
<p>30
E) none of the above</p>
<p>Answer to the first problem:</p>
<p>There are eight possible ways for three coin flips to turn out (2^3), but one of these is all tails, so there are 7 possible ways to get at least one head. There are 3 possible ways to get 2 heads (HHT, HTH, THH), so the answer is 3/7.</p>
<p>On the second, andy is right. f^2(x)=f(x/(x-1))=(x/(x-1))/((x/(x-1))-1),
but the denominator of this expands to (x-(x-1))/(x-1)=1/(x-1). Cancel the 1/(x-1) from top and bottom and you get f^2(x)=x/1=x. So f^3(x)=f(f^2(x))=f(x).</p>
<p>On the third, you know that there are at most 31 different sums. This is because there are only 32 (i.e., 2^5) subsets of the set of five coins, but one of those is the empty set and you are restricted by the terms of the problem to non-empty subsets ("one or more coins"). To find the exact number, you need to eliminate any duplicate sums, but you don't need to do that to solve the problem. Instead you can quickly rule out 32 and 120 (answer must be 31 or less) and 5 (any of the 5 coins taken alone is a distinct amount, and when you add a penny to any of the others you also get a new amount, so that makes at least nine). Now all you need to know is that you have at least 2 more distinct sums to eliminate 10 as an answer, and 15 cents (nickel and dime) and 30 cents (nickel and quarter) fill the bill nicely. So 10 can't be the answer either and you are left with (E).</p>