<li>Andy, Barbara, and Cindy are to have a bicycle race, henceforth to be called A, B, and C. The probability that A will win is 1/4, while the probability that B will win is 1/5. What is the probability that C will win if there will be exactly one winner.</li>
</ol>
<p>Now…I tried two different approaches and got two different answers. Both methods make sense to me, though the first way makes a little more sense.</p>
<p>Anyway, the first way is to add the probabilities of A and B together (1/4+1/5=9/20). Then the probability of C winning is 11/20.</p>
<p>The second method is to look at the probabilities of both A and B losing. (3/4 * 4/5=12/20=3/5.</p>
<p>Any thoughts?</p>
<li>How many different arrangements can be made by flying six flags at a time on a flagpole, if two pennants are red, three are blue, and one is white?</li>
</ol>
<p>I think this is the right way to do it, but I don’t understand why.</p>
<p>6comb2 * 4comb3 *1comb1 =60.</p>
<li>What is the number of ways that three identical TV sets and three identical record players can be arranged side by side?</li>
</ol>
<p>No idea how to do this one. My guess would be (3comb2)^2, but that’s just my guess. I don’t really understand why it would be 3 comb 2.</p>
<ol>
<li>I'm guessing 120.
(6!/(3x2))</li>
</ol>
<p>I think you're right (6!/(3!x2!)) = 60
For 3 I believe you do the same thing 6!/(3!x3!) = 20.
Could someone please confirm?</p>
<p>11/20 should be the answer to #1.
i understand ur 2nd solution,but theres just something wrong with it, idk why for now,i'll think about it when i get to school =D i hav alot of free time there.</p>
<p>ok so,as i was eating, i kinda fig out why #1's second solution is wrong
lets look at another example to refute that 2nd solution:
A , B ,C, chance of winning is 1/3 ,respectively.
so if C wants to win, then A and B gotta lose.
prob of A and B losing is 2/3 each.
u cannot 2/3 x 2/3 = 4/9
u cannot do this :s i dont know the fundamental reason for this,but i guess solution 2 works only when theres only 2 factors.
XD</p>
<p>... also, the prob of A losing ,2/3 does not indicate that C will win 1/3, it can be B 1/3 too, its just indicating that A will lose, but dk to which, B or C</p>
<p>I thought that the first method for #1 was right! I'm still confused as to why the second method doesn't work though...and school has already started for you? Mine doesn't start for a week.</p>
<p>Because multiplying the numbers together does not take into account the fact that there can only be a single winner.</p>
<p>Let's use the 1/3 probability proposed by Ren for the sake of simplicity.
I'll go draw some pictures explaining it.</p>
<p><a href="http://img103.imageshack.us/img103/213/82509659ol3.png%5B/url%5D">http://img103.imageshack.us/img103/213/82509659ol3.png</a></p>
<p>Multiplying the chances of them losing together is incorrect because it acts as if the chances of each event happening are independent of one another. The top left square in the picture is blank, which means neither A nor B lost. Since there can be only one winner, this cannot be true.</p>
<p>The chances of A winning are affected by the chances of B winning, and vice versa. You cannot multiply them together, because doing so assumes that they are independent actions. They are not because if A wins, B can no longer win. A limits B.</p>
<p>Thanks! That really cleared things up for me. Any thoughts on 2 and 3?</p>
<p>If all the flags were different, then the number of combinations is 6!.
But since they are not all different, you have to divide the answer. You divide by 2! for the red, 3! for the blue and 1! for the white.
So the answer is 6!/(3!2!). You were right on #2.</p>
<p>You can do the same thing for the third question.
The answer is 6!/(3!3!) = 20.</p>
