<p>A CD player chooses a track at random from three discs each with 20 tracks. What is the probability that it chooses track 2 of disc 2?</p>
<p>The answer is 1/60.</p>
<p>The explanation states that the track number is negligible-but what if the three different discs have the same track numbers? then wouldn't it be 1/3 x 1/60? </p>
<p>What do you think?</p>
<p>Where did you get this problem?</p>
<p>But uh, there are 60 tracks total right?, because there are 20 tracks per disc.</p>
<p>Track 2 is one track out of the 60 total tracks. </p>
<p>Did you get this wrong, or are you asking for clarification of your extension of the problem?</p>
<p>I don't understand your extension...:(</p>
<p>There are 60 total possibilities. Only one of which is track 2 or disk 2. Therefore its 1/60. This is rated easy, right?</p>
<p>My question is how would you go about answering this question and if you think the Q is misleading</p>
<p>The discs do have the same track numbers. If the disks were extensions of the previous (disk 1 holds 1 - 20, disk 2 hold 21 - 40, etc.), then it'd be stated.</p>
<p>Disk 1 {track 1, track 2, ... , track 20}
Disk 2 {track 1, track 2, ... , track 20}
Disk 3 {track 1, track 2, ... , track 20}</p>
<p>The question asks for track 2 of disk 2. </p>
<p>I don't know what you're asking lol.</p>
<p>You could say there's a 1/3 chance of choosing CD 2 and a 1/20 chance of picking track 2 from the 20 tracks on that CD, but you still get 1/3*1/20 = 1/60.</p>
<p>Oh crap. </p>
<p>Im gonna go get some sleep.</p>
<p>dont think into sat problems this much dude</p>