<p>I couldn't answer this question</p>
<p>Type of chocolate==========Percent COCOA BY WEIGHT
MILK========================25
DARK========================50
Bittersweet===================70</p>
<p>A website sells three types of chocolate bars of equal weight. If Serena orders two chocolate bars at random from the website, then melts them together, what is the probability that the resulting mix contains at least 50 percent cocoa by weight?</p>
<p>Thanks! :)</p>
<p>Unfortunately, you have or will fuind out why a few of us keep repeating to STAY AWAY from tests that are not published by ETS and the College Board. Not only are they rarely true the real tests in form and contents, but also useless because of careless errors. It is obvious that PR does not value the work of an editor. </p>
<p>This is why is a waste of time solving this type of misleading junk, let alone trying deciphering the work of a careless and amateurish author. </p>
<p>Read this past thread:</p>
<p><a href=“http://talk.collegeconfidential.com/sat-preparation/234343-how-do-you-do.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/234343-how-do-you-do.html</a></p>
<p>and</p>
<p><a href=“http://talk.collegeconfidential.com/sat-preparation/441485-need-help-math-question-not-bb.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/441485-need-help-math-question-not-bb.html</a></p>
<p>so Xiggi,
I have almost two weeks left before I take the SAT… I’m hoping for an 800 but I finished the blue book and I don’t feel like it is enough, what do you think? are there any relevant books to the SAT other than the blue book?</p>
<p>For tests, check the past editions of the BB2 (blue and red editions), TCB online course, and released QAS. </p>
<p>For strategies and support, the best avenue is this site.</p>
<p>I might be dumb right now, but this seems easy to me.</p>
<p>Assume each chocolate is 100g. Thus, any two chocolates have a weight of 200g. 50% Cocoa by weight is 100 g, which would mean that the combined amount of Cocoa needs to be > 100, making the probability 1/3, as only the combination of Bittersweet + Dark matches…</p>
<p>If I’m really making some horrible mistake, someone please correct me.</p>
<p>“which would mean that the combined amount of Cocoa needs to be > 100”</p>
<p>100 is at least 100! </p>
<p>The point is that the PR book indicates the answer is 1/2 in one place, gives an explanation for the answer to be 4/9 in another place, and nobody is the wiser as the correct answer might be different from the two suggested answers. </p>
<p>Just not worth the agony to figure out if PR can write or count correctly.</p>
<p>Hi there. Just because it’s from PR doesn’t mean it’s from an “amateurish/careless/misleading author”, let alone “junk”. The concepts are the same, and it’s just a matter of application. Dismissing a question you can’t do as “junk” is really unfair. #justsayin’ :)</p>
<p>In answer to your question, you first add the %tage of cocoa for all possible combinations of chocolate. So the possibilities are M+D, M+B, D+B. So let’s take M+D as an example. Their cocoa content is 25% + 50%, so it’s 75% total. But this is over 200%, so you have to convert it back to 100%. So 75/200x100% = 37.5%. </p>
<p>Apply this for the remaining 2 combinations and you’ll find that only M+D has a cocoa content below half. So, the answer is the probability is 2/3.</p>
<p>okay Xiggi. I’ll do what you said about practice and I’ll definitely ask here for what I don’t know. 
thanks for your support, man.</p>
<p>iatekatie… I’m sorry but 2/3 is incorrect.</p>
<p>Isn’t the answer just 2/3 * 2/3 = 4/9? I might be wrong, but that seems logical to me.</p>
<p>You pick the first bar, and that one has to be B or D: a chance of 2/3.
You pick the second bar, and that one also has to be B or D: a chance of 2/3.
Multiply: 2/3 * 2/3 = (2 * 2) / (3 * 3) = 4 / 9. </p>
<p>Is 4/9 the right answer, NaderMekadis?</p>
<p>absolutely right, dutchguy!
can you please correct me if I’m wrong? she only bought 2 out of 3 then 2/3 chance and then 2/3 chance multiplied by each other gives a 4/9… yeah that’s logical. :)</p>
<p>Hello,</p>
<p>I’m confused. In the original version of this problem, as presented in the links Xiggi provided, the “percent cocoa by weight” for the Milk Chocolate bars was 35%. In this new thread, it is now 25%. Which one is it?</p>
<p>Dennis</p>
<p>Whoa, careless mistake there thanks DutchGuy, you’re right, I forgot to multiply the two fractions together. Nadermedakis, don’t be as careless in your SAT 
but I do hope you understand the question now!</p>
<p>By the way, for the ones interested in this problem, I hope you’ll consider that the purchaser does not have 9 different opportunities to purchase two bars. She can only order in SIX different ways. In this case, ordering Milk and Dark is the same as Dark and Milk. You cannot have duplicates. See below for details. You can only have 3 combinations of the same bars and 3 combinations of mixed bars. Think about order and repetitions. </p>
<p>MM
MD
MB
DD
DB
BB</p>
<p>Quote:
*In answer to your question, you first add the %tage of cocoa for all possible combinations of chocolate. So the possibilities are M+D, M+B, D+B. So let’s take M+D as an example. Their cocoa content is 25% + 50%, so it’s 75% total. But this is over 200%, so you have to convert it back to 100%. So 75/200x100% = 37.5%. </p>
<p>Apply this for the remaining 2 combinations and you’ll find that only M+D has a cocoa content below half. So, the answer is the probability is 2/3.* </p>
<p>If you use the 25% number for M, the answer is not 2/3. However, it could be for a problem with 35 percent in M! </p>
<p>Quote:
*Isn’t the answer just 2/3 * 2/3 = 4/9? I might be wrong, but that seems logical to me.</p>
<p>You pick the first bar, and that one has to be B or D: a chance of 2/3.
You pick the second bar, and that one also has to be B or D: a chance of 2/3.
Multiply: 2/3 * 2/3 = (2 * 2) / (3 * 3) = 4 / 9. </p>
<p>Is 4/9 the right answer, NaderMekadis? * </p>
<p>Regardless of using 25% or 35%, the answer cannot be 4/9.</p>
<p>Quote:
*Whoa, careless mistake there thanks DutchGuy, you’re right, I forgot to multiply the two fractions together. Nadermedakis, don’t be as careless in your SAT but I do hope you understand the question now! * </p>
<p>See above. The error is not in multiplying the two fractions.</p>
<p>xiggi is right. 1/2 is the right answer, I believe.</p>
<ol>
<li>Using 25, 50 and 70</li>
</ol>
<p>Has to be at least 100/200 in combined weight.</p>
<p>MM 50
MD 75
MB 95
DD 100
DB 120
BB 140</p>
<p>Answer is 3/6 or 1/2</p>
<p>===============================</p>
<ol>
<li>Using 35, 50 and 70 as in the previously quoted problems</li>
</ol>
<p>Has to be at least 100/200 in combined weight.</p>
<p>MM 70
MD 85
MB 105
DD 100
DB 120
BB 140</p>
<p>Answer is 4/6 or 2/3</p>
<p>===============================
3. Using 35, 50 and 70 as in the quoted problems</p>
<p>BUT if PR meant that at least meant more than 100/200 in combined weight.</p>
<p>MM 70
MD 85
MB 105
DD 100
DB 120
BB 140</p>
<p>Answer is 3/6 or 1/2</p>
<p>What Xiggi is right about is that the question is not well-posed. I believe that you can correctly interpret it both ways, so that the answer is either 1/2 or 4/9. It depends on how you interpret what it means to select the candy randomly. Do you mean, “The store offers 6 ways to buy two bars: MM, MD, MB, DB, BB and DD – you pick one of the 6”. This interpretation leads to the answer 1/2.</p>
<p>But maybe you mean “The store keeps the bars in 3 bins and you have a spinner – spin the spinner and pick the bar that the arrow points to, then repeat.” Do it that way and the people who vote for 4/9 are correct.</p>
<p>So what’s the point? Well, one point is that this question is FLAWED. It could NEVER be on the SAT. Stick with college board material.</p>
<p>A second point is that probability problems can be truly vexing. Anyone want to talk about the Monty Hall paradox?</p>
<p>
</p>
<p>Obviously I have to agree with the second statement, as it was exact point in post 2: “This is why is a waste of time solving this type of misleading junk, let alone trying deciphering the work of a careless and amateurish author.” </p>
<p>However, I am not sure I can agree with the interpretation in the first part. “A website sells three types of chocolate bars of equal weight. If Serena orders two chocolate bars at random from the website” … I do not think that the website uses a spinner to fill that order, and there are only six ways to fill the random order. If I order two bags of Senseo Coffee at Amazon, it does not matter if the Decaf is on the bottom or the top of the package. ;)</p>
<p>Hello,</p>
<p>Several years ago, I used to do some freelance work as a proofreader and answer checker for rough drafts of new and upcoming editions of mathematics textbooks and workbooks.</p>
<p>The type of problem which used to cause the most arguments and disagreements, as to the correct answers and solutions, even among the main textbook writers themselves, were probability problems such as this one.</p>
<p>Dennis</p>
<p>Xiggi, I fully agree with you - screw PR!!! Hahaha just kidding. Yes you’re right, I forgot to include the possibility of picking the same type of chocolates. Thanks for pointing that out. You sure do read deeply into the question!</p>