<p>I did this probability question, below, and don't know why my way of answering the question is wrong. </p>
<p>Of 6 radios in a store, 2 are defective and 4 are not. A customer chooses 2 of these radios at random. If the first radio chosen is not defective what is the probability that the second radio chosen is also not defective? </p>
<p>answer is 3/5 </p>
<p>What I did was this: </p>
<p>Probability on first pick = 2/3
Probability on second pick = 3/5 </p>
<p>Since I took the question as: the probability of getting two defective radios in a row i.e. non defective and non defective</p>
<p>then: 2/3 X 3/5 = 2/5 </p>
<p>Why is my thinking wrong- the question to me is asking for successive trials?</p>
<p>It says the probability that the second radio won’t be defective, so you ONLY take into account that 3/5 radios left aren’t defective, you don’t take into account the initial probability that you choose a non defective radio. Read the question more clearly next time</p>
<p>What you did here was misread the question, they do not want the probability of two defective radios, they want you to assume that one has been picked and is not defective, all you need is the second probability. Just, the sat people being sneaky, it’s a really easy problem with a trick in it.
Sent from my DROID RAZR using CC</p>
<p>It’s a conditional probability problem…in this case, one of the non-defective radios is removed, so there are 2 defective and 3 non-defective remaining. Ans = 3/5.</p>
<p>if x=4 , which of the following is the greatest in value ?
(A) (x+1) (x+2)
(B) (X+1)(X-1)
(C) (X-2)(X+2)
(D) (x-2)(x+1)
(E) (x-4)(x+4)
I need the explanation pllz :(</p>
<p>All the factors in all 4 expressions are non-negative, so we look check the largest factors first. Well, x+4 is the largest, but the x-4 term makes it 0. The next two largest factors are x+2 and x+1 which both appear in A.</p>
<p>@ strategicfiasco Think about it logically. We start with 2 defective and 4 non-defective. We take out one that is not defective so we are left with 2 defective and 3 non-defective. The probability of choosing a non-defective is now 3/(3+2)=3/5</p>
<ol>
<li><p>A lot of folks consider it bad manners to hijack somebody else’s thread to change the subject. If you have your own unrelated question, you can start your own thread.</p></li>
<li><p>RandomHSer’s answer is way more complicated than necessary. They told you x = 4. Plug in 4 for x, and multiply.</p></li>
</ol>
<p>@strategicfiasco, conditional probability is in the form “Probability of X given Y occured” or something like that. This problem could be considered conditional probability.</p>
<p>I think rspence is right. When you have total 5 radios left out of which 2 are defective and 3 are not defective, there are 3/5 chances that you will choose non defective.</p>
<p>Okay thanks for all the help. From what I understand, correct me if I am wrong, I treated the question as if the two events - picking a radio that was not defective and then picking a second radio non defective - were independent events when in fact they are not and therefore we cannot multiply the two events together to get the probability that the question asks for. That is, since the probability of a second event occurring -non defective radio- changes to 3/5 from 2/3 then we have a conditional probability problem in which case we are dealing with dependent (given that probability) probabilities, meaning we use the formula: P(A and B) = P(A) · P(B|A)</p>
<p>Correct me if I’m wrong, but I have never seen a question on SAT that would require to combine two probabilities. It’s always a one step solution. Always think simple :)</p>
<p>^ there is this one:
CB book, 2nd edition, Test 7, pg 773, #15:
15. The Acme Plumbing Company will send a team of 3 plumbers to work on a certain job. The company has 4 experienced plumbers and 4 trainees. If a team consists of 1 experienced plumber and 2 trainees, how many different such teams are possible?</p>
<p>That question always got on my nerves… would still do it wrong again I think… (even my teacher did wrong the first time and was sure it was the cb that was mistaken… lol… i’ll never let him forget that :P)</p>
<p>Yeah yeah, i know… but the process… ya know the fact that u have to not simple multiply… because the order of trainees doesn’t count…
It’s very confusing… :/</p>
<p>Ah, I guess these are super rare. It doesn’t take much time writing down the options. I solved it in like 30 sec: ABCD are experienced plumbers and 1234 are trainees.</p>
<p>A 12/13/14/23/24/34 - 6 possible combinations for every experienced plumber
B (same combinations) - 6
C (same combinations) - 6
D (same combinations) - 6</p>
<p>so there are 24 possible combinations.</p>
<p>Unless you are very experienced and confident in solving these things mathematically, I strongly recommend to write down the possibilities. This is a much safer method. As you see, even teachers get tricked if they try to do it mathematically. SAT just wants us to know how to deal with simple every day situations. You don’t try to solve your every day matters with maths, I guess?</p>
<p>While studying for SAT I learnt one thing - think simple! There is nothing complicated about SAT. If you imagine that it is something complicated, your mind creates numerous traps for you. </p>
<p>Every time I do the test I tell myself “Everything has a simple solution. If you think it is complicated, cancel and think again!”.</p>
<p>AimingAt750, I used to use that method of listing but found I was too slow to list everything and always felt vulnerable listing stuff. The nPr and nCr methods, assuming you actually understand how to use it, is far quicker and more accurate in my opinion. That is only if you know what you are doing. If you don’t then you won’t know how to tackle a problem with aplomb</p>