-
Welcome to MSO!
We are an online community for current and prospective medical, dental and allied health students and early career professionals from Australia and New Zealand.Please read: About MSO | Annual Welcome and Important Information | MSO Rules
What's New
The following forums have recently been launched: -
Please consider registering on MSO. Benefits of registering are:
- Able to post and participate in the forum
- After 10 posts: Private Message Other Users
- After 25 posts: Access to the Chatbox
- After 100 posts: Custom user titles and Ad-free experience
If you would like to get involved with MSO or have ideas, suggestions, comments, criticisms or other feedback please Contact Us
Registered members with 100+ posts do not see Ads
UCAT: Decision Making
- Thread starter DrC
- Start date
Scorpion
Member
It's easier to understand the last part if you skim over the first bit again after reading it. Basically, your bag of letters was (referring to the 2nd paragraph) a bag of 80 letters with 0.4 x 80 = 32 vowels, more than (0.6 x 80)/2 = 24 which are either B, T, S, H, N, R or C, and just one of Z, X, J and Q. This information presented in the second paragraph is distracting and the first paragraph contains most of the necessary information.
So you started with this big bag of letters and you played Scrabble with them for a while, until now when you (according to 1st paragraph) have 20 letters left. Of these 20 letters, you know there are 0.4 x 20 = 8 vowels, plus 20 - 8 = 12 consonants, of which we know there is a B, T, S, H, N and P. That means there are 6 unknown letters in the bag, and it is with these 6 unknown letters we can determine which answer option is certainly true. To do this, we will try to find scenarios in which each statement is untrue (much like a conclusion Yes/No question):
1. The likelihood of James picking out a vowel is 0.4. There are 6 unknown letters. To maximise the probability of choosing a single consonant, lets say all 6 are Bs, plusthe other B we've already counted. That makes 7 Bs, less than the 8 vowels we have already counted. Therefore, it is more likely James will pick a vowel that any other single letter. As this is true, this is the answer to choose. If you go through the answers methodically, this shouldn't take too much time (and you don't even need to consider the 2nd paragraph or other answer options).
2. We don't know how many of the 8 vowels are As. We don't actually know if any vowels are As, in which case it would be much less likely for James to pick out an A than any one consonant.
3. If all of the rarest pieces have already been used (ie. taken from the 80 letter bag before there were only 20 letters left) there would be a 0% chance of picking out a rare piece.
4. 40% (8) of the remaining 20 letters are vowels. Anything that is not a vowel is a consonant, so 60% (12) of the remaining 20 letters are vowels. Therefore, the probabilities of picking a vowel and picking a consonant are not equal.
A1 just saw your post but figure I'll still post this as it's written now
Could someone please explain this to me. The irony of this statement is that UCAT gave the example of 'featherless birds' in their prior answer responses but didn't apply it to this one.
Am I simply missing something here, or have UCAT failed with their syllogism logic once again?
Am I simply missing something here, or have UCAT failed with their syllogism logic once again?
Attachments
This works. There cannot be an animal which is both "not a bird" and "does have feathers". So an animal must be at least one of "a bird" or "does not have feathers". A featherless bird falls into both categories, which fulfills "or".
So the statement is "An animal is a bird or it does not have feathers". In the instance of a featherless bird, we have An animal which is a bird but does not have feathers - thus fulfilling the category of an animal that is a bird and does not have feathers. Thereby, complying with both sides of the statement - an animal that "is a bird" and it "does not have feathers". I still don't understand....
It was established earlier in the thread that "or" can include "and". Basically, if you manage to colour in the whole Venn diagram with both possibilities, regardless of whether there is overlap, then it is true.
The only time you don't want any overlap is when the question specifically says "x is a or b but not both".
Do you have an example using this?
How would everyone respond to this statement?
"All teachers at this school are male. No teacher at this school play golf"
"Some males play golf"
I would answer yes - as I think this comes back to what A1 mentioned prior. Along the lines of UCAT 'taking the benefit of the doubt' of the possibility that it could be yes considering the minority of males at this school in comparison with the large set of "all males". Effectively, the sub-section of "males at this school" cannot be representative of the larger encompassing "males" in general. The wording of 'some' also implies a positive response (from me at least) tending towards a "Yes" than a "No".
Would be delighted to hear others' inputs on this!
"All teachers at this school are male. No teacher at this school play golf"
"Some males play golf"
I would answer yes - as I think this comes back to what A1 mentioned prior. Along the lines of UCAT 'taking the benefit of the doubt' of the possibility that it could be yes considering the minority of males at this school in comparison with the large set of "all males". Effectively, the sub-section of "males at this school" cannot be representative of the larger encompassing "males" in general. The wording of 'some' also implies a positive response (from me at least) tending towards a "Yes" than a "No".
Would be delighted to hear others' inputs on this!
Last edited:
How do you know males even play golf? It never gives info on males at all. There could be zero males that play golf.
Registered members with 100+ posts do not see Ads
I know. This is what I was wrestling with before in the thread. The UCAT response in their question banks follows this logic, however.
Look in this example at Statement Number 3. Then look at the response.
Attachments
threefivetwo
less gooo
The answer is no. You can't say some males play golf - the only sample of males given in the statement is the teachers at the school, where none of them play golf. There is no benefit of the doubt when the passage is clear-cut about it - benefit of the doubt applies more to qualifiers such as 'may' where the passage has supporting information.
The same could be argued for the example I just referenced - yet the answer is apparently "yes"
threefivetwo
less gooo
All the tourists at the resort are Europeans, representing a subset of the overall population of Europeans that exist. None of the tourists at the resort like trekking. Therefore, some Europeans don't like trekking. Some = not all.
versus
All teachers at the school are male, representing a subset of the overall population of males. No teachers at the school play golf. We cannot assume that some males play golf, as there isn't any information in the passage to support this - assuming this would be extrapolating beyond the confines of the passage. Therefore, the answer here is no.
Hope that clears it up
I think what OP is referring to is that in the European example, there is no information as to whether any Europeans like trekking at all, thus unable to fulfill with certainty the criteria "not all".All the tourists at the resort are Europeans, representing a subset of the overall population of Europeans that exist. None of the tourists at the resort like trekking. Therefore, some Europeans don't like trekking. Some = not all.
versus
All teachers at the school are male, representing a subset of the overall population of males. No teachers at the school play golf. We cannot assume that some males play golf, as there isn't any information in the passage to support this - assuming this would be extrapolating beyond the confines of the passage. Therefore, the answer here is no.
Hope that clears it up
It appears in this case UCAT has decided to use the definition of some as "at least one" rather than "at least one but not all". Regardless, the golf question is a definite no- you cannot even confirm "at least one", let alone "at least one but not all".
hi, im from the uk - but i was wondering if anyone had any tips for the yes or no questions? I tend to get all of the others right but these i really do struggle
also is it better to do the others first, then the yes and no last? I'm not too sure about what order to do them in
also is it better to do the others first, then the yes and no last? I'm not too sure about what order to do them in
Oh ok I think you meant 'all teachers at this school are male and all teachers play golf. 'Question is 'some males play golf.' In this case the answer is yes as although the subset of males: teachers all play golf, there could be a broader category of males who don't play golf, hence not 100% of males play golf.
i don't really have tips as i am not the best at it either
( but i heard that the yes no questions should be prioritised as they are worth 2 marks compared to the others that are worth 1Registered members with 100+ posts do not see Ads
Do keep in mind that they're worth two marks BECAUSE they're harder and more time-consuming, you're not exactly doubling your marks by doing them first. For me, some of the later questions in DM are basically free marks (probability, Venn diagrams and sometimes logic puzzles), so I used to do them first to make sure I'd taken care of all the low hanging fruit, then portion out my remaining time to the 2-mark syllogisms. peachesss experiment a bit to see what works best for you.i don't really have tips as i am not the best at it either
Ah yes, that clears it up, though It still seems unreasonable because all males could play golf, and given some≠all the conclusion couldn't be drawn. Regardless, I'll stick with their response demonstrated in their practice tests!
Hey everyone! could someone please help me with this question.
I think the answer should be C since after one white counter has been taken out their is a 2/3 chance of the counter being black and 1/3 chance of the counter being white. How is the chance of taking a white counter 0.5?
I think the answer should be C since after one white counter has been taken out their is a 2/3 chance of the counter being black and 1/3 chance of the counter being white. How is the chance of taking a white counter 0.5?
Attachments
Similar threads
Registered members with 100+ posts do not see Ads