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Can anyone answer this Statistics Question - derf1840 - 01-29-2008
I am having some difficulty answering this typwe of question. I was hoping someone could help??? ALSO: Are there many questions like this on the DSST stats exam????
1)
How many ways can 12 baseball cards be divided between Jon and Ryan so that One may get 9 and the other 3 baseball cards??????
and
2)
What is the probability of rolling at least one 1 in two throws of the dice. ( I think that I know the answer to this one). I am wondering if my study guide is wrong??
The possible answers to this questuion are 2/36, 11/35, 25/36, 18/36
I hope someone can answer these
Thanks for any help
Fred
Can anyone answer this Statistics Question - larry7crys - 01-30-2008
The DSST exam is full of these ?s. Try studying more on probablity and combinations. Do you have a calculator with these functions ie nCr and nPr? I'm sorry but I basically dumped all that info the day after the exam.
Can anyone answer this Statistics Question - gcalvin - 01-30-2008
1) This is just a simple combination problem in disguise, because any combination of three cards for Jon automatically leaves Ryan with the other nine. So it's just:
nCr = n! / ( r! (n - r)! )
where n = 12 and r = 3. So...
= 12! / ( 3! (12 - 3)! ) = 479,001,600 / ( 6 * 362880 )
= 479,001,600 / 2,177,280 = 220
If it's really worded exactly as you stated, then you have to multiply by two, for an answer of 440, because for each of the 220 combinations, it could be Jon with three cards and Ryan with nine, or Ryan with three cards and Jon with nine, and those are different ways of dividing the cards. But I would consider that in the realm of "trick questions" of the kind that rarely appear on CLEP, DANTES or ECE exams.
2) There are 36 possible combinations of two throws of a die, of which 11 have a 1.
1-1
1-2
1-3
1-4
1-5
1-6
2-1
3-1
4-1
5-1
6-1
So the answer is 11/36. Technically the "right" way to solve it is to figure the probability of it not happening, and subtract that result from one. So you multiply the probabilities of two "I didn't roll a one" throws (5/6 * 5/6) and subtract the result from one. (1 - 5/6*5/6 = 1 - 25/36 = 11/36)
The reason you have to do it that way, is that multiplying two probabilities gives the combined probability of both events happening -- an AND condition. But that doesn't work for an OR condition, so you have to convert that into (NOT A) AND (NOT B), and then subtract the result from 1 (absolute certainty). I hope that makes some sense. hilarious
-Gary-
Can anyone answer this Statistics Question - derf1840 - 01-30-2008
I got the same answer as you did fore the first question 220. However, the study guide I am using says the answer is [SIZE="5"]440[/SIZE]. Is the study guide wrong?? I just don't know how they came to that conclusion
By the way thanks for your assistance with these questions. If I have any more problems, I was hoping that maybe you could help me through it......It is really really appreciated.
Thanks again
Fred
Can anyone answer this Statistics Question - gcalvin - 01-30-2008
derf1840 Wrote:I got the same answer as you did fore the first question 220. However, the study guide I am using says the answer is [SIZE="5"]440[/SIZE]. Is the study guide wrong?? I just don't know how they came to that conclusionLike I said, that's really a trick question then. 220 is the total combinations for Jon to get three cards and Ryan to get nine cards. But since Jon could get nine and Ryan three instead, there are actually twice as many possibilities, so it's 440. I wouldn't expect anything that deceptive on your real exam.
By the way thanks for your assistance with these questions. If I have any more problems, I was hoping that maybe you could help me through it......It is really really appreciated.
Thanks again
Fred
-Gary-