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4!/4^4=3/32 (pls check again)=9% approx.PussInBoots wrote:I disagree
Total combinations: 4*4*4*4
Total combinations of different numbers: 4*3*2*1
Answer is 4! / 4^4 = 1/8, hence ~12%
xcusemeplz2009 wrote:4!/4^4=3/32 (pls check again)=9% approx.PussInBoots wrote:I disagree
Total combinations: 4*4*4*4
Total combinations of different numbers: 4*3*2*1
Answer is 4! / 4^4 = 1/8, hence ~12%
i do agree with this method
Guys, the two "methods" related above both lead to the same result:me10gmat800 wrote:xcusemeplz2009 wrote:4!/4^4=3/32 (pls check again)=9% approx.PussInBoots wrote:I disagree
Total combinations: 4*4*4*4
Total combinations of different numbers: 4*3*2*1
Answer is 4! / 4^4 = 1/8, hence ~12%
i do agree with this method
Have one doubt so allow me to reopen this discussion...
Is this one of the standard probability formulas/rules (4!/4^4 generalized form -> n!/n^n) that could be used in such type of questions?
I solved this q in similar fashion that automatically led to this rule/formula:
1st person able to choose in 4/4 ways, 2nd person 3/4, 3rd 2/4 and 4th 1/4.
Therefore the probability is 4/4 * 3/4 * 2/4 * 1/4 = 4! / 4^4 (rule)
Could someone explain why and when this rule could be used?
Thanks very much in advance.
thank u........for your brief explanation.........Geva Stern wrote:Guys, the two "methods" related above both lead to the same result:me10gmat800 wrote:xcusemeplz2009 wrote:4!/4^4=3/32 (pls check again)=9% approx.PussInBoots wrote:I disagree
Total combinations: 4*4*4*4
Total combinations of different numbers: 4*3*2*1
Answer is 4! / 4^4 = 1/8, hence ~12%
i do agree with this method
Have one doubt so allow me to reopen this discussion...
Is this one of the standard probability formulas/rules (4!/4^4 generalized form -> n!/n^n) that could be used in such type of questions?
I solved this q in similar fashion that automatically led to this rule/formula:
1st person able to choose in 4/4 ways, 2nd person 3/4, 3rd 2/4 and 4th 1/4.
Therefore the probability is 4/4 * 3/4 * 2/4 * 1/4 = 4! / 4^4 (rule)
Could someone explain why and when this rule could be used?
Thanks very much in advance.
1 * (3/4) * (2/4) * (1/4) = 3/32
rewrite the 1 as 4/4 (which is what FinanceBioE effectively did), and you get the famed
4! / 4^4 = 3/32.
the only difference is a matter of interpretation of the results: is 3/32 equal to 9% or 12%?
the answer is indeed 9%, but i can understand the confusion of those who went for 12: 4/32 is 1/8=12.5%, which is almost (but not quite) the same as the actual result of 3/32. The way to transition from 3/32 to percents is to remember the following rule:
to move from
Percent ---> fraction you divide by 100 Thus, 12% is equivalent to the fraction 12/100.
fraction ---> percent you do the opposite: multiply by 100. Thus, 3/32 is 3*100/32 "percents" = 300/32 %= slightly less than 10%, or 9% final.
As for whether this question is a standard probability formula: there are too many of them (and too many combi/probability questions that break them slightly) for you to rely on definite formulas, IMHO - that's what makes this topic so formidable, as each question has its own quirks. So, yes, any question exactly such as this one will conform to the n!/n^n formula, but there's really no substitute to reasoning through the question and building the relevant calculation for each individual case, i.e.: number of wanted outcomes = reduce one each choice = 4*3*2*1=4!
Total number of outcomes = 4 for each choice.
OR
probability for each event, as with the original reply (my own favored method here).
You got it. The idea is to break down the problem into a series of rolls, one step at a time: Not three dice, but first, second, third roll, and deal with each one separetly.dazzle wrote:Hi All,
I have a very basic question here. I am extremely poor with probability and combinations.
Can any provide a short explanation as to how the total no of combinations is n^n ?
i have seen a few problems " when 3 dice are rolled the total no of combinations are 6^3.
similarly in this case, it is 4^4.
Can someone pls explain why?
is the general rule something like , (possible combinations to be played with)^ (no of times the activity is done)??
ie In the 3 dice case, since each die has 6 combinations and it is rolled 3 times so it should be 6^3
Similarly in this problem since there are 4 nos (hence 4 combinations to be played with) ^ (no of times/ppl) ?
Appreciate any help in this regard.
Thanks
Indeed - to the letter.dazzle wrote:Thanks a bunch for explaining the concept Geva.
"What's the probability of getting "heads" on all three tosses of an even coin?"
For the first toss, we can have 2 outcomes. ie head or tail.
hence as per your explanation, we will have 2^3 = 8 total outcomes.
The possibility of HHH will therefore be 1/8
Hope my answer is correct .
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