Hi again,

So 4 people have the option of choosing 1,2,3 or 4

Therefore the first person can choose any number and the probability of that is 1

The second person can only choose 3 of the 4 numbers and the probability of that is 3/4

The third person can only choose 2 of the 4 numbers and the probability of that is 2/4

The final person can only choose 1 of the 4 numbers and the probability of that is 1/4

Therefore

1 * (3/4) * (2/4) * (1/4) = 3/32

~9%

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%

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%

4!/4^4=3/32 (pls check again)=9% approx.

i do agree with this method

Please, what's the answer? is it 9% correct?
Silvia

Could anyone please clarify what is the correct solution for this problem.

Sanjib please clarfiy the confusion.

What is the correct answer

xcusemeplz2009 wrote:

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%

4!/4^4=3/32 (pls check again)=9% approx.

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.

me10gmat800 wrote:

xcusemeplz2009 wrote:

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%

4!/4^4=3/32 (pls check again)=9% approx.

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.

Guys, the two "methods" related above both lead to the same result:
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).

Geva Stern wrote:

me10gmat800 wrote:

xcusemeplz2009 wrote:

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%

4!/4^4=3/32 (pls check again)=9% approx.

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.

Guys, the two "methods" related above both lead to the same result:
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).

thank u........for your brief explanation.........

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

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

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.
For each roll, figure out the number of possible outcomes. On a die, these are 1-6, so there are 6 possible options. So each of the rolls (1st, 2nd 3rd) has 6 options. The final outcome is 6*6*6, or 6^3.
Why do we multiply the number of each roll? For one, you can just take it as a given - in a string of events, multiply the numbers at the end.
The logic behind it is this: the first die roll has 6 options: 1-6
For each of these options, there are 6 options for the second roll:
If we rolled 1 on the first die, we have 6 options for the second: 1-1, 1-2, 1-3, 1-5, 1-6
If we rolled a 2 on the first die, again, 6 rolls: 2-1, 2-2, 2-3...etc.
Thus, 6 options for the first die, and 6 options for each of them on the second die come down to 6*6=36 options.
The third die presents another 6 options for each of these 6*6 combinations of the first 2, so again, we multiply times 6 again: 6*6*6=6^3.

Let's say that we want the probability of three different rolls on 6 dice. You already have the bottom of the fraction - the total number of possible combinations: 6^3. How many wanted outcomes?
Here, we again break down to one step at a time, and ask how many options are wanted outcomes for each roll:
For the first die, we don't really care what number comes up, so all 6 options are wanted outcomes.
For the second roll, we want a different number than the first: whatever we rolled on the first, we'd only have 5 options for a different number on the second.
For the third roll, we want a different number from the first two: only 4 remain.
Thus, the total number of wanted outcomes for different results on three dice is 6*5*4.
The total probability of this even will therefore be 6*5*4 / 6*6*6 = 20/36 = 5/9.

What's the probability of getting "heads" on all three tosses of an even coin?

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 .

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 .

Indeed - to the letter.
Let's mix things up a bit. What's the probability of getting two "heads" on three tosses? Anyone?