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Re: Repeated Outcomes
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04 Aug 2006, 16:40
CaptainZeefor700 wrote: This question drives me mad  I understand why the total number of outcomes is 81 and why the total number of outcomes when each secretary gets at least one report is 36, but when I am trying to use logic instead of formula, the former seems to be giving up. Please let me know where I am wrong.
So, my understanding is that the total number of outcomes when 4 reports (A,B,C,D) are assigned randomly among 3 secretaries (1st, 2nd, 3rd) is as follows:
1) all 4 reports are assigned to only one secretary  3 outcomes, plus
2) 3 reports out of four (4C3) are assigned to one secretary (*3) and the fourth is assigned to either secretary (*2) => 4C3*3*2 = 4*3*2=24 outcomes, plus
3)  two reports out of four (4C2) are assigned to a secretary (*3) and the other two are assigned like this: both to one secretary, both to the other secretary, one report per each secretary and vice versa, i.e. 4 options > (*4). To make the last one clearer, let's say when reports AB go to 1st sec., there are 4 outcomes: first â€“ reports CD both go to 2nd sec.; second  CD go both to 3rd sec.; third  report C goes to 2nd sec. and D to 3rd sec.; and the fourth outcome  C goes to 3rd sec. and D goes to 2nd sec. So, 4C2*3*4 = 6*3*4 = 72
Thus, we have 3+24+72 = way out of 81. I understand I have repetitive counts, but can't figure out where.
Thanks in advance.
Your problem lies with 3), I think. You definitely have some double counting problem there. For example, you would count "reports AB go to 1st and then CD go to 2nd" as one outcome, and "report CD go to 2nd and then AB go to 1st" as another outcome, although they are really the same thing. Let me know if I've confused you more.
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Re: Letters and Words
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04 Aug 2006, 16:54
CaptainZeefor700 wrote: HongHu wrote: Letters and Words
Vowels: A, U, A, I, A. Consonants: S, T, R, L
Example: From the word AUSTRALIA, a letter is taken at random and put back. Then a second letter is taken and put back. This process is repeated for four letters. What would be the possibility that two vowels and two consonants have been chosen?
This is a case of "with replacement": Total outcomes: 9^4 Outcomes with two vowels and two consonants: C(4,2)*5^2*4^2 You can calculate the probability from here.
Shouldn't this be just 5/9*5/9*4/9*4/9?
That's a great point. Remember you should still multiply it by C(4,2), since you don't know where the two vowels is going to be placed.
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I disagree with honghu solution
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10 Aug 2006, 14:52
Example:
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C Total outcome: 3^4=81 Total outcome that three secretary are assigned at least one report: We need to choose one secretary C(3,1) to be assigned of two reports C(4,2) and then the rest two secretary each to be assigned of one report P(2,2). C(3,1)*C(4,2)*P(2,2)=36 Probability = 36/81=4/9
This is my thinking how the solution should be:
For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C
Total outcome: 3^4=81
Total outcome that three secretary are assigned at least one report:
Chose 1 sectretary and chose 2 reports = C(3,1)*C(4*2)
then, chose 1 secretary(out of two) and 1 report(out of two) = C(2,1)*C(2,1)
then, chose 1 secretary(out of one) and 1 report(out of one) = C(1,1)*C(1,1)
net combinations where each secretary has atleast 1 report is multiple of all three above = C(3,1)*C(4*2)*C(2,1)*C(2,1)*C(1,1)*C(1,1)
3*6*2*2*1*1 = 72
So the probability is 72/81.
I would appreciate if HongHu or someone else can point out if there is any error in my solution and why.



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This is a common mistake. You have double counted your outcomes. Say if you have two people and two reports and you want to assign one report each person. According to your solution, you would first choose one person and one report, then you would assign the second report to the second person. Therefore your total outcome would be C(2,1)*C(2,1)*C(1,1)*C(1,1). However, the correct answer should be C(2,1).
With your solution, you would count A,1 and B,2 as one outcome, and B,2 and A,1 as another outcome. However, you can clearly see that these two outcomes are really one outcome.
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Re: I disagree with honghu solution
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11 Aug 2006, 16:33
AlHindi wrote: Example:
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C Total outcome: 3^4=81 Total outcome that three secretary are assigned at least one report: We need to choose one secretary C(3,1) to be assigned of two reports C(4,2) and then the rest two secretary each to be assigned of one report P(2,2). C(3,1)*C(4,2)*P(2,2)=36 Probability = 36/81=4/9
This is my thinking how the solution should be: For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C Total outcome: 3^4=81 Total outcome that three secretary are assigned at least one report: Chose 1 sectretary and chose 2 reports = C(3,1)*C(4*2) then, chose 1 secretary(out of two) and 1 report(out of two) = C(2,1)*C(2,1) then, chose 1 secretary(out of one) and 1 report(out of one) = C(1,1)*C(1,1)
net combinations where each secretary has atleast 1 report is multiple of all three above = C(3,1)*C(4*2)*C(2,1)*C(2,1)*C(1,1)*C(1,1) 3*6*2*2*1*1 = 72 So the probability is 72/81.
I would appreciate if HongHu or someone else can point out if there is any error in my solution and why.
My understanding is:
C(4,2) reports go to each of the 3 secretary (i.e 6*3 = 18)
Each of those 18 has four possible outcomes: the rest two reports go to only one secretary; the two reports go to another secretary; both secretaries have one report each; and then they swap reports. Out of these four outcomes only the last two are favorable, thus 18*2 = 36.
However, the same logic doesn't seem to be working when I try to get the total number of all outcomes... I tried to figure that out HongHu, but couldn't..



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How many of the questions discussed above are GMAT type of questions? To me, few seem to be. It would be better if we stuck to typical GMAT probability questions, no?



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Probability
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30 Aug 2006, 07:27
Hello Everyone,
Juts trying to improve my Probability knowledge. I have difficulties understanding the principle behind the following 2 questions:
Question1:
There are 3 green; 4 yellow; and 5 blue balls in a sack. 6 balls are taken at random without replacement. Find the probability to have 1 green; 2 yellow; and 3 blue balls.
To get this I use the (favourable outcome)/(total outcome) Principle with the help of the combination formula:
[1C3 (green)Ã— 2C4 (yellow)Ã—3C5 (blue)]/12C6
I want to use the same principle in the following question, but it does not to work. Anyone knows why?
There are 9 beads in a bag. 3 beads are red, 3 beads are blue, and 3 beads are black. If two beads are chosen at random, what is the probability that they are both blue?
My way of thought: (favourable outcome)/(total outcome) (3C2)/(9C6)
For the last question the formula to use is (3A2)/(9A6), but why?
Many thanks



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Joined: 12 Sep 2006
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Thanks for all the great posts so far!!!
I am pretty new to the forum but finding all the input very useful. Just came across a couple problems i couldn't solve. Any help is appreciated!!!
PS
1. From 6 positive numbers and 6 negative numbers, how many groups of 4 numbers, yielding a positive product, can be formed?
A. 30
B. 255
C. 625
D. 720
E. 960
DS
2. When flipping a coin, the probability that it lands heads up is not equal to the probability that it lands tails up. If the coin is flipped twice, what is the probability that it lands heads up just once?
1. For any flipping, the probability that the coin lands heads up is twice the probability that it lands tails up.
2. For any double flipping, the probability that the coin lands heads up on the first flip and tails up on the second flip is 2/9



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Combination Permutation
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09 Oct 2006, 03:50
I know that this area is my weak area in GMAT. Can anybody tell me where I can download or buy GMAT permutation & combinations questions?



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Combinations
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09 Oct 2006, 04:19
After research I have found theoretical explanation of the combinations concept I found in the net.



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permutations theoretical concept (sorry first time the file hasn't attached)



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How many different 6letter arrangements ca nbe formed from the word THIRST if T must always immediately precede H?



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permutations
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20 Oct 2006, 10:15
my answer is 6. Although I haven't used formulas. I base my decision on common sense.



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Rayn wrote: How many different 6letter arrangements ca nbe formed from the word THIRST if T must always immediately precede H?
There are two Ts and one H here. Which T needs to immediately precede H? Or does it matter?
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keeeeeekse wrote: Thanks for all the great posts so far!!!
I am pretty new to the forum but finding all the input very useful. Just came across a couple problems i couldn't solve. Any help is appreciated!!!
PS 1. From 6 positive numbers and 6 negative numbers, how many groups of 4 numbers, yielding a positive product, can be formed?
A. 30 B. 255 C. 625 D. 720 E. 960
Step 1: {Positive product} = {choose 0, 2, 4 positive numbers} ={0 positive} U {2 positive} U {4 positive} and they are disjoint Step 2: = 6C0*6C4 + 6C2*6C2 + 6C4*6C0 = 15 + 225 + 15 = answer B (and to make it a probability problem we would divide this by 12C4) keeeeeekse wrote: DS 2. When flipping a coin, the probability that it lands heads up is not equal to the probability that it lands tails up. If the coin is flipped twice, what is the probability that it lands heads up just once?
1. For any flipping, the probability that the coin lands heads up is twice the probability that it lands tails up.
Step 1: P(H) = 2*P(T) and P(H) + P(T) = 1 which implies P(H) = 2/3 Step 2: (Binomial) Let X= # heads. P(X=1) = 2C1*(2/3)^1*(1/3)^1 = 4/9 keeeeeekse wrote: 2. For any double flipping, the probability that the coin lands heads up on the first flip and tails up on the second flip is 2/9
Cute.. This means P(H)*P(T) = 2/9 but you can't get the P(H). Fortunately in the binomial formula above, all you need is the P(H)*P(T) which means the answer is the same = 4/9



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HongHu wrote: Rayn wrote: How many different 6letter arrangements ca nbe formed from the word THIRST if T must always immediately precede H? There are two Ts and one H here. Which T needs to immediately precede H? Or does it matter?
How would you differentiate one T from the other?
Count (A U B) = Count(A) + Count(B)  Count(A n B) [n= intersect]
# arrangements = # arrangement with first T preceding H + # arrangements with second T preceding H  # words that are the same
so treat the TH as 1 letter
= 5! + 5!  #words that are the same
so treat TTH as 1 letter
= 5! + 5!  4!
= 120 + 120  24 = 216



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Re: Probability
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01 Nov 2006, 07:02
fmeinsen wrote: Hello Everyone,
Juts trying to improve my Probability knowledge. I have difficulties understanding the principle behind the following 2 questions:
Question1: There are 3 green; 4 yellow; and 5 blue balls in a sack. 6 balls are taken at random without replacement. Find the probability to have 1 green; 2 yellow; and 3 blue balls.
To get this I use the (favourable outcome)/(total outcome) Principle with the help of the combination formula: [1C3 (green)Ã— 2C4 (yellow)Ã—3C5 (blue)]/12C6
I want to use the same principle in the following question, but it does not to work. Anyone knows why?
There are 9 beads in a bag. 3 beads are red, 3 beads are blue, and 3 beads are black. If two beads are chosen at random, what is the probability that they are both blue?
My way of thought: (favourable outcome)/(total outcome) (3C2)/(9C6) For the last question the formula to use is (3A2)/(9A6), but why?
Many thanks
Usually people would write, e.g., 3C1 for "3 choose 1" instead of 1C3 but maybe this is some English thing.
These problems do work the same way. The denominator in the second problem is 9C2 because you are choosing only 2 beads. Thus, the answer is 3C2*3C0*3C0/9C2 = 3/36 = 1/12



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Re: I disagree with honghu solution
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01 Nov 2006, 07:15
CaptainZeefor700 wrote: AlHindi wrote: Example:
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C Total outcome: 3^4=81 Total outcome that three secretary are assigned at least one report: We need to choose one secretary C(3,1) to be assigned of two reports C(4,2) and then the rest two secretary each to be assigned of one report P(2,2). C(3,1)*C(4,2)*P(2,2)=36 Probability = 36/81=4/9
This is my thinking how the solution should be: For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C Total outcome: 3^4=81 Total outcome that three secretary are assigned at least one report: Chose 1 sectretary and chose 2 reports = C(3,1)*C(4*2) then, chose 1 secretary(out of two) and 1 report(out of two) = C(2,1)*C(2,1) then, chose 1 secretary(out of one) and 1 report(out of one) = C(1,1)*C(1,1)
net combinations where each secretary has atleast 1 report is multiple of all three above = C(3,1)*C(4*2)*C(2,1)*C(2,1)*C(1,1)*C(1,1) 3*6*2*2*1*1 = 72 So the probability is 72/81.
I would appreciate if HongHu or someone else can point out if there is any error in my solution and why. My understanding is: C(4,2) reports go to each of the 3 secretary (i.e 6*3 = 18) Each of those 18 has four possible outcomes: the rest two reports go to only one secretary; the two reports go to another secretary; both secretaries have one report each; and then they swap reports. Out of these four outcomes only the last two are favorable, thus 18*2 = 36. However, the same logic doesn't seem to be working when I try to get the total number of all outcomes... I tried to figure that out HongHu, but couldn't..
Wow, you guys are really messed up on this problem. So everyone is comfortable with the binomial distribution. If the problem said there are two secretaries and 4 reports what is the probability each of the secretaries got 2 reports, everyone would immediately write down 4C2*(1/2)^2*(1/2)^2 = 6/16 = 3/8. So now we have a trinomial problem (and more generally a multinomial problem) that we just do the same way except that thing in the front is no longer a simple combination, it is
4!/(1!*1!*2!). So the answer is
4!/(1!*1!*2!)*(1/3)^1*(1/3)^1*(1/3)^2



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Re: Binomial Distribution
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04 Dec 2006, 00:53
HongHu wrote: Binomial Distribution
In each individual test, the probability of A happening is p and not happening is 1p. What is the probability of A happening exactly k times in n repeated tests?
Formula: C(n,k) * p^k * (1p) ^ nk
Example: When a coin has tossed, it will show a head at 50% probability and a tail at 50% probability. What is the probability of getting two heads among four tosses? C(4, 2) * (1/2) ^2 * (1/2) ^2 = 6/16 = 3/8
Example: The probability of raining is 0.3 and not raining is 0.7. What is the probability of getting three days of rains among seven days? C(7, 3)*0.3^3*0.7^4
Example: The probability of a new born baby is a boy is 50% and that of a new born baby is a girl is 50%. Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?
Probability that at least two babies are boys = Probability that two babies are boys + probability that three babies are boys + ... + Probability that ten babies are boys (also) = 1 Probability that non are boys  probability that only one is a boy Choose the easier route P=1C(10,0)*0.5^0*0.5^10C(10,1)*0.5^1*0.5^9 =10.5^1010*0.5^10 =111*0.5^10
May be a silly question and it's early!! But what do you mean by C(4, 2). Could you list out the calc?



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Updated on: 13 Dec 2006, 16:09
PS: To yield a (+) product, you just need one even number in any group of four; that is, you need these combos:
1 even, 3 odd
2 even, 2 odd
3 even, 1 odd
4 even, 0 odd
You can notice that all these combos amount to the total 
0 even, 4 odd = 6C0*6C4
The total number of combos is 12C4. Therefore, the number of combos of even products of 4 numbers is
12C4  6C0*6C4
DS: We have p(H) <> p(T). For 2 flips and 1 head up,
prob = 2C1*pH^1*pT^1. = 2*pH*pT.
From 1: pH = 2*pT; pH + pT = 1 => pH = 2/3, pT = 1/3.
From 2: pH*pT = 2/9.
We see that from either 1 or 2 we can solve the problem, then the answer is D.
keeeeeekse wrote: Thanks for all the great posts so far!!!
I am pretty new to the forum but finding all the input very useful. Just came across a couple problems i couldn't solve. Any help is appreciated!!!
PS 1. From 6 positive numbers and 6 negative numbers, how many groups of 4 numbers, yielding a positive product, can be formed?
A. 30 B. 255 C. 625 D. 720 E. 960
DS 2. When flipping a coin, the probability that it lands heads up is not equal to the probability that it lands tails up. If the coin is flipped twice, what is the probability that it lands heads up just once?
1. For any flipping, the probability that the coin lands heads up is twice the probability that it lands tails up.
2. For any double flipping, the probability that the coin lands heads up on the first flip and tails up on the second flip is 2/9
Originally posted by Andr359 on 13 Dec 2006, 15:59.
Last edited by Andr359 on 13 Dec 2006, 16:09, edited 1 time in total.







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