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Re: TOUGH & TRICKY SET Of PROBLEMS
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15 Jan 2011, 14:13
yogesh1984 wrote: Bunuel wrote: 4. ADDITION PROBLEM: AB + CD = AAA, where AB and CD are twodigit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?
(A) 1 (B) 3 (C) 7 (D) 9 (E) Cannot be determined
AB and CD are two digit integers, their sum can give us only one three digit integer of a kind of AAA it's 111. So, A=1. 1B+CD=111 C can not be less than 9, because no to digit integer with first digit 1 (mean that it's<20) can be added to two digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111). C=9
Answer: D. Bunuel I have doubt here consider following AB as 25 CD 86 which sums to 111 and ly if we consider AB as 15 & CD as 96 then also we end up with 111 so I am doubtfull with the answer provided above....(please do correct if Am missing something ) SOLUTION: AB and CD are two digit integers, their sum can give us only one 3igit integer of a kind of AAA: 111. So, A=1 > 1B+CD=111. C can not be less than 9, because no 2digit integer with first digit 1 (mean that it's <20) can be added to 2digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111) > C=9. Answer: D. So, as A=1 then AB can not be 25. Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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20 Feb 2011, 17:24
Bunuel wrote: SOLUTION: 2. THE PRICE OF BUSHEL: The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of \(5x\) cents per day while the price of wheat is decreasing at a constant rate of \(\sqrt{2}*xx\) cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?
(A) $4.50 (B) $5.10 (C) $5.30 (D) $5.50 (E) $5.60
Note that we are not asked in how many days prices will cost the same.
Let \(y\) be the # of days when these two bushels will have the same price.
First let's simplify the formula given for the rate of decrease of the price of wheat: \(\sqrt{2}*xx=1.41xx=0.41x\), this means that the price of wheat decreases by \(0.41x\) cents per day, in \(y\) days it'll decrease by \(0.41xy\) cents;
As price of corn increases \(5x\) cents per day, in \(y\) days it'll will increase by \(5xy\) cents;
Set the equation: \(320+5xy=5800.41xy\), solve for \(xy\) > \(xy=48\);
The cost of a bushel of corn in \(y\) days (the # of days when these two bushels will have the same price) will be \(320+5xy=320+5*48=560\) or $5.6.
Answer: E. Bunuel, While solving for xy i get xy = 56 5xy  0.41xy = 580320 4.59xy = 260 xy = 56.6 cents Let me know if i missed something!



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Re: TOUGH & TRICKY SET Of PROBLEMS
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20 Feb 2011, 17:38
mbafall2011 wrote: Bunuel wrote: SOLUTION: 2. THE PRICE OF BUSHEL: The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of \(5x\) cents per day while the price of wheat is decreasing at a constant rate of \(\sqrt{2}*xx\) cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?
(A) $4.50 (B) $5.10 (C) $5.30 (D) $5.50 (E) $5.60
Note that we are not asked in how many days prices will cost the same.
Let \(y\) be the # of days when these two bushels will have the same price.
First let's simplify the formula given for the rate of decrease of the price of wheat: \(\sqrt{2}*xx=1.41xx=0.41x\), this means that the price of wheat decreases by \(0.41x\) cents per day, in \(y\) days it'll decrease by \(0.41xy\) cents;
As price of corn increases \(5x\) cents per day, in \(y\) days it'll will increase by \(5xy\) cents;
Set the equation: \(320+5xy=5800.41xy\), solve for \(xy\) > \(xy=48\);
The cost of a bushel of corn in \(y\) days (the # of days when these two bushels will have the same price) will be \(320+5xy=320+5*48=560\) or $5.6.
Answer: E. Bunuel, While solving for xy i get xy = 56 5xy  0.41xy = 5803204.59xy = 260 xy = 56.6 cents Let me know if i missed something! From 320+5xy=5800.41xy > 5xy +0.41xy=580320 > xy=48.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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05 Sep 2011, 06:26
fluke wrote: The following answers may not be correct. If you find some discrepancy in the solution, please PM me or post it under the same topic. I took a cursory look into the OA's but am yet to match all my answers and compare the logic. Bunuel wrote: 5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a
minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20
Let A's speed: A m/s Let B's speed: B m/s 1/10th of a minute = 6sec 1/30th of a minute = 2sec Total distance of the race track = 480m First heat; A's distance = 480m A's speed = A m/s A's time = t secs B's distance = 48048=432m B's speed = B m/s B's time = t+6 secs A's time = B's time  6 (if A took 1000 seconds; B took 1006 seconds) \(\frac{480}{A}=\frac{432}{B}6\)  1st (\(rate*time=distance \quad or \quad time,t=D/r\)) Second heat; A's distance = 480m A's speed = A m/s A's time = t secs B's distance = 480144=336m B's speed = B m/s B's time = t2 secs A's time = B's time + 2 \(\frac{480}{A}=\frac{336}{B}+2\)  2nd Using 1st and 2nd, \(\frac{432}{B}6=\frac{336}{B}+2\) \(\frac{432}{B}\frac{336}{B}=8\) \(\frac{96}{B}=8\) \(B=12\) B's speed : 12m/s Ans: "A" I dont have any Question on calculations made above. But i do have a clarification to ask. While calculating the time of B , it appears that we have not accounted the time taken by B to cover the first 48m(first heat) or 144m (second heat). ( As if the start time taken into consideration is the point from when both A and B are in action ). Why is that we are going by this. I mean why are we not considering the head start time provided to B. Please correct me if i missed anything.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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05 Sep 2011, 10:55
vinodmallapu wrote: fluke wrote: The following answers may not be correct. If you find some discrepancy in the solution, please PM me or post it under the same topic. I took a cursory look into the OA's but am yet to match all my answers and compare the logic. Bunuel wrote: 5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a
minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20
Let A's speed: A m/s Let B's speed: B m/s 1/10th of a minute = 6sec 1/30th of a minute = 2sec Total distance of the race track = 480m First heat; A's distance = 480m A's speed = A m/s A's time = t secs B's distance = 48048=432m B's speed = B m/s B's time = t+6 secs A's time = B's time  6 (if A took 1000 seconds; B took 1006 seconds) \(\frac{480}{A}=\frac{432}{B}6\)  1st (\(rate*time=distance \quad or \quad time,t=D/r\)) Second heat; A's distance = 480m A's speed = A m/s A's time = t secs B's distance = 480144=336m B's speed = B m/s B's time = t2 secs A's time = B's time + 2 \(\frac{480}{A}=\frac{336}{B}+2\)  2nd Using 1st and 2nd, \(\frac{432}{B}6=\frac{336}{B}+2\) \(\frac{432}{B}\frac{336}{B}=8\) \(\frac{96}{B}=8\) \(B=12\) B's speed : 12m/s Ans: "A" I dont have any Question on calculations made above. But i do have a clarification to ask. While calculating the time of B , it appears that we have not accounted the time taken by B to cover the first 48m(first heat) or 144m (second heat). ( As if the start time taken into consideration is the point from when both A and B are in action ). Why is that we are going by this. I mean why are we not considering the head start time provided to B. Please correct me if i missed anything. When A gives B a head start of 48 m, it means B starts from 48 m ahead of the start line while A starts from the start line. They both start the race at the same time and run till A reaches the finish line. Thereafter, only B runs for 1/10th of a minute. There is no head start time provided to B, only head start distance.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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05 Sep 2011, 11:04
Bunuel wrote: 5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a
minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20
Alternative Approach here: A starts from the start line and runs till the finish line in both the races. He would take the same time in both the cases then. On the other hand, in the first race, B takes 6 secs more than A while in the second race, B takes 2 secs less than A. So there is a time difference of 8 secs in the time taken by B in the two cases. B travels (144  48 =) 96 m less in the second race and taken 8 secs less in the second race. This means that in the first race, B runs a distance of 96m in 8 secs. So speed of B = 96/8 = 12 m/sec
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Re: TOUGH & TRICKY SET Of PROBLEMS
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27 Feb 2012, 09:11
flokki wrote: Bunuel wrote: SOLUTION OF 810 8. THE AVERAGE TEMPERATURE: The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?
A. 20 B. 25 C. 40 D. 45 E. 75
Average=50, Sum of temperatures=50*5=250 As the min temperature is 45, max would be 2504*45=70 > The range=70(max)45(min)=25
Answer: B.
Isnt the correct answer A. it says 45 is the lowest temperature. Doest that statement imply that all other numbers have to be greater than 45. Hence at least 45,00001?No, it doesn't. There can be more than 1 temperatures equal to 45. COMPLETE SOLUTION: Given: T(min)=45 and average=50 > sum of temperatures=50*5=250. We want to maximize the range > in order to maximize the range we need to maximize the highest temperature > in order to maximize the highest temperature we should make all other temperatures as low as possible, so equal to 45 (lower limit) > T(max)=2504*45=70 > Range=T(max)T(min)=7045=25. Answer: B. Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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27 Mar 2012, 07:11
Hi Bunuel, I understand the below solution. However, I was wondering if it could be solved using combination approach. Please tell me where I am going wrong. Probability = favorable/ total outcomes. Total outcomes = 9C2 (i.e total ways if drawing 2 balls) Favorable = 3C2*4C1 Hence Probability = 3C2*4C1/9C2 = 1/6. which is completely wrong. Please reply. Thanks H Quote: 6. PROBABILITY OF DRAWING: A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn? (A) 2/27 (B) 1/9 (C) 1/3 (D) 4/27 (E) 2/9
This is with replacement case (and was solved incorrectly by some of you):
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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27 Mar 2012, 09:08
imhimanshu wrote: Hi Bunuel, I understand the below solution. However, I was wondering if it could be solved using combination approach. Please tell me where I am going wrong. Probability = favorable/ total outcomes. Total outcomes = 9C2 (i.e total ways if drawing 2 balls) Favorable = 3C2*4C1 Hence Probability = 3C2*4C1/9C2 = 1/6. which is completely wrong. Please reply. Thanks H Quote: 6. PROBABILITY OF DRAWING: A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn? (A) 2/27 (B) 1/9 (C) 1/3 (D) 4/27 (E) 2/9
This is with replacement case (and was solved incorrectly by some of you):
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D. Since we have the case with replacement then we can not use \(C^2_9\) for it, because it gives total # of ways to select 2 different balls out of 9 without replacement. If you wan to solve this question with combinations approach you still should consider two scenarios: \(P(RW)=2*\frac{C^1_3*C^1_2}{C^1_9*C^1_9}=\frac{4}{27}\), we are multiplying by 2 for the same reason: there are two possible wining scenarios RW and WR.. Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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03 Apr 2012, 10:04
theamwan wrote: Bunuel wrote: SOLUTION: 3. LEAP YEAR: How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?
A. 1 B. 2 C. 3 D. 4 E. 5
Probability of a randomly selected person NOT to be born in a leap year=3/4 Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1 9/16=7/16<1/2 So, we are looking for such n (# of people), when 1(3/4)^n>1/2 n=3 > 127/64=37/64>1/2
Thus min 3 people are needed.
Answer: C. Hi Bunuel  Regarding this question, I believe the answer should be 2 i.e. (B). It is because the probability of a person's birth year to be a leap yr = 1/2, this is identical to a rainy/nonrainy day situation, where a year is either a leap or normal year. Case I  No of people = 1 Probability of atleast 1 person's birth yr falling on a leap year = 1/2 = 50% .....INCORRECT Case II  No of people = 2 Probability of none of guys' birth year falling on a leap year = 1/2*1/2 = 1/4 Probability of atleast one of them was born on a leap year = 1  1/4 = 3/4 = 75% .....CORRECT Hence (B). What do you think? Thanks! That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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03 Apr 2012, 10:17
Bunuel wrote: That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.
But the question doesn't say that it's a set of 4 consecutive years. What if the years chosen are 2011, 2010, 2009, 2007? IMO, a person's birth year to fall on a leap year should be a binary value  Y or N, hence a 50% probability.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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03 Apr 2012, 10:32
theamwan wrote: Bunuel wrote: That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.
But the question doesn't say that it's a set of 4 consecutive years. What if the years chosen are 2011, 2010, 2009, 2007? IMO, a person's birth year to fall on a leap year should be a binary value  Y or N, hence a 50% probability. Again that's not correct. Let me ask you a question: what is the probability that a person in born on Monday? Is it 1/2? No, it's 1/7.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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29 Jun 2012, 20:10
Bunuel wrote: SOLUTION: 6. PROBABILITY OF DRAWING: A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn? (A) 2/27 (B) 1/9 (C) 1/3 (D) 4/27 (E) 2/9
This is with replacement case (and was solved incorrectly by some of you):
\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D. The question says that red and white balls are selected in two Successive draws. Doesn't this imply that white is selected AFTER red? Thus no need for x2? Thanks, Diana



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Re: TOUGH & TRICKY SET Of PROBLEMS
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30 Jun 2012, 04:17
dianamao wrote: Bunuel wrote: SOLUTION: 6. PROBABILITY OF DRAWING: A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn? (A) 2/27 (B) 1/9 (C) 1/3 (D) 4/27 (E) 2/9
This is with replacement case (and was solved incorrectly by some of you):
\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D. The question says that red and white balls are selected in two Successive draws. Doesn't this imply that white is selected AFTER red? Thus no need for x2? Thanks, Diana No, in that case we would be asked "what is the the probability of the first ball being red and the second ball being white?"
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Re: TOUGH & TRICKY SET Of PROBLEMS
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16 Dec 2012, 22:30
Bunuel wrote: SOLUTION: 6. PROBABILITY OF DRAWING: A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn? (A) 2/27 (B) 1/9 (C) 1/3 (D) 4/27 (E) 2/9
This is with replacement case (and was solved incorrectly by some of you):
\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D. Hi Bunuel I guess the question should be reworded to restrict the number of draws to 2.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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16 Dec 2012, 23:08
geneticsgene wrote: Bunuel wrote: SOLUTION: 6. PROBABILITY OF DRAWING: A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn? (A) 2/27 (B) 1/9 (C) 1/3 (D) 4/27 (E) 2/9
This is with replacement case (and was solved incorrectly by some of you):
\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D. Hi Bunuel I guess the question should be reworded to restrict the number of draws to 2. I think we are given that: "What is the probability of drawing a red and a white ball in two successive draws..."
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Re: TOUGH & TRICKY SET Of PROBLEMS
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20 Dec 2012, 16:26
I think i made mistake since 6 can be also a probable number but is not a multiple of 4..............Bunuel can u pls help me solve this one using multiple principle



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Re: TOUGH & TRICKY SET Of PROBLEMS
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21 Dec 2012, 04:40
Archit143 wrote: I think i made mistake since 6 can be also a probable number but is not a multiple of 4..............Bunuel can u pls help me solve this one using multiple principle If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?A. 25% B. 50% C. 62.5% D. 72.5% E. 75% \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; (Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.) Now, in EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8=0.625. Answer: C. Similar question: divisibleby12probability121561.htmlHope it helps.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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25 Jan 2013, 01:22
Bunuel wrote: angel2009 wrote: Bunuel wrote: SOLUTION: 6. PROBABILITY OF DRAWING: A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn? (A) 2/27 (B) 1/9 (C) 1/3 (D) 4/27 (E) 2/9
This is with replacement case (and was solved incorrectly by some of you):
P=2*3/9*2/9=4/27
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D. I've some problem to understand this ; as the two events are mutually exclusive , independent. So why not the answer is 3/9*2/9 = 2/27 Winning scenario consists of TWO cases RW and WR. Probability of each case is: 3/9*2/9, so 3/9*2/9+3/9*2/9=2*3/9*2/9. Only 3/9*2/9 would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters. Hey can you please help with the solution using combination formula P(1R1W) = 3C1 x 2C1 / (9C2) = 1/6 WHY? what am i doing wrong?




Re: TOUGH & TRICKY SET Of PROBLEMS &nbs
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