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Re: New Set of Mixed Questions!!!
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17 Jan 2014, 04:57
Bunuel wrote: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8
Nice question Bunuel!!!! +1 for it. Here is my take. We have 5 red marbles and 3 red marbles. We just want that in the 7th pick blue marble should not appear. That indirectly means Red marble should appear. We can visualize it as arrangement of 8 marbles (of which 5 are red and 3 are blue) in such a way that 7th marble will be red. We can fix 1 red marble at 7th place and can arrange remaining marbles (Total 7 = 4 Red + 3 Blue) as \(\frac{7!}{4!3!}\) = This is our desired outcome \(Probability = \frac{Desired Outcomes}{Total Outcomes}\) Desired Outcomes = \(\frac{7!}{4!3!}\) = 5 Total outcomes = Arrangement of all 8 marbles ( 8 = 5 Red + 3 Blue) at 8 places = \(\frac{8!}{5!3!}\) = 8 Probability = \(\frac{5}{8}\) = D
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Re: New Set of Mixed Questions!!!
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01 Jun 2014, 01:29
Bunuel wrote: 9. If x and y are integers and x + y = 12, which of the following must be true? A. Both x and y are negative B. xy > 0 C. If y < 0, then x > 0 D. If y > 0, then x < 0 E. x  y > 0
Look at option D: if y is positive, then x must be negative in order the sum of x and y to be negative.
Answer: D. I think C & D are right. If x is +ve i.e. 12 and y is ve i.e. 24 then sum of x and y will be negative. Please explain Neha



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Re: New Set of Mixed Questions!!!
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Re: New Set of Mixed Questions!!!
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30 Oct 2014, 20:30
Bunuel wrote: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8
Basically we need to find the probability that the seventh marble drawn is red (so not blue).
Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8.
The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Answer: D. The question is asking for the probability of 7th marble without replacement. So the probability will change with every draw first 5/8 second 4/7 What I am missing here ? Thanks in advance.



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Re: New Set of Mixed Questions!!!
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31 Oct 2014, 04:15
vvd wrote: Bunuel wrote: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8
Basically we need to find the probability that the seventh marble drawn is red (so not blue).
Now, the initial probability of drawing red marble is 5/8. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is 5/8.
The same for blue marble: the probability of drawing blue marble is 3/8, the probability that for instance the 8th marble drawn is blue is still 3/8. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Answer: D. The question is asking for the probability of 7th marble without replacement. So the probability will change with every draw first 5/8 second 4/7 What I am missing here ? Thanks in advance. I've tried to explain this several times in this thread. For example, here: newsetofmixedquestions150204120.html#p1311823, and here: newsetofmixedquestions150204120.html#p1312115If you still have doubts please check similar questions: aboxcontains3yellowballsand5blackballsonebyone90272.htmlabagcontains3whiteballs3blackballs2redballs100023.htmleachoffourdifferentlockshasamatchingkeythekeys101553.htmlif40peoplegetthechancetopickacardfromacanister97015.htmlamedicalresearchermustchooseoneof14patientsto127396.htmlabagcontains3whiteballs3blackballs2redballs100023.htmlHope this helps.
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Re: New Set of Mixed Questions!!!
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22 Nov 2014, 07:01
Bunuel wrote: I. n^2 divided by 4 yields the reminder of 1 > odd^2 divided by 4 always yields the reminder of 1. So, this statement must be true.
Can someone explain that rule please?



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Re: New Set of Mixed Questions!!!
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05 Dec 2014, 04:20
srikirs007 wrote: BONUS QUESTION: 11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8
SOLUTION: A marble can be drawn from the bowl in 8 ways. A red marble can drawn in 5 ways. So the probability that a marble is NOT blue is 5/8 My Doubt is : The fact that marbles are drawn without replacement does not change anything? I have my doubts on this one .. can anyone please explain!
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Re: New Set of Mixed Questions!!!
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18 Jan 2015, 05:51
Bunuel wrote: 9. If x and y are integers and x + y = 12, which of the following must be true? A. Both x and y are negative B. xy > 0 C. If y < 0, then x > 0 D. If y > 0, then x < 0 E. x  y > 0
Look at option D: if y is positive, then x must be negative in order the sum of x and y to be negative.
Answer: D. Hi, By this logic, then shouldn't even Answer C be ok? Can you please explain why D satisfies the "must" rule and C does not? Thank you, TO



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Re: New Set of Mixed Questions!!!
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18 Jan 2015, 06:33
thorinoakenshield wrote: Bunuel wrote: 9. If x and y are integers and x + y = 12, which of the following must be true? A. Both x and y are negative B. xy > 0 C. If y < 0, then x > 0 D. If y > 0, then x < 0 E. x  y > 0
Look at option D: if y is positive, then x must be negative in order the sum of x and y to be negative.
Answer: D. Hi, By this logic, then shouldn't even Answer C be ok? Can you please explain why D satisfies the "must" rule and C does not? Thank you, TO Given x + y = 12: the sum of two numbers is negative. (C) says if y is negative, then x is positive. Well, that's not necessarily true: if y is negative, x can be positive, negative, or 0. For example: y = 20, x = 8. y = 10, x = 2. y = 12, x = 0. (D) says if y is positive, then x is negative. Now, if y is positive, then for the sum of x and y to be negative (12), x must be negative. If it's not, the sum of x and y would be positive, not negative 12. Does this make sense?
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Re: New Set of Mixed Questions!!!
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29 Feb 2016, 08:51
"Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. 7/8 B. 3/4 C. 2/3 D. 5/8 E. 3/8" Bunuel. Can you help me in this? I tried the below way. What is wrong in that? As per the question, the Seventh marble is Red. That means any of the below combo is possible RRRRBBRB  Here  (5/8)*(4/7)*(3/6)*(2/5)*(3/4)*(2/3)*(1/2)*1  What is wrong in this approach. Kindly help?
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Re: New Set of Mixed Questions!!!
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01 Feb 2017, 19:16
Bunuel wrote: 9. If x and y are integers and x + y = 12, which of the following must be true? A. Both x and y are negative B. xy > 0 C. If y < 0, then x > 0 D. If y > 0, then x < 0 E. x  y > 0
Look at option D: if y is positive, then x must be negative in order the sum of x and y to be negative.
Answer: D. Dear Bunuel, What if C. If \(y < 0\), then \(x > 0\), \(x = 12\), \(y = 24\), \(x + y = 12  24 = 12\)?
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Re: New Set of Mixed Questions!!!
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02 Feb 2017, 05:14
ziyuenlau wrote: Bunuel wrote: 9. If x and y are integers and x + y = 12, which of the following must be true? A. Both x and y are negative B. xy > 0 C. If y < 0, then x > 0 D. If y > 0, then x < 0 E. x  y > 0
Look at option D: if y is positive, then x must be negative in order the sum of x and y to be negative.
Answer: D. Dear Bunuel, What if C. If \(y < 0\), then \(x > 0\), \(x = 12\), \(y = 24\), \(x + y = 12  24 = 12\)? C is not always true. y < 0 does not necessarily mean that x > 0. For example, consider y = 1 and x = 11.
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Re: New Set of Mixed Questions!!!
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