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# New Set of Mixed Questions!!!

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Re: New Set of Mixed Questions!!!  [#permalink]

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17 Jan 2014, 04:57
2
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!!!  [#permalink]

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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.

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!!!  [#permalink]

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01 Jun 2014, 01:33
nehamodak 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.

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

C is not always true: if y is negative it's not necessary x to be positive. Consider y=-1 and x=-11.

But D must be true: if y is positive, then x must be negative in order the sum of x and y to be negative.

Hope it's clear.
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Re: New Set of Mixed Questions!!!  [#permalink]

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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).

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 ?

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Joined: 02 Sep 2009
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Re: New Set of Mixed Questions!!!  [#permalink]

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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).

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 ?

I've tried to explain this several times in this thread. For example, here: new-set-of-mixed-questions-150204-120.html#p1311823, and here: new-set-of-mixed-questions-150204-120.html#p1312115

If you still have doubts please check similar questions:
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html

Hope this helps.
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Re: New Set of Mixed Questions!!!  [#permalink]

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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!!!  [#permalink]

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22 Nov 2014, 07:06
1
oss198 wrote:
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?

Not much there to explain...

Odd^2 divided by 4 always yields the reminder of 1:
1^2 = 1 divided by 4 yields the reminder of 1;
3^2 = 9 divided by 4 yields the reminder of 1;
5^2 = 25 divided by 4 yields the reminder of 1;
7^2 = 49 divided by 4 yields the reminder of 1;
...

Or: odd^2 = (2x + 1)^2 = 4x^2 + 4x + 1 --> divided by 4 yields the reminder of 1

Hope it's clear.
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Re: New Set of Mixed Questions!!!  [#permalink]

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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!!!  [#permalink]

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05 Dec 2014, 04:27
1
aashu4uiit wrote:
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!

I tried to explain this several times on this thread. For example, here: new-set-of-mixed-questions-150204-120.html#p1311823, and here: new-set-of-mixed-questions-150204-120.html#p1312115

If you still have doubts please check similar questions:
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html

Hope this helps.
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Re: New Set of Mixed Questions!!!  [#permalink]

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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.

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!!!  [#permalink]

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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.

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!!!  [#permalink]

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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!!!  [#permalink]

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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.

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!!!  [#permalink]

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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.

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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13 May 2018, 17:56
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