Total way to select (9 good & 1 bad pen) with 4 pen defective
G*G*G*D*G*G*G*G*G*G => 9*8*7*1*6*5*4*3*2*1 => 9!
Total way with no pre condition 10!

Probability=> 9!/10! => 1/10

First three pens must be non-defective.
Probability to choose them:
9/10 * 8/9 * 7/8 ( we are not replacing the pens)

Fourth pen is defective:
1/7 is probability to choose it.

Total probability = 9/10*8/9*7/8*1/7 = 1/10

Hence Option (A) is our answer.

Best,
Gladi

Posted from my mobile device
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The question asks us to find P(4th pen selected is defective)

Here's all you need to know:

When we are selecting items without replacement, P(1st selection has some attribute) = P(2nd selection has that attribute) = P(3rd selection has that attribute) = P(4th selection has that attribute) . . . etc

So, in this question, P(1st pen selected is defective) = P(2nd pen selected is defective) = P(3rd pen selected is defective) = P(4th pen selected is defective) . . . etc

P(1st pen selected is defective) = 1/10, so P(4th pen selected is defective) = 1/10 as well.


ASIDE: This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something (often either work, like getting wood for the fire, or dumb, like eating something that shouldn't be eaten).

So, someone would hold up n pieces of grass (for n guys), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.

There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece were minimized since every person before him had a chance of drawing the short piece before it got to his turn.

The truth of the matter is that each of the n guys had a 1/n chance of selecting the shortest piece, regardless of the order in which they selected.

The same applies to the original question here.

Cheers,
Brent
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Nice, you got it...

"When we are selecting items without replacement, P(1st selection has some attribute) = P(2nd selection has that attribute) = P(3rd selection has that attribute) = P(4th selection has that attribute) . . . etc"


In fact , the attributes will be exactly same with replacement also.


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I cannot really understand this explanation . can you please explain ?

Hii
Since we do not know the outcome of the first three draws, we have to take all different cases of first three draws, which on solving, gives the same answer. This approach is clearly explained by Gladiator59.

BEST APPROACH: By using Logic, In fact, the first three draw (if we don't know its outcome) will not effect the 4th draw, and all the draws have equal chance probability.
Hence if the question had been find the probability of getting defective pen in 10th draw, we would get same answer = 1/10.

There is a similar question , but since here we know the outcome of first 3 draws, it definitely changes the probability of 4th draw.
Only one of ten pens in a box is defective. Three pens are
Only one of ten pens in a box is defective. Three pens are randomly drawn from the box, one at a time. If none of those three pens is defective, what is the probability that the next pen randomly drawn is defective?

(A) 1/10
(B) 1/7
(C) 7/10
(D) 6/7
(E) 9/10

Some more question based on this logic: https://gmatclub.com/forum/users-self-m ... l#p1979430

1) If from a well-shuffled deck of 52 cards, five cards are drawn at random one by one with replacement. What is the probability of fifth card being spade?
A) 1/3
B) 1/4
C) 2/5
D) 3/13
E) 3/26

2) If from a well-shuffled deck of 52 cards, five cards are drawn at random one by one WITHOUT replacement. What is the probability of fifth card being spade?
A) 1/3
B) 1/4
C) 2/5
D) 3/13
E) 3/26





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There is point E inside Square ABCD. A portion or number of portions are created inside the square ABCD such that both \(\angle\)AED and \(\angle\)BEC are acute angle in that portion/portions.

Find the approx percentage of area of portion/portions to total area of Square ABCD.

[A]20%
[B]30%
[C]40%
[D]50%
[E]60%

Answer should be 20%.

See my approach.

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If E is a point inside the Square ABCD, what is the approx probability that both the triangles AED and BEC are acute angled triangle?
A. 1/3
B. 1/4
C. 1/5
D. 2/3
E. 3/7

Source:- Self made

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Only Case 6 is favourable case

Area of shaded area = Area of square - 2 * Area of semicircle with diameter equal to side of square
= \(a^2\) - \(2*\frac{1}{2}*\pi*(\frac{a}{2})^2\)
= \(a^2\) -\(\pi*\frac{a^2}{4}\)
Total Area = Area of square =\(a^2\)
probability that both the triangles AED and BEC are acute angled triangle =\(\frac{Area of shaded area}{Total Area}\)
=\(\frac{(4-\pi)}{4}\)=0.215, Taking \(\pi\)=3.14
So OA=C

[A]\(\frac{1}{5}\)
[B]\(\frac{1}{4}\)
[C]\(\frac{1}{3}\)
[D]\(\frac{1}{2}\)
[E]\(\frac{4}{5}\)

This question can be easily answered using ball parking.
See approach.


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OA:C

Concept used:
1) Diagonal divides the square into 2 triangles of equal area
2) Median divides the triangle into 2 triangles of equal area


let the Area of Square ABCD be 100
Area of triangle ABD = \(\frac{100}{2}\) =50
in triangle ABD, PD is median, So area of triangle BDP = 50/2 =25
Now similarly, Area of triangle BDS = 25
Area of Shaded region is less than Sum of Area of triangle BDS & Atea od triangle BDP
= Area of shaded region < 25+25
= Area of shaded region < 50... (eq 1)

Similarly, it is clear from FIgure,
Area of triangle BDM>area of triangle BMP
hence, Area of triangle BDM> (area of triangle BDP)/2 = 25/2=12.5
Similarly, Area of triangle BND > 12.5
Area of shaded region >12.5+12.5 =25
Area of shaded region >25 ... (eq2)

From eq 1 & 2. 25<Area of shaded region<50
or, \(\frac{25}{100}\) <\(\frac{Area of shaded region}{Area of Square}\)<\(\frac{50}{100}\)
Hence required fraction is between 1/4 and 1/2.
So we get from the options , Fraction =\(\frac{1}{3}\)





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tyson1629


Area of square = \(a^2\)
Area of shaded area (rhombus) =\(\frac{{a^2}}{{3}}\)

Ratio= \(\frac{1}{3}\)

Sent from my iPhone using GMAT Club Forum mobile app

[b]Kashpia

your solution has not been uploaded.
I think that there is some problem with GC app, specifically while attaching image.

A has thrice as many cakes as B has. How many cakes does A have?

1) A and B together have as many cakes as L
2) B has 1 cake less than A

Source: Self made
I do not have an OA for this, will be posting my solution and doubts in the next post.
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1) Clearly not sufficient.

2) Gives me the following equations \(A=3B, B=A-1\). On solving I get the number of cakes for A as \(3/2\)

Doubt Since the number of cakes has to be an integer, and 2) gives me a non-integer answer, should the answer be E or B?

PS: This is a self made question, made it to clarify this doubt, can such questions show up on the GMAT?
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