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

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

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%

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

OA:C

tyson1629


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

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

[b]Kashpia

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

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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