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The figure below shows two T-shaped cardboard pieces, both to be folde

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Bunuel
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The figure below shows two T-shaped cardboard pieces, both to be folde [#permalink] The figure below shows two T-shaped cardboard pieces, both to be folde   16 Dec 2020, 05:00
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The figure below shows two T-shaped cardboard pieces, both to be folded into a pair of cube-shaped dice.



On a fair throw of both dice, what is the probability that NEITHER die will show either a solid white or solid black surface facing up?

(A) 1/6
(B) 1/5
(C) 1/4
(D) 1/3
(E) 2/5


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2020-12-15_15-58-39.png
2020-12-15_15-58-39.png [ 8.03 KiB | Viewed 3361 times ]
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Re: The figure below shows two T-shaped cardboard pieces, both to be folde [#permalink] The figure below shows two T-shaped cardboard pieces, both to be folde   16 Dec 2020, 05:13
Total no. of possible options = 36
Possible times when both dices will have dual color(black+white) side = 6 [Neither dice will have a solid surface facing up]
Therefore, probability = favourable/total = 6/36 = 1/6

Hence, answer choice A
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Re: The figure below shows two T-shaped cardboard pieces, both to be folde [#permalink] The figure below shows two T-shaped cardboard pieces, both to be folde   31 Dec 2020, 17:23
I'm confused on this. Bunuel Can you post the solution?

Aren't there 7 possible scenarios of all black and white squares?
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The figure below shows two T-shaped cardboard pieces, both to be folde [#permalink] The figure below shows two T-shaped cardboard pieces, both to be folde   15 Feb 2021, 10:07
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The figure below shows two T-shaped cardboard pieces, both to be folded into a pair of cube-shaped dice.



On a fair throw of both dice, what is the probability that NEITHER die will show either a solid white or solid black surface facing up?

(A) 1/6
(B) 1/5
(C) 1/4
(D) 1/3
(E) 2/5


Show Spoiler
Attachment:
2020-12-15_15-58-39.png
Not having solid colors i.e. neither black nor white.
For 1st dice not having solid color(neither of black and white) = 3/6
For 2nd dice not having solid color = 2/6

P = 3/6 * 2/6 = 6/36 = 1/6.

Answer A.
Note: We are getting rid of even those cases in which one of them has either of the two solid colors i.e. either 1st has solid black and 2nd has solid white or 1st has solid white and 2nd has solid black color.
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Re: The figure below shows two T-shaped cardboard pieces, both to be folde [#permalink] The figure below shows two T-shaped cardboard pieces, both to be folde   17 Feb 2021, 22:20
IMO, the first die and the second die have different number of solid Black surfaces. hence, needs to be deal indivisually.

die 1: 1- p(getting solid black or solid white)
1-1/2 { 1/2= (2/6+1/6) <<(2 black surface out of 6 and 1 white surface out of 6 total surface)
=1/2

die 2: 1-2/3 {2/3= 3/6+1/6} << 3 black and one white surface out of 6)
=1/3

P(neither gives in solid black/white) = 1/2*1/3=1/6 Answer.
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Re: The figure below shows two T-shaped cardboard pieces, both to be folde [#permalink] The figure below shows two T-shaped cardboard pieces, both to be folde   30 Jul 2023, 05:44
Take Cardboard piece 1,
Probability of not a solid black or solid white P1 = 3/6=1/2

Take Cardboard piece 2,
Probability of not a solid black or solid white P2= 2/6=1/3

Required probability = Probability neither cardboard piece show solid black or solid white = P1*P2= (1/2)*(1/3)=(1/6)

Hence A
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Re: The figure below shows two T-shaped cardboard pieces, both to be folde [#permalink]
30 Jul 2023, 05:44
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