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

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The figure below shows two T-shaped cardboard pieces, both to be folde [#permalink]
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Bunuel wrote:
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

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.

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]
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]
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
Re: The figure below shows two T-shaped cardboard pieces, both to be folde [#permalink]
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