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A 6-sided die has 3 black sides and 3 white sides. If the die is throw

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A 6-sided die has 3 black sides and 3 white sides. If the die is throw  [#permalink]

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New post 05 Mar 2019, 02:55
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

81% (01:13) correct 19% (01:26) wrong based on 26 sessions

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Re: A 6-sided die has 3 black sides and 3 white sides. If the die is throw  [#permalink]

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New post 05 Mar 2019, 03:42
A 6-sided die has 3 black sides and 3 white sides. If the die is thrown 4 times, what is the probability that, on at least one of the throws, the die will land with a black side up?

Solution:
Probability of at least one of the throws that the die will land with a black side up = 1- (Probability of having white side up all four times)

P(At least once black up) = 1- (3*3*3*3)/(6*6*6*6) => 1- (1/16) => 15/16
Answer must be 'E'
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A 6-sided die has 3 black sides and 3 white sides. If the die is throw  [#permalink]

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New post 05 Mar 2019, 03:59
Bunuel wrote:
A 6-sided die has 3 black sides and 3 white sides. If the die is thrown 4 times, what is the probability that, on at least one of the throws, the die will land with a black side up?


A. 1/16

B. 3/16

C. 1/2

D. 9/16

E. 15/16


Probability of a throw showing white at the top = 3/6 = 1/2
Probability of a throw showing Black at the top = 3/6 = 1/2

METHOD-1

Probability of die lying with black side up atleast once in 4 times = 1 - (probability of die to lie at white face up all 4 times)

i.e. Probability of die lying with black side up atleast once in 4 times \(= 1 - (\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}) = \frac{15}{16}\)

Answer: Option E

METHOD-2

Probability of 1 out of 4 throw to land on black side = \((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) * 4C1
Probability of 2 out of 4 throw to land on black side = \((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) * 4C2
Probability of 3 out of 4 throw to land on black side = \((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) * 4C3
Probability of 4 out of 4 throw to land on black side = \((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) * 4C4

i.e. Total Probability = (4C1 + 4C2 + 4C3 + 4C4) *\((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) = \(\frac{15}{16}\)

Answer: Option E
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Re: A 6-sided die has 3 black sides and 3 white sides. If the die is throw  [#permalink]

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New post 05 Mar 2019, 08:09
P of each black side = 3/6 ; 1/2
and P of each white side = 3/6 ; 1/2
so die will land with a black side = 1- white side P
1- ( 1/2)^4 = 15/16
IMO E

Bunuel wrote:
A 6-sided die has 3 black sides and 3 white sides. If the die is thrown 4 times, what is the probability that, on at least one of the throws, the die will land with a black side up?


A. 1/16

B. 3/16

C. 1/2

D. 9/16

E. 15/16
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Re: A 6-sided die has 3 black sides and 3 white sides. If the die is throw  [#permalink]

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New post 05 Mar 2019, 08:19
IMO E .

event with white side up = 3*3*3*3 = 81
total events = 6*6*6*6

Probab of white side up = (3*3*3*3)/(6*6*6*6) = 1/16

So atleast one black side up = 1-1/16 = 15/16
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Re: A 6-sided die has 3 black sides and 3 white sides. If the die is throw   [#permalink] 05 Mar 2019, 08:19
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