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

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

by BTGmoderatorDC » Sat Apr 27, 2019 5:43 pm

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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

OA E

Source: Princeton Review
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by Jay@ManhattanReview » Mon Apr 29, 2019 9:41 pm
BTGmoderatorDC 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

OA E

Source: Princeton Review
Since the 6-sided die has 3 black sides and 3 white sides, i.e., an equal number of black and white sides, the probability of getting a black side up = 1/2

Instead of calculating the probability that, on at least one of the throws, the die will land with a black side up, let's calculate the probability that, on none of the throws, the die land with a black side up. Once we get this, we will deduct the result from 1 to get the required answer.

Probability that, on none of the throws, the die land with a black side up = 1/2 * 1/2 *1/2 *1/2 = 1/16

Thus,

Probability that, on at least one of the throws, the die will land with a black side up = 1 - 1/16 = 15/16

The correct answer: E

Hope this helps!

-Jay
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by swerve » Thu May 02, 2019 9:41 am
\(P\) of each black side \(= \frac{3}{6} = \frac{1}{2}\)

and \(P\) of each white side \(= \frac{3}{6}=\frac{1}{2}\)

so die will land with a black side \(= 1- \text{white side } P\)

\(1- \left( \frac{1}{2}\right)^4 = \frac{15}{16}\)
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by Scott@TargetTestPrep » Thu May 02, 2019 4:20 pm
BTGmoderatorDC 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

OA E

Source: Princeton Review

We see that P(black) = 3/6 = 1/2 and P(white) = 3/6 = 1/2. We can use the formula:

P(at least one black face comes up in 4 throws) = 1 - P(no black faces come up in 4 throws)

In other words,

P(at least one black face comes up in 4 throws) = 1 - P(all white faces come up in 4 throws)

Therefore,

P(at least one black face comes up in 4 throws) = 1 - (1/2)^4 = 1 - 1/16 = 15/16.

Answer: E

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by deloitte247 » Sat May 04, 2019 9:49 pm
$$\Pr obability\ of\ a\ throw\ showing\ white\ at\ the\ top=\frac{3}{6}=\frac{1}{2}$$
$$\Pr obability\ of\ a\ throw\ showing\ black\ at\ the\ top=\frac{3}{6}=\frac{1}{2}$$
If the die is thrown 4 times, the probability that, on at least one of the throws, the die will land with a black = 1 - (Pr. of the die due to land with a white face up at all 4 times) $$=1-\left(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\right)$$
$$=1-\frac{1}{16}$$
$$=\frac{\left(16-1\right)}{16}$$
$$=\frac{15}{16}\ \ \ \ \ \ \ \ \ \left(option\ E\right)$$
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