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

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

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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Source: Beat The GMAT — Problem Solving | Sat Apr 27, 2019 5:43 pm

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

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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swerve: Legendary Member; Posts: 2579; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

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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Scott@TargetTestPrep: GMAT Instructor; Posts: 8211; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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

Scott Woodbury-Stewart
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deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

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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