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

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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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