## For one roll of a certain die, the probability of rolling a two is $$1/6$$. If this die is rolled $$4$$ times, which of

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### For one roll of a certain die, the probability of rolling a two is $$1/6$$. If this die is rolled $$4$$ times, which of

by BTGmoderatorLU » Sat Jun 10, 2023 3:04 pm

00:00

A

B

C

D

E

## Global Stats

Source: Manhattan Prep

For one roll of a certain die, the probability of rolling a two is $$1/6$$. If this die is rolled $$4$$ times, which of the following is the probability that the outcome will be a two at least $$3$$ times?

A. $$\left(\dfrac{1}{6}\right)^4$$

B. $$2\left(\dfrac{1}{6}\right)^3+\left(\dfrac{1}{6}\right)^4$$

C. $$3\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)+\left(\dfrac{1}{6}\right)^4$$

D. $$4\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)+\left(\dfrac{1}{6}\right)^4$$

E. $$6\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)+\left(\dfrac{1}{6}\right)^4$$

The OA is D

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### Re: For one roll of a certain die, the probability of rolling a two is $$1/6$$. If this die is rolled $$4$$ times, which

by Asimo3089 » Tue Jun 13, 2023 9:28 am
BTGmoderatorLU wrote:
Sat Jun 10, 2023 3:04 pm
Source: Manhattan Prep

For one roll of a certain die, the probability of rolling a two is $$1/6$$. If this die is rolled $$4$$ times, which of the following is the probability that the outcome will be a two at least $$3$$ times?

A. $$\left(\dfrac{1}{6}\right)^4$$

B. $$2\left(\dfrac{1}{6}\right)^3+\left(\dfrac{1}{6}\right)^4$$

C. $$3\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)+\left(\dfrac{1}{6}\right)^4$$

D. $$4\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)+\left(\dfrac{1}{6}\right)^4$$

E. $$6\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)+\left(\dfrac{1}{6}\right)^4$$

The OA is D
Dice roll, two chance $$1/6$$, four times. Find chance of getting two at least three times.

To solve, we choose option with two three times, any outcome on fourth roll, and also possibility of two four times. Correct option: D.

Mathematically, we have $$4\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)+\left(\dfrac{1}{6}\right)^4$$ as the chance of getting two at least three times out of four rolls.

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