A fair die is rolled 4 times. What is the probability that a 3 is obtained on at least one of the rolls? Round your answer to three decimal places.

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Let $$ p $$ be the probability of obtaining a 3 on one roll. Since the die is fair,

$$
p = \frac{1}{6}
$$

The probability of not obtaining a 3 on one roll is

$$
1 - p = \frac{5}{6}
$$

When rolling the die 4 times, the probability that 3 is not obtained in any of the rolls is

$$
\left(\frac{5}{6}\right)^4 = \frac{625}{1296}
$$

Thus, the probability that a 3 is obtained on at least one roll is the complement:

$$
1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{1296 - 625}{1296} = \frac{671}{1296}
$$

Converting this to a decimal:

$$
\frac{671}{1296} \approx 0.518
$$

So, the probability that a 3 is obtained on at least one of the rolls is approximately $$0.518$$.
The probability of rolling at least one 3 in 4 rolls is approximately 0.518.

A fair die is rolled 4 times. What is the probability that a 3 is obtained on at least one of the rolls? Round your answer to three decimal places.

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