To calculate the probability that a 5 is obtained on at least one of the rolls when a fair die is rolled 4 times, we can use the complement rule.

Step 1: Calculate the probability of not getting a 5 on a single roll.

Since there are 6 faces on a fair die, and only one of them is a 5, the probability of not getting a 5 on a single roll is:

$P(\text{not a 5}) = \frac{5}{6}$

Step 2: Calculate the probability of not getting a 5 on all 4 rolls.

The rolls are independent of each other, so the probability of not getting a 5 on all 4 rolls is:

$P(\text{not a 5 in 4 rolls}) = \left(\frac{5}{6}\right)^4$

$P(\text{not a 5 in 4 rolls}) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296} \approx 0.482$

Step 3: Use the complement rule to find the probability of getting at least one 5.

The probability of getting at least one 5 is the complement of the probability of not getting a 5 in all 4 rolls:

$P(\text{at least one 5}) = 1 - P(\text{not a 5 in 4 rolls})$

$P(\text{at least one 5}) = 1 - 0.482 = 0.518$

Final Answer:

The probability of getting a 5 on at least one of the 4 rolls is approximately 0.518.

Do you want more details on how to handle similar probability questions?

Here are 5 related questions:

1. What is the probability of getting at least one 6 when rolling a fair die 4 times?
2. How do you calculate the probability of getting exactly one 5 in 4 rolls?
3. What is the probability of getting no 5s in 5 rolls of a fair die?
4. How would the probability change if the die were biased?
5. How do you calculate the probability of getting a specific sequence of outcomes in multiple die rolls?

Tip: The complement rule is often the easiest method when asked for "at least one" in probability problems!